The fundamental theorem of finite semidistributive lattices

TL;DR

This paper establishes a fundamental theorem characterizing finite semidistributive lattices via a set with additional structure, linking lattice theory with combinatorics and geometry.

Contribution

It introduces a new fundamental theorem for finite semidistributive lattices, connecting them to structured sets and geometric interpretations.

Findings

Characterization of finite semidistributive lattices through admissible subsets

Clarification of lattice constructions like canonical join representations

Extension of results to infinite lattices

Abstract

We prove a Fundamental Theorem of Finite Semidistributive Lattices (FTFSDL), modelled on Birkhoff's Fundamental Theorem of Finite Distributive Lattices. Our FTFSDL is of the form "A poset L is a finite semidistributive lattice if and only if there exists a set Sha with some additional structure, such that L is isomorphic to the admissible subsets of Sha ordered by inclusion; in this case, Sha and its additional structure are uniquely determined by L." The additional structure on Sha is a combinatorial abstraction of the notion of torsion pairs from representation theory and has geometric meaning in the case of posets of regions of hyperplane arrangements. We show how the FTFSDL clarifies many constructions in lattice theory, such as canonical join representations and passing to quotients, and how the semidistributive property interacts with other major classes of lattices. Many of our results also apply to infinite lattices.

Equations

$$X^{\perp }=\{y\in \text{Sha}:x\nrightarrow y\forall x\in X\}\text{mbox}and{}^{\perp }X=\{y\in \text{Sha}:y\nrightarrow x\forall x\in X\}.$$

$$x\mapsto (\{j\in \mathrm{JIrr}(L):j\le x\},{\kappa }^d\left(\{m\in \mathrm{MIrr}(L):m\ge x\}\right))$$

$$x\mapsto (\{j\in {\mathrm{JIrr}}^c(L):j\le x\},{\kappa }^d\left(\{m\in {\mathrm{MIrr}}^c(L):m\ge x\}\right))$$

$$X^{\perp}=\Sha\setminus\bigl{(}\operatorname{down}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{$\hookrightarrow$}}}}(\operatorname{down}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{$\twoheadrightarrow$}}}}X)\bigr{)}\quad and\quad^{\perp}X=\Sha\setminus\bigl{(}\operatorname{up}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{$\twoheadrightarrow$}}}}(\operatorname{up}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{$\hookrightarrow$}}}}X)\bigr{)}.$$

$$\kappa (T(x),T(x{)}^{\perp })=({}^{\perp }F(x),F(x))\text{and}{\kappa }^d({}^{\perp }F(x),F(x))=(T(x),T(x{)}^{\perp }).$$

$$x=\bigvee \{j\in \mathrm{JIrr}(L):j\le x\}=\bigwedge \{m\in \mathrm{MIrr}(L):m\ge x\}.$$

$$x\mapsto (\{j\in \mathrm{JIrr}(L):j\le x\},{\kappa }^d\left(\{m\in \mathrm{MIrr}(L):m\ge x\}\right))$$

$$x=\bigvee \{j\in {\mathrm{JIrr}}^c(L):j\le x\}=\bigwedge \{m\in {\mathrm{MIrr}}^c(L):m\ge x\}.$$

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$$\begin{array}[]{|r||c|c|c|c|c|c|c|c|c|c|}\hline\cr&x^{--}&x^{-}&x^{\circ}&x^{+}&x^{++}&y^{--}&y^{-}&y^{\circ}&y^{+}&y^{++}\\ \hline\cr\hline\cr x^{++}&&&&&1^{++}&&&&x^{++}&x^{++}\\ \hline\cr x^{+}&&&&1^{++}&x^{+}&&&&x^{+}&x^{+}\\ \hline\cr x^{\circ}&&&1^{++}&1^{\circ}&1^{\circ}&&&x^{++}&x^{\circ}&x^{\circ}\\ \hline\cr x^{-}&&1^{++}&1^{-}&1^{-}&1^{-}&x^{++}&x^{++}&x^{-}&x^{-}&x^{-}\\ \hline\cr x^{--}&1^{++}&x^{--}&x^{--}&x^{--}&x^{--}&x^{--}&x^{--}&x^{--}&x^{--}&x^{--}\\ \hline\cr\end{array}$$

$${}^{\triangle}\!\bigl{(}U^{\triangle}\bigr{)}={}^{\perp}\bigl{(}U^{\triangle}\bigr{)}\cap X={}^{\perp}{\bigl{(}}U^{\triangle}\cup Y{\bigr{)}}={}^{\perp}{\Big{(}}(U^{\perp}\cap X)\cup Y{\Big{)}}.$$

$$\mathrm{Del}(X,c)=X\setminus \{x\in X:x\to c\}={}^{\perp }(Y\cup \{c\})$$

$${}_{\Theta }=\{x\in \text{Sha}:(T_{\ast }(x),T_{\ast }(x{)}^{\perp })\not\equiv (T(x),T(x{)}^{\perp })\mathrm{mod}\Theta \}.$$

$${}_{\Theta }=\{x\in \text{Sha}:({}^{\perp }F(x),F(x))\not\equiv ({}^{\perp }{F}^{\ast }(x),F^{\ast }(x))\mathrm{mod}\Theta \}.$$

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$$x\mapsto (\{j\in \mathrm{JIrr}(L):j\le x\},\{m\in \mathrm{MIrr}(L):m\ge x\})$$

$$1$$

$$R\mapsto (\{\Sigma \in \text{Sha}(\mathcal{A},B):J(\Sigma )\le R\},\left(\{\Sigma \in \text{Sha}(\mathcal{A},B):M(\Sigma )\ge R\}\right))$$

$$(X,Y)\mapsto \underset{\Sigma \in X}{\bigwedge }J(\Sigma )=\underset{\Sigma \in Y}{\bigwedge }M(\Sigma ).$$

Full text

Nathan Reading was partially supported by the National Science Foundation under Grant Number DMS-1500949. David Speyer was partially supported by the National Science Foundation under Grant Number DMS-1600223. Hugh Thomas was partially supported by an NSERC Discovery Grant and the Canada Research Chairs program.

1. 1 Introduction
2. 2 The fundamental theorem
3. 3 The infinite case
4. 4 Factorization systems for intervals
5. 5 Covers and canonical join representations
6. 6 Quotient Lattices
7. 7 Relation to other types of finite lattices
8. 8 Motivating examples

1. Introduction

A lattice is a partially ordered set such that every pair $x,y$ of elements has a meet (greatest lower bound) $x\wedge y$ and join (least upper bound) $x\vee y$. A lattice is distributive if the meet operation distributes over the join operation and the join distributes over the meet. (These two distributivity conditions are equivalent.)

The simplest class of examples of distributive lattices are the lattices of downsets in a fixed poset $P$. A downset (or order ideal) in $P$ is a subset $D$ of $P$ such that if $x\in D$ and $y\leq x$ then $y\in D$. Let $P$ be a poset and let $\operatorname{Downsets}(P)$ be the set of downsets in $P$, partially ordered by containment. It is readily verified that the union of two downsets is a downset and that the intersection of two downsets is a downset. As a consequence, $\operatorname{Downsets}(P)$ is a distributive lattice. Birkhoff [6] showed that every finite distributive lattice is of this form $\operatorname{Downsets}(P)$. This result is often called Birkhoff’s Representation Theorem, but since that name also refers to other theorems in universal algebra, we follow [25] in calling it the Fundamental Theorem of Finite Distributive Lattices (FTFDL).

An element $j$ of a lattice $L$ is join-irreducible if, for all finite subsets $X$ of $L$, if $j=\bigvee X$ then $j\in X$. Equivalently, $j\in L$ is join-irreducible if $j$ is not minimal in $L$ and cannot be written as $x\vee y$ for $x,y<j$. In a finite lattice $L$, this is equivalent to saying that $j$ covers exactly one element of $L$; we will denote that unique element by $j_{*}$. We write $\operatorname{JIrr}(L)$ for the set of join-irreducible elements of $L$, with the partial order induced from $L$. Similarly, an element $m$ is meet-irreducible if, for all finite subsets $X$ of $L$, if $m=\bigwedge X$ then $m\in X$. Equivalently, $m$ is not maximal and we cannot write $m=x\wedge y$ for $m<x$, $y$ or, equivalently in the finite case, if $m$ is covered by exactly one element $m^{*}$ of $L$. We write $\operatorname{MIrr}(L)$ for the set of meet-irreducible elements of $L$, again with the induced partial order.

Theorem 1.1** (FTFDL).**

A finite poset $L$ is a distributive lattice if and only if it is isomorphic to $\operatorname{Downsets}(P)$ for some finite poset $P$. In this case, $P$ is isomorphic to $\operatorname{JIrr}(L)$. The map $x\mapsto\{j\in\operatorname{JIrr}:j\leq x\}$ is an isomorphism from $L$ to $\operatorname{Downsets}(\operatorname{JIrr}(L))$, with inverse $X\mapsto\bigvee X$.

The main result of this paper is a theorem, analogous to Theorem 1.1, that characterizes a larger class of lattices: finite semidistributive lattices (defined below). Two important examples of semidistributive lattices are the weak order on a finite Coxeter group and the lattice of torsion classes of a finite-dimensional algebra (see Section 8.2). These two examples have recently been connected by a series of papers [18, 14, 10] relating the torsion classes of Dynkin type preprojective algebras to weak orders of the corresponding Coxeter groups. Moreover, the Cambrian lattice, which describes the structure of the corresponding cluster algebra, is a semidistributive lattice described as a quotient of weak order, and can also be described as a lattice of torsion classes.

The language and terminology of this paper deliberately echoes these motivating examples: Finite semidistributive lattices are realized in terms of binary relations on a set (the Cyrillic letter “sha”), suggestive of the shards [20, 21, 22] that govern much of the lattice theory of the weak order on a finite Coxeter group. The relations ${\mathord{\to}}$, ${\mathord{\hookrightarrow}}$, and ${\mathord{\twoheadrightarrow}}$ echo the structure of an abelian category. We discuss these motivating examples in more detail in Section 8.

A lattice $L$ is join semidistributive if whenever $x,y,z\in L$ satisfy $x\vee y=x\vee z$, they also satisfy $x\vee(y\wedge z)=x\vee y$. This is equivalent to the following condition: If $X$ is a nonempty finite subset of $L$ such that $x\vee y=z$ for all $x\in X$, then $\left(\bigwedge_{x\in X}x\right)\vee y=z$. The lattice is meet semidistributive if the dual condition $(x\wedge y=x\wedge z)\implies(x\wedge(y\vee z)=x\wedge y)$ holds. Equivalently, if $X$ is a nonempty finite subset of $L$ such that $x\wedge y=z$ for all $x\in X$, then $\left(\bigvee_{x\in X}x\right)\wedge y=z$. The lattice is semidistributive if it is both join semidistributive and meet semidistributive. The notion of semidistributivity goes back to Jónsson [15], who in particular showed that free lattices are semidistributive.

In a finite semidistributive lattice, for each join-irreducible element $j$, the set $\{y:j\wedge y=j_{*}\}$ has a maximum element, which we call $\kappa(j)$. For each meet-irreducible element $m$, the set $\{x:m\vee x=m\}$ has a minumum element $\kappa^{d}(m)$. The existence of these elements characterizes semidistributivity of finite lattices. The maps $\kappa$ and $\kappa^{d}$ are inverse bijections between $\operatorname{JIrr}(L)$ and $\operatorname{MIrr}(L)$ (see Theorem 2.28 and preceding references).

Given a (binary) relation ${\mathord{\to}}$ on a set and a subset $X\subseteq\Sha$, we define

[TABLE]

A maximal orthogonal pair is a pair $(X,Y)$ of subsets of with $Y=X^{\perp}$ and $X={}^{\perp}Y$. If $(X,Y)$ and $(X^{\prime},Y^{\prime})$ are maximal orthogonal pairs, then $X\subseteq X^{\prime}$ if and only if $Y\supseteq Y^{\prime}$. The lattice of maximal orthogonal pairs for ${\mathord{\to}}$ is the set $\operatorname{Pairs}({\mathord{\to}})$ of maximal orthogonal pairs, partially ordered by containment in the first component, or equivalently reverse containment in the second component. As we will discuss further in Section 7.3, all lattices can be described as lattices of maximal orthogonal pairs from some relation, and in many ways. We will now describe conditions on $(\Sha,{\mathord{\to}})$ which imply that $\operatorname{Pairs}({\mathord{\to}})$ is semidistributive, and such that each finite semidistributive lattice has a unique such representation.

Let be a set and let ${\mathord{\to}}$ be a reflexive relation on . We use ${\mathord{\to}}$ to define two other relations, ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ on . Define $x\twoheadrightarrow y$ iff for all $y\to z$, we also have $x\to z$. Dually, define $x\hookrightarrow y$ if and only if for all $z\to x$, we also have $z\to y$. Each of ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ is obviously a preorder (reflexive and transitive, but perhaps not antisymmetric). We call $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ the factorization of ${\mathord{\to}}$ and write $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})=\operatorname{Fact}({\mathord{\to}})$.

Similarly, we define an operation $\operatorname{Mult}$ called multiplication that takes an ordered pair of preorders on to a reflexive relation on . Namely, define $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ to be the relation ${\mathord{\to}}$ where $x\to z$ if and only if there exists $y\in\Sha$ such that $x\twoheadrightarrow y\hookrightarrow z$.

We suggest pronouncing ${\mathord{\to}}$, ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ as “to”, “onto” and “into”. The relations $X\to Y$, $X\twoheadrightarrow Y$ and $X\hookrightarrow Y$ should roughly be thought of as analogous to “there is a nonzero map from $X$ to $Y$”, “there is a surjection from $X$ onto $Y$” and “there is an injection from $X$ into $Y$” in some category to be defined later. For precise statements along these lines, see Section 8.2; we warn the reader that the precise interpretation of ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ is more subtle that the rough statement here. The definitions of $\operatorname{Fact}$ and $\operatorname{Mult}$ are easy to motivate in the context of this analogy.

We define a factorization system to be a tuple $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ such that ${\mathord{\to}}$, ${\mathord{\hookrightarrow}}$, and ${\mathord{\twoheadrightarrow}}$ are relations on a set having $\operatorname{Fact}({\mathord{\to}})=({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ and ${\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})={\mathord{\to}}}$. Since, in a factorization system, ${\mathord{\to}}$ and $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ determine each other, any condition on a factorization system can be thought of either as a condition on ${\mathord{\to}}$ or a condition on $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. A factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is called finite if the set is finite.

We say that a factorization system obeys the order condition if we do not have $x\twoheadrightarrow y\twoheadrightarrow x$ or $x\hookrightarrow y\hookrightarrow x$ with $x\neq y$; in other words, the order condition states that the preorders ${\mathord{\hookrightarrow}}$ and ${\mathord{\twoheadrightarrow}}$ are partial orders. We will say that a factorization system obeys the brick condition if we do not have $x\twoheadrightarrow y\hookrightarrow x$ with $x\neq y$. The name “brick condition” comes from the notion of a “brick” in representation theory, which is a module $X$ such that any nonzero map $X\to X$ is an isomorphism. If we had $X\twoheadrightarrow Y\hookrightarrow X$ for some $Y\not\cong 0,X$, then the composite map would be neither [math] nor an isomorphism; our brick condition rules out the combinatorial analogue of this. See Section 8.2 for more precise statements. We will say that a factorization system is two-acyclic if it obeys the order condition and the brick condition.

The first main result of the paper is the Fundamental Theorem of Finite Semidistributive Lattices. If $L$ is a finite semidistributive lattce, we define three relations on $\operatorname{JIrr}(L)$: $i\to_{L}j$ if and only if $i\not\leq\kappa(j)$ in $L$, $i\twoheadrightarrow_{L}j$ if and only if $i\geq j$ in $L$, and $i\hookrightarrow_{L}j$ if and only if $\kappa(i)\geq\kappa(j)$ in $L$.

Theorem 1.2** (FTFSDL).**

A finite poset $L$ is a semidistributive lattice if and only if it is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for a finite two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. In this case, $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ and $(\operatorname{JIrr}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$ are isomorphic. The map

[TABLE]

is an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\to}}_{L})$, with inverse $(X,Y)\mapsto\bigvee X=\bigwedge(\kappa(Y))$.

In the course of proving Theorem 1.2, we will prove some more general theorems that apply to some infinite lattices. We therefore discuss some definitional distinctions which are trivial for finite lattices. A complete lattice is a poset $L$ such that every subset $X$ of $L$ has a meet (greatest lower bound) $\bigwedge X$ and join (least upper bound) $\bigvee X$. The definition of a lattice given above is weaker: It is equivalent to requiring that $\bigwedge X$ and $\bigvee X$ exist for all finite subsets $X$. In particular, all finite lattices are complete.

An element $j$ of a complete lattice $L$ is completely join-irreducible if $j=\bigvee X$ implies $j\in X$ for subsets $X\subseteq L$. Equivalently, $j$ is completely join-irreducible if and only if there exists an element $j_{*}$ such that $x<j$ if and only if $x\leq j_{*}$. (In an infinite lattice, this is a stronger condition than simply requiring that $j$ covers exactly one element.) The element $j_{*}$ is $\bigvee\{x\in L:x<j\}$. (Recall that $j$ is join-irreducible if $j=\bigvee X$ implies $j\in X$ for finite subsets $X\subseteq L$. Thus every completely join-irreducible element is join-irreducible, but not vice versa.) A completely meet-irreducible element $m$ is defined dually, and we write $m^{*}$ for the element $\bigwedge\{x\in L:x>m\}$ with the property that $x>m$ if and only if $x\geq m^{*}$. We write $\operatorname{JIrr}^{c}$ and $\operatorname{MIrr}^{c}$ for the sets of completely join-irreducible elements and completely meet-irreducible elements.

A complete lattice $L$ is generated by completely join-irreducible elements if each element of $L$ can be written as a (possibly infinite) join of completely join-irreducible elements. Equivalently, for all $x\in L$, we have $x=\bigvee_{j\leq x,\ j\in\operatorname{JIrr}^{c}(L)}j$. We define generated by completely meet-irreducible elements dually, and we say that a lattice is meet and join generated by irreducibles if it satisfies both properties.

We will call a lattice $L$ a $\kappa$-lattice if it is complete, if it is meet and join generated by irreducibles, and if there are inverse bijections $\kappa:\operatorname{JIrr}^{c}(L)\to\operatorname{MIrr}^{c}(L)$ and $\kappa^{d}:\operatorname{MIrr}^{c}(L)\to\operatorname{JIrr}^{c}(L)$ such that $\kappa(j)$ is the maximum element of $\{y:j\wedge y=j_{*}\}$ and $\kappa^{d}(m)$ is the minimum element of $\{x:m\vee x=m^{*}\}$. It is a known result that a finite lattice is semidistributive if and only if it is a $\kappa$-lattice; we will reprove this as Corollary 2.29. We say that a $\kappa$-lattice $L$ is well separated if whenever $z_{1}\not\leq z_{2}$, there is some $j\in\operatorname{JIrr}^{c}(L)$ with $z_{1}\geq j$ and $\kappa(j)\geq z_{2}$. (Note that the converse is clear: If there is some $j$ with $z_{1}\geq j$ and $\kappa(j)\geq z_{2}$ then $z_{1}\not\leq z_{2}$, as otherwise we would have $\kappa(j)\geq z_{2}\geq z_{1}\geq j$, which is absurd.) In any $\kappa$-lattice, we define relations ${\mathord{\to}}_{L}$, ${\mathord{\twoheadrightarrow}}_{L}$, and ${\mathord{\hookrightarrow}}_{L}$ on $\operatorname{JIrr}^{c}(L)$ just as we did in the finite case: $i\to_{L}j$ if and only if $i\not\leq\kappa(j)$, $i\twoheadrightarrow_{L}j$ if and only if $i\geq j$ in $L$, and $i\hookrightarrow_{L}j$ if and only if $\kappa(i)\geq\kappa(j)$ in $L$.

We will prove the following theorem, which is a generalization of the Fundamental Theorem of Finite Semidistributive Lattices to well separated $\kappa$-lattices.

Theorem 1.3**.**

A (not necessarily finite) poset $L$ is a well separated $\kappa$-lattice if and only if it is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for a two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. If so, $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is isomorphic to $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$, and the map

[TABLE]

is an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\to}}_{L})$, with inverse $(X,Y)\mapsto\bigvee X=\bigwedge(\kappa(Y))$.

Section 2 contains the proof of Theorem 1.3, and Section 2.5 shows how the Fundamental Theorem of Finite Semidistributive Lattices (Theorem 1.2) follows as a special case. In Section 3, we place Theorem 1.3 in a broader context by discussing various conditions on a complete lattice that are equivalent, in the finite case, to semidistributivity. In particular, we provide counterexamples to conceivable versions of FTFSDL that concern completely semidistributive infinite lattices.

Suppose $L$ is a finite semidistributive lattice, realized as the maximal orthogonal pairs for a two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. It is apparent that any interval $[x,y]$ in $L$ is also a finite semidistributive lattice. In Section 4, given an interval $[x,y]$ in $\operatorname{Pairs}({\mathord{\to}})$, we explicitly construct a two-acyclic factorization system (defined on a subset of ) whose lattice of maximal orthogonal pairs is isomorphic to $[x,y]$. Our construction works for any lattice of the form $\operatorname{Pairs}({\mathord{\to}})$, for a two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ (which need not be finite).

In Section 5, we describe cover relations in the lattice of maximal orthogonal pairs, giving a particularly satisfying answer in the finite case. A major tool for the study of finite semidistributive lattices is the notion of canonical join representation of elements. Using the results on covers, we describe canonical join representations in the finite case in terms of the relation ${\mathord{\to}}$.

It is known that every quotient of a finite semidistributive lattice is semidistributive. (See for example [10, Lemma 2.14(b)] for a proof and for discussion of the infinite case.) In Section 6, we describe all quotients of a finite semidistributive lattice in terms of the FTFSDL. We define a relation $\mathord{\rightsquigarrow}$ on and prove:

Theorem 1.4**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a finite two-acyclic factorization system and $\Che\subseteq\Sha$ is a $\mathord{\rightsquigarrow}$-upset. Then the restriction of $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ to is a two-acyclic factorization system and the map $(X,Y)\mapsto(X\cap\Che,Y\cap\Che)$ is a surjective lattice homomorphism between the corresponding lattices of maximal orthogonal pairs. Every lattice quotient of $\operatorname{Pairs}({\mathord{\to}})$ arises in this way for a unique $\mathord{\rightsquigarrow}$-upset .

Theorem 1.4 allows us to give a precise description of the lattice $\operatorname{Con}(\operatorname{Pairs}({\mathord{\to}}))$ of congruences of any finite semidistributive lattice. As discussed in Section 6, the congruence lattice of a finite lattice $L$ can be understood in terms of a preorder on $\operatorname{JIrr}(L)$ called the forcing preorder.

Corollary 1.5**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a finite two-acyclic factorization system. The map $x\mapsto(T(x),T(x)^{\perp})$ is an isomorphism from the transitive closure of $\mathord{\rightsquigarrow}$ on to the forcing preorder on $\operatorname{JIrr}(\operatorname{Pairs}({\mathord{\to}}))$. This map induces an isomorphism from the poset of $\mathord{\rightsquigarrow}$-downsets under containment to the congruence lattice $\operatorname{Con}(\operatorname{Pairs}({\mathord{\to}}))$.

We prove Theorem 1.4 and Corollary 1.5 by generalizing them to the infinite case, with appropriate conditions on the two-acyclic factorization systems and congruences (Theorem 6.2 and Corollary 6.14).

Section 7 relates the FTFSDL to work on other classes of finite lattices, namely distributive lattices, congruence uniform lattices, general finite lattices, and extremal lattices. Section 8 concludes the paper by discussing the motivating examples, namely posets of regions of tight hyperplane arrangements and lattices of torsion classes of finite-dimensional algebras.

2. The fundamental theorem

In this section, we prove the FTFSDL by proving its generalization Theorem 1.3. We will separate out the following two auxiliary theorems, one of which is a piece of Theorem 1.3, and the other of which is a useful fact about $\kappa$-lattices that are not necessarily well separated.

Theorem 2.1**.**

If $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a (not necessarily finite) two-acyclic factorization system, then $\operatorname{Pairs}({\mathord{\to}})$ is a well separated $\kappa$-lattice. Furthermore, writing $L$ for $\operatorname{Pairs}({\mathord{\to}})$, the system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is isomorphic to $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$.

Theorem 2.2**.**

If $L$ is a $\kappa$-lattice, then the map

[TABLE]

is an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\to}}_{L})$, with inverse $(X,Y)\mapsto\bigvee X=\bigwedge(\kappa(Y))$. The system $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$ has all of the properties of a two-acyclic factorization system except that possibly $\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})\subsetneq{\mathord{\to}}_{L}$.

After some preliminary material in Sections 2.1 and 2.2, we prove Theorem 2.1 in Section 2.3. We prove Theorem 2.2 in Section 2.4, where we also complete the proof of Theorem 1.3. Finally, in Section 2.5, we prove the Fundamental Theorem of Finite Semidistributive Lattices (Theorem 1.2) by showing that it is the finite case of Theorem 1.3.

2.1. Conventions about partially ordered sets

In a two-acyclic factorization system, the relations ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ are partial orders. When we use the language of partial orders to describe them, we consider $x$ to be greater than or equal to $y$ if $x\twoheadrightarrow y$ or $x\hookrightarrow y$ respectively. For example, if we say “$x$ is an ${\mathord{\twoheadrightarrow}}$-minimal element of $S$”, we mean that, for any $x^{\prime}\in S$ other than $x$, we do not have $x\twoheadrightarrow x^{\prime}$. Throughout the paper, if we say that a set $P$ is a partially ordered set under the relation $R$ and if $R$ is some sort of arrow pointing from $x$ to $y$, then we intend $x\ R\ y$ to correspond to $x\geq y$.

For a directed graph with vertex set $V$, we define a downset of $V$ to be a set $D\subseteq V$ such that $x\in D$ and $x\to y$ implies $y\in D$; we define $U$ to be an upset if $y\in U$ and $x\to y$ implies $x\in U$. We allow ourselves to use this language even when the relation ${\mathord{\to}}$ is not transitive and not antisymmetric. The downsets of ${\mathord{\to}}$ are the same as the downsets of the transitive closure of ${\mathord{\to}}$, and are in natural bijective correspondence with the downsets of the graph formed by collapsing each strongly connected component of ${\mathord{\to}}$ to a point.

In a partially ordered set $P$, we say that an element $x$ covers another element $y$, written $x\gtrdot y$, if $x>y$ and there does not exist any $z$ with $x>z>y$. The Hasse diagram of $P$ is the graph with vertex set $P$ and an edge from $x$ to $y$ if $x\gtrdot y$.

2.2. Fundamentals of factorization systems

We now prove some basic facts about the operations $\operatorname{Fact}$ and $\operatorname{Mult}$, introduced in the introduction.

Proposition 2.3**.**

Suppose ${\mathord{\to}}$ is a reflexive relation, and ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ are reflexive and transitive relations, on a set .

Given two pairs of reflexive and transitive relations $({\mathord{\twoheadrightarrow}}_{1},{\mathord{\hookrightarrow}}_{1})\subseteq({\mathord{\twoheadrightarrow}}_{2},{\mathord{\hookrightarrow}}_{2})$, then $\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{1},{\mathord{\hookrightarrow}}_{1})\subseteq\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{2},{\mathord{\hookrightarrow}}_{2})$. 2. 2.

$({\mathord{\twoheadrightarrow}}\cup{\mathord{\hookrightarrow}})\subseteq\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. 3. 3.

$\operatorname{Mult}(\operatorname{Fact}({\mathord{\to}}))\subseteq{\mathord{\to}}$. 4. 4.

$\operatorname{Fact}(\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}}))\supseteq({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. 5. 5.

$\operatorname{Mult}(\operatorname{Fact}(\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})))=\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$.

The containment symbol between ordered pairs stands for containment in each component.

Proof.

Property 1 is obvious.

For Property 2, write ${\mathord{\to}}$ for $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. Suppose $x\twoheadrightarrow y$. By definition of multiplication, since ${\mathord{\hookrightarrow}}$ is reflexive (and thus $y\hookrightarrow y$), we have $x\to y$. Similarly, if $x\hookrightarrow y$, then since $x\twoheadrightarrow x$, we have $x\to y$.

For Property 3, write $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ for $\operatorname{Fact}({\mathord{\to}})$ and write ${\mathord{\to}}^{\prime}$ for $\operatorname{Mult}(\operatorname{Fact}({\mathord{\to}}))$. If $x\to^{\prime}y$, then there exists $z\in\Sha$ such that $x\twoheadrightarrow z\hookrightarrow y$. By Property 2, we have $z\to y$, and therefore also $x\to y$ by the definition of $\operatorname{Fact}({\mathord{\to}})$.

To prove Property 4, we write ${\mathord{\to}}$ for $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, and write $({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$ for $\operatorname{Fact}(\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}}))$. We must show that, if $x\twoheadrightarrow y$, then $x\twoheadrightarrow^{\prime}y$. Unpacking the definition of ${\mathord{\twoheadrightarrow}}^{\prime}$, we must show that, if $y\to z$, then $x\to z$. Since $y\to z$, there exists $w$ with $y\twoheadrightarrow w\hookrightarrow z$. Then by transitivity of ${\mathord{\twoheadrightarrow}}$, we have $x\twoheadrightarrow w\hookrightarrow z$, so $x\to z$. We have shown that $x\twoheadrightarrow^{\prime}y$. The proof for ${\mathord{\hookrightarrow}}$ and ${\mathord{\hookrightarrow}}^{\prime}$ is similar.

For Property 5, we have $\operatorname{Mult}(\operatorname{Fact}(\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})))\subseteq\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ by Property 3. Combining Properties 1 and 4, we have the opposite containment as well. ∎

Proposition 2.4**.**

Let ${\mathord{\to}}$ be a relation on and let $\operatorname{Fact}({\mathord{\to}})=({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. Let $(T,F)$ be a maximal orthogonal pair for ${\mathord{\to}}$. Then $T$ is a downset for ${\mathord{\twoheadrightarrow}}$ and $F$ is an upset for ${\mathord{\hookrightarrow}}$. In other words, if $x\twoheadrightarrow x^{\prime}$ and $x\in T$ then $x^{\prime}\in T$ and, if $y^{\prime}\hookrightarrow y$ and $y\in F$ then $y^{\prime}\in F$.

Proof.

We prove the claim about $T$; the claim about $F$ is analogous. Let $x\twoheadrightarrow x^{\prime}$ and suppose that $x\in T$. We must show that, if $x\not\to y$ then $x^{\prime}\not\to y$. The contrapositive, that if $x^{\prime}\to y$ then $x\to y$, is the definition of ${\mathord{\twoheadrightarrow}}$. ∎

Remark 2.5**.**

The letters $T$ and $F$ suggest “Torsion” and “torsion-Free”, terminology from the theory of torsion pairs. See Section 8.2.

Proposition 2.6**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system. If $x\to y\hookrightarrow x$ or $x\twoheadrightarrow y\to x$, then $x=y$.

Proof.

We prove the first statement; the second is similar. Let $x\to y\hookrightarrow x$. Since ${\mathord{\to}}=\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, there is some $w$ with $x\twoheadrightarrow w\hookrightarrow y\hookrightarrow x$. By transitivity of ${\mathord{\hookrightarrow}}$, we have $x\twoheadrightarrow w\hookrightarrow x$ so, by the brick property $x=w$. Now we have $x=w\hookrightarrow y\hookrightarrow x$ so, by the order property, $x=y$ as well. ∎

Our definitions have a symmetry under reversing arrows, which we spell out in the easy proposition below.

Proposition 2.7**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a factorization system. Define ${\mathord{\to}}^{\mathrm{op}}$ by $y\to^{\mathrm{op}}x$ if $x\to y$ and similarly define $x\hookrightarrow^{\mathrm{op}}y$ if $y\hookrightarrow x$ and $x\twoheadrightarrow^{\mathrm{op}}y$ if $y\twoheadrightarrow x$. Then $(\Sha,\to^{\mathrm{op}},{\mathord{\hookrightarrow}}^{\mathrm{op}},{\mathord{\twoheadrightarrow}}^{\mathrm{op}})$ is a factorization system; to be explicit, ${\mathord{\hookrightarrow}}^{\mathrm{op}}$ has taken on the role ${\mathord{\twoheadrightarrow}}$ and vice versa. The system $(\Sha,\to^{\mathrm{op}},{\mathord{\hookrightarrow}}^{\mathrm{op}},{\mathord{\twoheadrightarrow}}^{\mathrm{op}})$ is two-acyclic if and only if $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is. The map $(X,Y)\mapsto(Y,X)$ is an anti-isomorphism from $\operatorname{Pairs}({\mathord{\to}}^{\mathrm{op}})$ to $\operatorname{Pairs}({\mathord{\to}})$.

In a factorization system, the pair $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ determines ${\mathord{\to}}$. It is often practical to construct examples of factorization systems by giving $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ and defining ${\mathord{\to}}=\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. It is therefore useful to state the hypotheses of being two-acyclic and factorization in terms solely of the partial orders ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$. After introducing some notation, we do this:

Given partial orders ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ on and a subset $X\subseteq\Sha$, we write \operatorname{down}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}X for the set of elements weakly below elements of $X$ in the sense of ${\mathord{\twoheadrightarrow}}$. Similarly, we write \operatorname{up}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}X, \operatorname{down}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}X and \operatorname{up}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}X.

Proposition 2.8**.**

Given two partial orders ${\mathord{\hookrightarrow}}$ and ${\mathord{\twoheadrightarrow}}$ on a set , the tuple $(\Sha,\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}}),{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system if and only if the following conditions hold.

1. (i)

There do not exist distinct $x$ and $y$ in ** with $x\hookrightarrow y$ and $y\twoheadrightarrow x$. 2. (ii)

$x\twoheadrightarrow y$* if and only if \operatorname{down}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}\operatorname{down}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}\{y\}\subseteq\operatorname{down}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}\operatorname{down}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}\{x\} for all $x,y\in\Sha$.* 3. (iii)

$x\hookrightarrow y$* if and only if \operatorname{up}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}\operatorname{up}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}\{x\}\subseteq\operatorname{up}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}\operatorname{up}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}\{y\} for all $x,y\in\Sha$.*

If the conditions hold, then for any $X\subseteq\Sha$, the sets $X^{\perp}$ and ${}^{\perp}X$ are described by

[TABLE]

Proof.

We write ${\mathord{\to}}$ for $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. Then \operatorname{down}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}(\operatorname{down}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}X)={\{y:\exists x\in X,x\to y\}} and \operatorname{up}_{\raisebox{7.0pt}{\scalebox{0.8}{\rotatebox{-90.0}{\twoheadrightarrow}}}}(\operatorname{up}_{\raisebox{7.0pt}[7.0pt][2.0pt]{\scalebox{0.75}{\rotatebox{-90.0}{\hookrightarrow}}}}X)=\{y:\exists x\in X,y\to x\}. Conditions (ii) and (iii) amount to the assertion that $\operatorname{Fact}(\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}}))=({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. Thus $(\Sha,\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}}),{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a factorization system if and only if (ii) and (iii) hold. The order condition holds in any case, because ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ are partial orders. Condition (i) is a restatement of the brick condition. The final statement is clear. ∎

2.3. From factorization systems to $\kappa$-lattices

In this section, we prove Theorem 2.1. We begin by quickly showing that $\operatorname{Pairs}({\mathord{\to}})$ is a complete lattice for any binary relation ${\mathord{\to}}$.

Given a binary relation ${\mathord{\to}}$ on a set , and a subset $X\subseteq\Sha$, we define $\overline{X}$ to be ${}^{\perp}(X^{\perp})$. We call a set $X$ closed if $\overline{X}=X$. We quote the following standard results from Birkhoff [7]. Note that Birkhoff works more generally with two sets and a relation between them (a situation which will occur for us in Section 7.3 ) and that his relation $\rho$ corresponds to our relation $\not\to$.

Proposition 2.9** ([7], Theorem V.19).**

For any binary relation ${\mathord{\to}}$, the map $X\mapsto\overline{X}$ is a closure operator on . In other words,

• •

$X\subseteq\overline{X}$,

• •

$\overline{\overline{X}}=\overline{X}$* for all $X\subseteq\Sha$, and*

• •

$X\subseteq Y$* implies $\overline{X}\subseteq\overline{Y}$ for all $X,Y\subseteq\Sha$.*

Proposition 2.10** ([7], Theorems V.1 and V.2).**

Let $X\mapsto\overline{X}$ be any closure operator on a set . Then the containment order on closed subsets of is a complete lattice. The meet operation is $\cap$.

We also make some easy observations.

Proposition 2.11**.**

If ${\mathord{\to}}$ is a binary relation on a set , then $(X,Y)\mapsto X$ is an isomorphism from $\operatorname{Pairs}({\mathord{\to}})$ to the complete lattice of closed sets under containment order. The inverse is $X\mapsto(X,X^{\perp})$.

Proposition 2.12**.**

If ${\mathord{\to}}$ is a binary relation on , and $S\subseteq\Sha$, then ${}^{\perp}S$ is closed.

Proof.

We must show that $\overline{{}^{\perp}S}=S$. Expanding the definition of closure, this says that {}^{\perp}S={}^{\perp}{\big{(}}{\big{(}}{}^{\perp}S{\big{)}}^{\perp}{\big{)}}. This is standard, see for example [7], the Corollary before Theorem V.19. ∎

We will often identify $\operatorname{Pairs}({\mathord{\to}})$ with the poset of closed sets and specify elements $(X,Y)$ of $\operatorname{Pairs}({\mathord{\to}})$ by giving only the closed set $X$.

We now show that $\operatorname{Pairs}({\mathord{\to}})$ is a $\kappa$-lattice when $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system. To this end, we study completely join-irreducible elements and completely meet-irreducible elements in $\operatorname{Pairs}({\mathord{\to}})$. For $x\in\Sha$, define $T(x)=\{x^{\prime}:x\twoheadrightarrow x^{\prime}\}$ and $F(x)=\{x^{\prime}:x^{\prime}\hookrightarrow x\}$.

Proposition 2.13**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system. For any $x\in\Sha$, we have $T(x)={}^{\perp}(\{x\}^{\perp})$ and $F(x)=({}^{\perp}\{x\})^{\perp}$. Moreover, $x$ is uniquely determined by $T(x)$ and is uniquely determined by $F(x)$.

Proof.

Checking that $T(x)$ is the closure $\overline{\{x\}}:={}^{\perp}(\{x\}^{\perp})$ is a matter of unfolding definitions: We have $x^{\prime}\in\overline{\{x\}}$ if and only if, for all $y\in\Sha$, if $x\not\to y$ then $x^{\prime}\not\to y$. Taking the contrapositive, we have $x^{\prime}\in\overline{\{x\}}$ if and only if $x^{\prime}\to y$ implies $x\to y$, which is the definition of $x\twoheadrightarrow x^{\prime}$. If $T(x_{1})=T(x_{2})$, then $x_{1}\twoheadrightarrow x_{2}$ and $x_{2}\twoheadrightarrow x_{1}$, so the order condition implies that $x_{1}=x_{2}$. We have checked both assertions about $T(x)$; the assertions about $F(x)$ are proved similarly. ∎

We can now describe the completely join-irreducible elements and completely meet-irreducible elements of $\operatorname{Pairs}({\mathord{\to}})$. We define $T_{*}(x)=T(x)\setminus\{x\}$ and $F^{*}(x)=F(x)\setminus\{x\}$.

Proposition 2.14**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system.

$\operatorname{JIrr}^{c}(\operatorname{Pairs}({\mathord{\to}}))=\{(T(x),T(x)^{\perp}):x\in\Sha\}$. The unique element covered by $(T(x),T(x)^{\perp})$ is $(T_{*}(x),T_{*}(x)^{\perp})$. 2. 2.

$\operatorname{MIrr}^{c}(\operatorname{Pairs}({\mathord{\to}}))=\{({}^{\perp}F(x),F(x)):x\in\Sha\}$. The unique element covering $({}^{\perp}F(x),F(x))$ is $({}^{\perp}F^{*}(x),F^{*}(x))$.

Proof.

We check the claim about completely join-irreducible elements; the claim about completely meet-irreducible elements is dual.

The set $T(x)$ is closed by Proposition 2.13. If $T^{\prime}$ is a closed set with $T^{\prime}\subset T(x)$ and $T^{\prime}\not\subseteq T_{*}(x)$, then $x\in T^{\prime}$, and thus $T(x)\subseteq T^{\prime}$ by Proposition 2.4. That is, every closed set strictly contained in $T(x)$ is contained in $T_{*}(x)$, so we can complete the proof by showing that $T_{*}(x)$ is closed. Since closure preserves containment and $T(x)$ is closed, the closure of $T_{*}(x)$ is either $T_{*}(x)$ or $T(x)$. To see that $\overline{T_{*}(x)}\neq T(x)$, note that $x\in T_{*}(x)^{\perp}$ by Proposition 2.6. So $x\not\in{}^{\perp}(T_{*}(x)^{\perp})=\overline{T_{*}(x)}$ and we deduce that $T_{*}(x)$ is closed.

Conversely, suppose that $T$ is a completely join-irreducible closed set. By Proposition 2.4, $T$ is a downset of ${\mathord{\twoheadrightarrow}}$, so $T=\bigcup_{x\in T}T(x)$. Each $T(x)$ is closed, so $T=\bigvee_{x\in T}T(x)$. By the definition of complete join-irreducibility, $T=T(x)$ for some $x\in T$, as desired. ∎

Thus in a two-acyclic factorization system, $(T(x),T(x)^{\perp})\leftrightarrow({}^{\perp}F(x),F(x))$ is a bijection between $\operatorname{JIrr}^{c}(\operatorname{Pairs}({\mathord{\to}}))$ and $\operatorname{MIrr}^{c}(\operatorname{Pairs}({\mathord{\to}}))$. We will show that these bijections yield maps $\kappa$ and $\kappa^{d}$ manifesting that $\operatorname{Pairs}({\mathord{\to}})$ is a $\kappa$-lattice.

Proposition 2.15**.**

Let $x\in\Sha$. Then ${}^{\perp}F(x)$ is the maximum element in the set of closed sets $T$ obeying $T(x)\cap T=T_{*}(x)$.

Proof.

We first check that $T(x)\cap({}^{\perp}F(x))=T_{*}(x)$. Since $x\in F(x)$, we have $x\not\in{}^{\perp}F(x)$, so we only need to show that $T_{*}(x)\subseteq{}^{\perp}F(x)$. That is, given $y\in\Sha$ with $x\twoheadrightarrow y$ and $x\neq y$ and $z\in\Sha$ with $z\hookrightarrow x$, we need to show that $y\not\to z$. But if $y\to z$, then there exists $w\in\Sha$ such that $y\twoheadrightarrow w\hookrightarrow z$. By the transitivity and antisymmetry of ${\mathord{\twoheadrightarrow}}$, we have $x\twoheadrightarrow w$ and $x\neq w$. By the transitivity of ${\mathord{\hookrightarrow}}$, we have $w\hookrightarrow x$. But then $x\twoheadrightarrow w\hookrightarrow x$, contradicting the brick condition. By this contradiction, we conclude that $T(x)\cap({}^{\perp}F(x))=T_{*}(x)$.

It remains to show that any closed set $T$ with $T(x)\cap T=T_{*}(x)$ has $T\subseteq{}^{\perp}F(x)$. The hypothesis that $T(x)\cap T=T_{*}(x)$ can be restated as $T_{*}(x)\subseteq T$ and $x\not\in T$.

Suppose for the sake of contradiction that there is some $u\in T$ and $u\not\in{}^{\perp}F(x)$. Thus $u\to v\hookrightarrow x$ for some $v$, so there exists $y$ such that $u\twoheadrightarrow y\hookrightarrow v$. Since $T$ is an order ideal for ${\mathord{\twoheadrightarrow}}$, we have $y\in T$ and, by transitivity of ${\mathord{\hookrightarrow}}$, we have $y\hookrightarrow x$.

Now, $x\not\in T$ and $T$ is closed, so there is some $z\in T^{\perp}$ such that $x\to z$. Thus, for some $w$, we have $x\twoheadrightarrow w\hookrightarrow z$, and in particular $w\to z$. If $w\neq x$ then $w\in T_{*}(x)\subseteq T$, but then the facts $w\to z$ and $z\in T^{\perp}$ contradict each other. Therefore, we must have $x=w$, so that $x\hookrightarrow z$. We also showed above that $y\hookrightarrow x$, so now we conclude that $y\hookrightarrow z$ and thus $y\to z$. Since $y\in T$, this contradicts $z\in T^{\perp}$. We conclude that $T\subseteq{}^{\perp}F(x)$, and the proof is complete. ∎

Writing $F^{*}(x)$ for $F(x)\setminus\{x\}$, the following proposition is dual to Proposition 2.15.

Proposition 2.16**.**

Let $x\in\Sha$. Then $T(x)$ is the minimum element in the set of closed sets $T$ obeying ${}^{\perp}F(x)\vee T={}^{\perp}F^{*}(x)$.

Propositions 2.15 and 2.16 combine to prove the following piece of Theorem 2.1.

Proposition 2.17**.**

If $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system, then $\operatorname{Pairs}({\mathord{\to}})$ is a $\kappa$-lattice. Specifically,

[TABLE]

To prove that $\operatorname{Pairs}({\mathord{\to}})$ is well separated, we first point out a simple lemma.

Lemma 2.18**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system and let $(X_{1},Y_{1})$ and $(X_{2},Y_{2})\in\operatorname{Pairs}({\mathord{\to}})$. If $X_{1}\cap Y_{2}=\emptyset$ then $(X_{1},Y_{1})\leq(X_{2},Y_{2})$.

Proof.

We show the contrapositive. Suppose that $X_{1}\not\subseteq X_{2}$. Then there is some $p\in X_{1}\setminus X_{2}$. Since $p\not\in X_{2}$, there is some $r\in Y_{2}$ with $p\to r$. Factor this arrow as $p\twoheadrightarrow q\hookrightarrow r$. Then $q\in X_{1}\cap Y_{2}$. ∎

Proposition 2.19**.**

If $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system, then $\operatorname{Pairs}({\mathord{\to}})$ is well separated.

Proof.

For $z_{1}$ and $z_{2}\in\operatorname{Pairs}({\mathord{\to}})$ with $z_{1}\not\leq z_{2}$, we need to show that there is a completely join irreducible element $j$ of $\operatorname{Pairs}({\mathord{\to}})$ with $z_{1}\geq j$ and $\kappa(j)\geq z_{2}$. Let $z_{1}=(X_{1},Y_{1})$ and $z_{2}=(X_{2},Y_{2})$. By the contrapositive of Lemma 2.18, the assumption that $z_{1}\not\leq z_{2}$ implies that $X_{1}\cap Y_{2}\neq\emptyset$; let $q\in X_{1}\cap Y_{2}$. Put $j=(T(q),T(q)^{\perp})$. By Proposition 2.14, $j$ is completely join irreducible; by Proposition 2.15, $\kappa(j)=({}^{\perp}F(q),F(q))$. Because $X_{1}$ and $Y_{2}$ are a ${\mathord{\twoheadrightarrow}}$-downset and an ${\mathord{\hookrightarrow}}$-upset respectively, we have $X_{1}\supseteq T(q)$ and $F(q)\subseteq Y_{2}$, so $z_{1}\geq j$ and $\kappa(j)\geq z_{2}$. ∎

To complete the proof of Theorem 2.1, we now show that, if $L\cong\operatorname{Pairs}({\mathord{\to}})$ for a two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, then $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is isomorphic to $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$, for ${\mathord{\to}}_{L}$, ${\mathord{\twoheadrightarrow}}_{L}$, and ${\mathord{\hookrightarrow}}_{L}$ as defined just before Theorem 1.3.

Suppose that $L\cong\operatorname{Pairs}({\mathord{\to}})$ for some two-acyclic factorization system. From Proposition 2.14 we have bijections $\Sha\leftrightarrow\operatorname{JIrr}^{c}(L)\leftrightarrow\operatorname{MIrr}^{c}(L)$ given by $x\leftrightarrow(T(x),{}^{\perp}T(x))\leftrightarrow(F(x)^{\perp},F(x))$. We now establish that these bijections turn ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ into ${\mathord{\twoheadrightarrow}}_{L}$ and ${\mathord{\hookrightarrow}}_{L}$. Since ${\mathord{\to}}=\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ and ${\mathord{\to}}_{L}=\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$, this shows that ${\mathord{\to}}={\mathord{\to}}_{L}$ as well.

Proposition 2.20**.**

For $x_{1}$ and $x_{2}\in\Sha$, we have $x_{1}\twoheadrightarrow x_{2}$ if and only if we have $(T(x_{1}),{}T(x_{1})^{\perp}){\mathord{\twoheadrightarrow}}_{L}(T(x_{2}),{}T(x_{2})^{\perp})$ and $x_{1}\hookrightarrow x_{2}$ if and only if we have $({}^{\perp}F(x_{1}),F(x_{1})){\mathord{\hookrightarrow}}_{L}({}^{\perp}F(x_{2}),F(x_{2}))$.

Proof.

We prove the claim about ${\mathord{\twoheadrightarrow}}$; the claim about ${\mathord{\hookrightarrow}}$ is analogous. Unwinding definitions, we must show that $x_{1}\twoheadrightarrow x_{2}$ if and only if $\{x^{\prime}:x_{1}\twoheadrightarrow x^{\prime}\}\supseteq{\{x^{\prime}:x_{2}\twoheadrightarrow x^{\prime}\}}$. This holds because ${\mathord{\twoheadrightarrow}}$ is a partial order. ∎

This completes the proof of Theorem 2.1, which is part of Theorem 1.3.

2.4. From $\kappa$-lattices to factorization systems

In this section, we prove Theorem 2.2 and complete the proof of Theorem 1.3.

Proposition 2.21**.**

Suppose $L$ is a $\kappa$-lattice. Then $\operatorname{Fact}({\mathord{\to}}_{L})=({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$ and $\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})\subseteq{\mathord{\to}}_{L}$.

Proof.

There are two facts to verify. First, we want to show that $i\twoheadrightarrow_{L}j$ if and only if whenever $j\to_{L}\ell$, we also have $i\to_{L}\ell$. That is, we want $i\geq j$ if and only if $(j\not\leq\kappa(\ell))\implies(i\not\leq\kappa(\ell))$ for all $\ell\in\operatorname{JIrr}^{c}(L)$. Taking the contrapositive, we want $i\geq j$ if and only if $\{\ell\in\operatorname{JIrr}^{c}:i\leq\kappa(\ell)\}\supseteq\{\ell\in\operatorname{JIrr}^{c}:j\leq\kappa(\ell)\}$. The forward implication is obvious, and, since the image of $\kappa$ is $\operatorname{MIrr}^{c}$, the reverse follows because in a $\kappa$-lattice, every element is the meet of the set of completely meet irreducible elements above it.

Second, we want to show that $i\hookrightarrow_{L}j$ if and only if whenever $\ell\to_{L}i$, we also have $\ell\to_{L}j$. The argument is dual. We see that $\operatorname{Fact}({\mathord{\to}}_{L})=({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$. Proposition 2.3.3 thus implies that $\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})\subseteq{\mathord{\to}}_{L}$. ∎

We now check that $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$ is two-acyclic. We first check the order condition, and then the brick condition.

Proposition 2.22**.**

Suppose $L$ is a $\kappa$-lattice. If $x\twoheadrightarrow_{L}y\twoheadrightarrow_{L}x$ or $x\hookrightarrow_{L}y\hookrightarrow_{L}x$, then $x=y$.

Proof.

The definition of $x\twoheadrightarrow_{L}y\twoheadrightarrow_{L}x$ is that $x\leq y\leq x$, so this follows because $\leq$ is a partial order. Similarly, $x\hookrightarrow_{L}y\hookrightarrow_{L}x$ means $\kappa(x)\leq\kappa(y)\leq\kappa(x)$ so $\kappa(x)=\kappa(y)$, and the map $\kappa$ is invertible so $x=y$. ∎

Proposition 2.23**.**

Suppose $L$ is a $\kappa$-lattice. If $x\twoheadrightarrow_{L}y\hookrightarrow_{L}x$, then $x=y$.

Proof.

The hypotheses of the proposition are that $x\geq y$ and $\kappa(y)\geq\kappa(x)$. If $x>y$ then $x_{*}\geq y$ so $\kappa(y)\geq\kappa(x)\geq x_{*}\geq y$, contradicting that $\kappa(y)\not\geq y$. We deduce that $x$ is not strictly greater than $y$, so $x=y$. ∎

It remains to construct an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\to}}_{L})$. We define $J_{x}={\{j\in\operatorname{JIrr}^{c}(L):j\leq x\}}$ and $M_{x}=\{m\in\operatorname{MIrr}^{c}(L):m\geq x\}$.

Proposition 2.24**.**

Suppose $L$ is a $\kappa$-lattice. The map $x\mapsto(J_{x},\kappa^{d}(M_{x}))$ is an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\to}}_{L})$.

Proof.

Since each $x\in L$ has $x=\bigvee J_{x}$, the map $x\mapsto(J_{x},\kappa^{d}(M_{x}))$ is an isomorphism from $L$ to its image, where the image is partially ordered by containment in the first entry and equivalently reverse containment in the second entry. Thus it remains only to show that the image is the set of maximal orthogonal pairs for ${\mathord{\to}}_{L}$.

For $x\in L$ the set $J_{x}^{\perp}$ is $\{j\in\operatorname{JIrr}^{c}(L):i\leq\kappa(j),\,\forall i\in J_{x}\}$, which equals $\{j\in\operatorname{JIrr}^{c}(L):x\leq\kappa(j)\}=\kappa^{d}(\{m\in\operatorname{MIrr}^{c}(L):x\leq m\})=\kappa^{d}(M_{x})$. Also, ${}^{\perp}(\kappa^{d}(M_{x}))=\{i\in\operatorname{JIrr}^{c}(L):i\leq m,\,\forall m\in M_{x}\}$, which equals ${\{i\in\operatorname{JIrr}^{c}(L):i\leq x\}}=J_{x}$. We see that $(J_{x},\kappa^{d}(M_{x}))$ is a maximal orthogonal pair.

On the other hand, suppose $X$ and $Y$ are subsets of $\operatorname{JIrr}^{c}(L)$ such that $(X,Y)$ is a maximal orthogonal pair for ${\mathord{\to}}_{L}$ and write $x=\bigvee X$. Then $Y$ equals $X^{\perp}$, which equals $\{j\in\operatorname{JIrr}^{c}(L):i\leq\kappa(j),\,\forall i\in X\}=\{j\in\operatorname{JIrr}^{c}(L):x\leq\kappa(j)\}=\kappa^{d}(M_{x})$. Computing ${}^{\perp}(\kappa^{d}(M_{x}))$ as in the previous paragraph, we see that $X=J_{x}$. ∎

This completes the proof of Theorem 2.2. The following proposition completes the proof of Theorem 1.3.

Proposition 2.25**.**

If $L$ is a well separated $\kappa$-lattice, then $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$ is a two-acyclic factorization system and $L$ is isomorphic to $\operatorname{Pairs}({\mathord{\to}}_{L})$.

Proof.

Since $L$ is a $\kappa$-lattice, Theorem 2.2 says that $L\cong\operatorname{Pairs}({\mathord{\to}}_{L})$ and that $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$ obeys all properties of a two-acyclic factorization system except possibly that $\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})={\mathord{\to}}_{L}$. We now address that last point. Let $i$ and $j$ be completely join-irreducible elements with $i\to_{L}j$ or, in other words, $i\not\leq\kappa(j)$. Since $L$ is well separated, there is a join irreducible element $\ell$ such that $i\geq\ell$ and $\kappa(\ell)\geq\kappa(j)$. In other words, $i\twoheadrightarrow_{L}\ell\hookrightarrow_{L}j$. ∎

We pause here to give an alternate characterization of ${\mathord{\to}}_{L}$ for later use.

Lemma 2.26**.**

Let $L$ be a $\kappa$-lattice and let $i,j\in\operatorname{JIrr}^{c}(L)$. Then $i\to_{L}j$ if and only if $i\vee j_{*}\geq j$, where $j_{*}$ is the unique element of $L$ covered by $j$.

Proof.

Since $j_{*}\leq\kappa(j)$, we have $i\leq\kappa(j)$ if and only if $i\vee j_{*}\leq\kappa(j)$. Since $j\not\leq\kappa(j)$, if $i\vee j_{*}\leq\kappa(j)$, then $i\vee j_{*}\not\geq j$. On the other hand, if $i\vee j_{*}\not\leq\kappa(j)$, then since $j_{*}\leq i\vee j_{*}$, by definition of $\kappa$, we have $i\vee j_{*}\geq j$. ∎

2.5. Finite semidistributive lattices

In this section, we recall and prove some basic facts about finite lattices that show that FTFSDL is a special case of Theorem 1.3. We also use Proposition 2.8 to give a formulation of the FTFSDL which references only ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$, and not ${\mathord{\to}}$.

Proposition 2.27**.**

If $L$ is a finite lattice and $x\in L$, then

[TABLE]

Proof.

We argue the first equality; the other is dual. It is enough to show that $x$ is the join of a set of join-irreducible elements. If $x$ is not join-irreducible, then it is the join of a set $S$ of elements strictly lower than $x$ in $L$. By induction in $L$, each element of $S$ is the join of a set of join-irreducible elements, and thus $x$ is also. ∎

We use the following well-known characterization of semidistributivity in finite lattices. See, for example, [1, Theorem 3-1.4] or [11, Theorem 2.56]. (Note that these sources disagree on which map is called $\kappa$ and which is called $\kappa^{d}$.) Since the proof is short, we include it.

Theorem 2.28**.**

Suppose $L$ is a finite lattice.

$L$* is meet semidistributive if and only if for every $j\in\operatorname{JIrr}(L)$, the set ${\{x\in L:j\wedge x=j_{*}\}}$ has a maximum element $\kappa(j)$.* 2. 2.

$L$* is join semidistributive if and only if for each $m\in\operatorname{MIrr}(L)$, the set ${\{x\in L:m\vee x=m^{*}\}}$ has a minimum element $\kappa^{d}(m)$.*

If $L$ is semidistributive then $\kappa$ is a bijection from $\operatorname{JIrr}(L)$ to $\operatorname{MIrr}(L)$ with inverse $\kappa^{d}$.

Proof.

Suppose $L$ is meet semidistributive. Suppose $j$ is a join-irreducible element. Meet semidistributivity implies that the join $\bigvee\{y:j\wedge y=j_{*}\}$ is itself an element of $\{y:j\wedge y=j_{*}\}$. Thus this set has a maximum element (i.e., an element greater than all the other elements in the set). We denote it $\kappa(j)$.

Conversely, suppose that ${\{x\in L:j\wedge x=j_{*}\}}$ has a maximum element $\kappa(j)$ for every $j\in\operatorname{JIrr}(L)$. Let $x$, $y$, and $z$ be elements of $L$ with $x\wedge y=x\wedge z$. In any case, $x\wedge(y\vee z)\geq x\wedge y$. If $x\wedge(y\vee z)>x\wedge y$, let $j$ be minimal among elements that are $\leq x\wedge(y\vee z)$ and $\not\leq x\wedge y$. (It is in choosing this minimal element $j$ that we make use of the hypothesis of finiteness.) If $j$ covers two distinct elements $k_{1}$ and $k_{2}$, then $j=k_{1}\vee k_{2}$, but $k_{1}\vee k_{2}\leq x\wedge y$ since minimality of $j$ implies $k_{1}\leq x\wedge y$ and $k_{2}\leq x\wedge y$. Thus $j$ is join-irreducible. Minimality of $j$ also implies that $j_{*}\leq x\wedge y\leq y$. Similarly, $j_{*}\leq z$. However, $y\vee z\geq x\wedge(y\vee z)\geq j$, contradicting the existence of $\kappa(j)$. We conclude that $x\wedge(y\vee z)=x\wedge y$.

We have established the first numbered assertion; the second is dual.

We next check that the map $\kappa$ takes $\operatorname{JIrr}$ to $\operatorname{MIrr}$. If $\kappa(j)$ is not meet-irreducible, let $X$ be a set of elements with $\bigwedge X=\kappa(j)$ but $\kappa(j)\not\in X$. Then every element $x\in X$ has $x>\kappa(j)$, and thus $x\geq j$. Thus $\kappa(j)=\bigwedge X\geq j$, contradicting the definition of $\kappa(j)$. We conclude that $\kappa(j)\in\operatorname{MIrr}(L)$.

Finally, we must check that $\kappa$ and $\kappa^{d}$ are inverse. Write $\kappa(j)^{*}$ for the unique element covering $\kappa(j)$. If $L$ is semidistributive and $j\in\operatorname{JIrr}(L)$, then by definition of $\kappa(j)$, we have $j\leq\kappa(j)^{*}$ and $j_{*}\leq\kappa(j)$. Thus $j$ is a minimal element of ${\{x\in L:\kappa(j)\vee x=\kappa(j)^{*}\}}$, so that $\kappa^{d}(\kappa(j))=j$. The dual argument shows that $\kappa^{d}$ maps $\operatorname{MIrr}(L)$ to $\operatorname{JIrr}(L)$ and that $\kappa(\kappa^{d}(m))=m$ for all $m\in\operatorname{MIrr}(L)$. ∎

Proposition 2.27 and Theorem 2.28 combine to establish the following statement.

Corollary 2.29**.**

A finite lattice is a $\kappa$-lattice if and only if it is semidistributive.

Proposition 2.30**.**

If $L$ is a finite $\kappa$-lattice, then $L$ is well separated.

Proof.

Suppose $z_{1}\not\leq z_{2}$ in $L$. The set $\{x\in L:x\leq z_{1},\,x\not\leq z_{2}\}$ is not empty, because it contains $z_{1}$. Thus the set has a minimal element $j$. The element $j$ is join-irreducible; if $j=\bigvee X$ with $j\not\in X$, then $z_{2}$ is an upper bound for $X$, yielding the contradiction $j\leq z_{2}$. Also $j_{*}\leq z_{2}$, so $j\wedge z_{2}=j_{*}$, and thus $\kappa(j)\geq z_{2}$. ∎

Corollary 2.29, Proposition 2.30, and the finite case of Theorem 1.3 combine to prove Theorem 1.2, the Fundamental Theorem of Finite Semidistributive Lattices.

Using Proposition 2.8, we can restate the FTFSDL referring only to ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$. We can define $\operatorname{Pairs}(\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}}))$ directly as the set of pairs $(X,Y)$ with $Y=X^{\perp}$ and $X={}^{\perp}Y$ in the sense of Proposition 2.8, partially ordered by containment in the first component, or equivalently reverse containment in the second component.

Theorem 2.31** (FTFSDL, restated).**

A finite poset $L$ is a semidistributive lattice if and only if it is isomorphic to $\operatorname{Pairs}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ for partial orders ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ on a set , satisfying the conditions of Proposition 2.8. In this case, $(\Sha,{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is isomorphic to $(\operatorname{JIrr}(L),{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$, where $i\twoheadrightarrow_{L}j$ if and only if $i\geq j$ in $L$ and $i\hookrightarrow_{L}j$ if and only if $\kappa(i)\geq\kappa(j)$ in $L$. The map

[TABLE]

is an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$, with inverse $(X,Y)\mapsto\bigvee X=\bigwedge(\kappa(Y))$.

3. The infinite case

In this section, we discuss the infinite case further. We begin by explaining some choices we have made in the infinite case.

One choice we have made is to only consider lattices that are complete. This is for two reasons: First, one of our main motivations is the lattice of torsion classes (Section 8.2) for a finite-dimensional algebra, and this lattice is complete. Second, we have seen in Proposition 2.11 that for any binary relation ${\mathord{\to}}$ on a set , the lattice $\operatorname{Pairs}({\mathord{\to}})$ is complete.

We have similarly focused our attention on completely join-irreducible and completely meet-irreducible elements. Crucial here is the fact that $j$ is completely join-irreducible if and only if there is an element $j_{*}$ such that $x<j$ if and only if $x\leq j_{*}$. If $j$ is join-irreducible but not completely so, there is no such an element.

3.1. Conditions on infinite complete lattices

We now discuss some conditions on complete lattices, each of which, in the finite case, is either equivalent to semidistributivity or trivially true. The main theorem of this section describes the relationships between these conditions. In Section 3.2 we prove the main theorem, and in Section 3.3, we present numerous counterexamples to show that the claims of the theorem cannot be strengthened.

We begin with some discreteness conditions, which are easily seen to hold in all finite lattices. Recall that a complete lattice $L$ is meet and join generated by irreducibles if for all $x\in L$, we have $x=\bigvee_{j\leq x,\ j\in\operatorname{JIrr}^{c}(L)}j=\bigwedge_{m\geq x,\ m\in\operatorname{MIrr}^{c}(L)}m$. We say that $L$ is cover-separated, if, for all $w<z$, there exist $x$ and $y$ with $w\leq x\lessdot y\leq z$. We say that meets in $L$ are cover-determined if whenever $w<z$ and $w\wedge u<z\wedge u$, there is a cover $w\leq x\lessdot y\leq z$ with $x\wedge u<y\wedge u$. We say that joins in $L$ are cover-determined if the dual condition holds.

Remark 3.1**.**

The definition of a cover-separated lattice was introduced in [10], under the name “arrow-separated.” The phrases “joins are cover-determined” and “meets are cover-determined” are also related to terminology in [10], specifically the notion of cover-determined congruences that we will use later in Section 6.

We continue by describing some conditions that, in the finite case, are equivalent to semidistributivity. Recall that $L$ is join semidistributive if, for any nonempty finite subset $X$ of $L$ such that $x\vee y=z$ for all $x\in X$, we have ${\left(\bigwedge_{x\in X}x\right)\vee y=z}$. A lattice $L$ is completely join semidistributive if it is complete and if, for every nonempty subset $X$ of $L$ such that $x\vee y=z$ for all $x\in X$, we have ${\left(\bigwedge_{x\in X}x\right)\vee y=z}$. Complete meet semidistributivity is defined dually, and $L$ is called completely semidistributive if it is completely join semidistributive and completely meet semidistributive. Our interest in complete semidistributivity arises in part from the lattice of torsion classes for a finite-dimensional algebra, as discussed in Section 8.2.

Recall that a $\kappa$-lattice is a complete lattice $L$ that is meet and join generated by irreducibles, and has special bijections $\kappa$ and $\kappa^{d}$ between $\operatorname{JIrr}^{c}(L)$ and $\operatorname{MIrr}^{c}(L)$. Recall also that Corollary 2.29 says that a finite lattice is a $\kappa$-lattice if and only if it is semidistributive. Finally, recall that a $\kappa$-lattice is well separated if whenever $z_{1}\not\leq z_{2}$, there exists $j\in\operatorname{JIrr}^{c}(L)$ with $z_{1}\geq j$ and $\kappa(j)\geq z_{2}$.

The following theorem relates all of these conditions on a complete lattice.

Theorem 3.2**.**

Let $L$ be a complete lattice. The implications shown by solid arrows hold without additional hypotheses. The dashed implications hold under the additional hypothesis that $L$ is completely semidistributive.

$L$* is a well separated $\kappa$-lattice*

$L\cong\operatorname{Pairs}({\mathord{\to}})$* for a two-acyclic factorization system*

Joins and meets in $L$ are cover-determined

$L$* is a $\kappa$-lattice*

$L$* is meet and join generated by irreducibles*

$L$* is cover-separated*

We also remind the reader of Theorem 2.2: If $L$ is a $\kappa$-lattice, then $L\cong\operatorname{Pairs}({\mathord{\to}}_{L})$ and $\operatorname{Pairs}({\mathord{\to}}_{L})$ obeys all the conditions of a two-acyclic factorization system except that some ${\mathord{\to}}$ arrows may not factor as an ${\mathord{\twoheadrightarrow}}$ followed by an ${\mathord{\hookrightarrow}}$. We have not incorporated the statement $L\cong\operatorname{Pairs}({\mathord{\to}}_{L})$ into the diagram because ${\mathord{\to}}_{L}$ is only defined when $L$ is a $\kappa$-lattice.

Remark 3.3**.**

We regard completely semidistributive lattices obeying the conditions in the top row as the “good” lattices. All finite semidistributive lattices are in this class, as are lattices of torsion classes for finite-dimensional algebras, and we hope to construct lattices completing weak order on infinite Coxeter groups which will likewise obey these conditions.

Unfortunately, in the infinite case, none of the conditions we have mentioned imply semidistributivity (complete or otherwise). See Example 3.20 for a truly frustrating counterexample. We do not have a good replacement hypothesis which would imply semidistributivity.

3.2. Proof of Theorem 3.2

We now establish the implications of Theorem 3.2. We begin with the arrows which do not require complete semidistributivity. We already established in Theorem 1.3 that $L$ is a well separated $\kappa$-lattice if and only if it is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for a two-acyclic factorization system $(\Sha,\to,{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$.

Proposition 3.4**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system. Then joins and meets in $\operatorname{Pairs}({\mathord{\to}})$ are cover-determined.

Proof.

We prove the statement for meets. Let $X<Y$ in $\operatorname{Pairs}({\mathord{\to}})$ and let $Z\in\operatorname{Pairs}({\mathord{\to}})$ be such that $X\wedge Z<Y\wedge Z$. We first reduce to the case where $Z\leq Y$. Suppose we knew the theorem in this case, and put $Z^{\prime}=Y\wedge Z$. We have $X\wedge Z^{\prime}=X\wedge Y\wedge Z=X\wedge Z$ and $Y\wedge Z^{\prime}=Y\wedge Y\wedge Z=Y\wedge Z$, so the hypothesis $X\wedge Z^{\prime}<Y\wedge Z^{\prime}$ is satisfied and we have $Z^{\prime}\leq Y$. If we can find a cover $X<C_{1}\lessdot C_{2}<Y$ with $C_{1}\wedge Z^{\prime}<C_{2}\wedge Z^{\prime}$ then, since $C_{j}\wedge Z^{\prime}=(C_{j}\wedge Z)\wedge Y$, we will also have $C_{1}\wedge Z<C_{2}\wedge Z$. This completes our reduction.

With the reduction made, all of the objects $X\wedge Z$, $Y\wedge Z$, $X$, $Y$ and $Z$ lie in the interval $[X\wedge Z,Y]$. By Theorem 4.3, this interval also corresponds to a two-acyclic factorization system. We therefore may assume that $X\wedge Z=(\emptyset,\Sha)$ and $Y=(\Sha,\emptyset)$. The hypothesis $X\wedge Z<Y\wedge Z$ now simplifies to $Z\neq(\emptyset,\Sha)$. We identify each pair with its first element, so our hypotheses now are that $X$, $Z$ are closed sets with $X\cap Z=\emptyset$ and $Z\neq\emptyset$. We want to show that there is a cover $X\leq C_{1}\lessdot C_{2}$ with $C_{1}\cap Z\subsetneq C_{2}\cap Z$.

Since $Z\neq\emptyset$, we can find $p\in Z$. Since $X\cap Z=\emptyset$, we have $p\not\in X$ and thus there is $r\in X^{\perp}$ with $p\to r$. Factor this arrow as $p\twoheadrightarrow q\hookrightarrow r$. Since $Z$ is an ${\mathord{\twoheadrightarrow}}$-downset and $X^{\perp}$ is an ${\mathord{\hookrightarrow}}$-upset, we have $q\in X^{\perp}\cap Z$. Recall the notations $F(q)=\{s\in\Sha:s\hookrightarrow q\}$ and $F^{*}(q)=F(q)\setminus\{q\}$. We have $X\subseteq{}^{\perp}F(q)\lessdot{}^{\perp}F^{*}(q)$.

Clearly, $q\not\in{}^{\perp}F(q)$ so $q\not\in{}^{\perp}F(q)\cap Z$. Since $F(q)\setminus F^{*}(q)=\{q\}$, we have $q\in{}^{\perp}F^{*}(q)$, so $q\in{}^{\perp}F^{*}(q)\cap Z$. This shows that ${}^{\perp}F(q)\cap Z\neq{}^{\perp}F^{*}(q)\cap Z$, so ${}^{\perp}F^{*}(q)\lessdot{}^{\perp}F(q)$ is the desired cover. ∎

Proposition 3.5**.**

If either joins in $L$ are cover-determined or meets in $L$ are cover-determined, then $L$ is cover-separated.

Proof.

We consider the case that joins are cover-determined; the case of meets is dual. Let $w<z$ and let [math] be the minimum element of $L$. Then $w\vee 0<z\vee 0$, so there must be a cover $w\leq x\lessdot y\leq z$ with $x\vee 0<y\vee 0$. In particular, there is a cover separating $w$ and $z$. ∎

The remaining undashed implications say that a well separated $\kappa$-lattice is a $\kappa$-lattice and that a $\kappa$-lattice is meet and join generated by irreducibles. Both of these implications are true by definition.

We now turn to the implications which require complete semidistributivity. We begin with the bottom row.

Proposition 3.6**.**

Let $L$ be a completely semidistributive lattice which is join and meet generated by irreducibles. Then $L$ is a $\kappa$-lattice.

Proof.

This is proved exactly as in Theorem 2.28. ∎

Before proving the other leftward implications, we need the following lemma. This result is essentially [10, Proposition 2.20].

Lemma 3.7**.**

If $L$ is a completely semidistributive lattice and $u\lessdot v$ in $L$, then the set $\{t\in L:t\vee u=v\}$ contains a minimum element $\ell$, which is completely join-irreducible and has $\ell_{*}\leq u$.

Proof.

Let $T=\{t\in L:t\vee u=v\}$. Since $L$ is completely semidistributive, the element $\ell=\bigwedge T$ has $\ell\vee u=v$, so it is the desired minimum element of $T$. Suppose $\ell=\bigvee S$ for some set $S\subseteq L$. We must show that $\ell\in S$. In order to do this, we will show that $\bigvee(S\setminus\{\ell\})\leq u$; since $\ell\not\leq u$, this shows that $\bigvee(S\setminus\{\ell\})\neq\bigvee S$.

If $s\in S\setminus\{\ell\}$, then $u\leq s\vee u\leq\ell\vee u=v$ so, since $u\lessdot v$, we must have $s\vee u=u$ or $s\vee u=v$. If $s\vee u=v$ then $s$ is an element of $T$ that is strictly less than $\ell=\bigwedge T$. By this contradiction, we see that $s\vee u=u$. We have shown that $s\leq u<\ell$ for every $s\in S\setminus\{\ell\}$ and thus $\bigvee(S\setminus\{\ell\})\leq u$ as promised. We have shown that $\ell$ is completely join-irreducible.

We have $u\leq\ell_{*}\vee u\leq\ell\vee u=v$, but the second inequality must be strict by the definition of $\ell$. Since $u\lessdot v$, we see that $\ell_{*}\vee u=u$, so that $\ell_{*}\leq u$. ∎

Remark 3.8**.**

The map in Lemma 3.7 from covers of a complete semidistributive lattice to completely join-irreducible elements is known as the join-irreducible labelling. It is well-known in the finite case. An analogous result in the representation-theoretic setting was established in [5] (combining Theorems 1.0.2, 1.0.3, and 1.0.5).

Proposition 3.9**.**

If $L$ is a completely semidistributive, cover-separated lattice and $x\in L$, then

[TABLE]

Proof.

We prove that $x=\bigvee\{j\in\operatorname{JIrr}^{c}(L):j\leq x\}$. The other equality is dual. Let $x^{\prime}=\bigvee\{j\in\operatorname{JIrr}^{c}(L):j\leq x\}$. Then $x^{\prime}\leq x$, so suppose for the sake of contradiction that $x^{\prime}<x$. Since $L$ is cover-separated, there exist $u$ and $v$ with $x^{\prime}\leq u\lessdot v\leq x$. Let $\ell$ be the minimum element of $\{t:t\vee u=v\}$, which exists and is completely join-irreducible by Lemma 3.7.

Since $\ell\vee u=v\neq u$, we see that $\ell\not\leq u$ and hence $\ell\not\leq x^{\prime}$. But $\ell$ is completely join-irreducible, so we have contradicted the definition of $x^{\prime}$ as ${x^{\prime}=\bigvee\{j\in\operatorname{JIrr}^{c}:j\leq y\}}$. We must therefore reject the supposition that ${x^{\prime}<x}$. ∎

Our final task is to prove that all the conditions on the top row of Theorem 3.2 are equivalent in the completely semidistributive case. We will need another lemma.

In a $\kappa$-lattice $L$, suppose $i,j\in\operatorname{JIrr}^{c}(L)$. If there exists $\ell\in\operatorname{JIrr}^{c}(L)$ with $i\geq\ell$ and $\kappa(\ell)\geq\kappa(j))$, then $i\not\leq\kappa(j)$. (If $i\leq\kappa(j)$, then $\ell\leq i\leq\kappa(j)\leq\kappa(\ell)$, contradicting the definition of $\kappa(\ell)$.) We say that $L$ is weakly separated if the converse holds: If $i,j\in\operatorname{JIrr}^{c}(L)$ have $i\not\leq\kappa(j)$, then there exists $\ell\in\operatorname{JIrr}^{c}(L)$ with $i\geq\ell$ and $\kappa(\ell)\geq\kappa(j))$.

Lemma 3.10**.**

A $\kappa$-lattice $L$ is well separated if and only if it is weakly separated.

Proof.

Clearly any well separated $\kappa$-lattice is weakly separated. We show the converse. Suppose $L$ is weakly separated and $z_{1}\not\leq z_{2}$ in $L$. We will show that there exists $\ell\in\operatorname{JIrr}^{c}(L)$ with $z_{1}\geq\ell$ and $\kappa(\ell)\geq z_{2}$. Since $L$ is join generated by irreducibles (as part of the definition of a $\kappa$-lattice), there exists $i\in\operatorname{JIrr}^{c}(L)$ with $i\leq z_{1}$ but $i\not\leq z_{2}$. Since $i\not\leq z_{2}$ and because $L$ is meet generated by irreducibles, there exists $j\in\operatorname{JIrr}^{c}(L)$ such that $z_{2}\leq\kappa(j)$ and $i\not\leq\kappa(j)$. By weak separation, there exists $\ell\in\operatorname{JIrr}^{c}(L)$ with $i\geq\ell$ and $\kappa(\ell)\geq\kappa(j)$. Thus $z_{1}\geq i\geq\ell$ and $\kappa(\ell)\geq\kappa(j)\geq z_{2}$ as desired. ∎

Proposition 3.11**.**

Suppose $L$ is a completely semidistributive lattice. If either joins are cover-determined or meets are cover-determined, then $L$ is a well separated $\kappa$-lattice.

Proof.

We consider the case that joins are cover-determined; the case of meets is dual. The implications we have already established show that $L$ is a $\kappa$-lattice. It remains to establish well-separation. By Lemma 3.10, it is enough to establish weak separation.

Suppose $i,j\in\operatorname{JIrr}^{c}(L)$ have $i\not\leq\kappa(j)$. Then $i\vee\kappa(j)>\kappa(j)$ and $(i\wedge\kappa(j))\vee\kappa(j)=\kappa(j)$. Since joins are cover-determined in $L$, there exist $u,v\in L$ with $i\wedge\kappa(j)\leq u\lessdot v\leq i$ and $u\vee\kappa(j)\neq v\vee\kappa(j)$. By Lemma 3.7, the set $\{s\in L:s\vee u=v\}$ has a minimum element $\ell$, which is completely join-irreducible and has $\ell_{*}\leq u$. We have $\ell\leq v\leq i$.

If $\kappa(\ell)\not\geq\kappa(j)$, then $\kappa(j)\vee\ell_{*}\geq\ell$ by the definition of $\kappa(\ell)$. Thus $\kappa(j)\vee u\geq\ell$ as well. Since $\ell\vee u=v$, also $\kappa(j)\vee v=\kappa(j)\vee\ell\vee u$, which equals $\kappa(j)\vee u$ because $\kappa(j)\vee u\geq\ell$. But we already know $u\vee\kappa(j)\neq v\vee\kappa(j)$, and by this contradiction, we see that $\kappa(\ell)\geq\kappa(j)$. We have proven that $L$ is weakly separated, as desired. ∎

3.3. Counterexamples

We now present a number of counter-examples, to sharpen the distinctions between the various conditions on complete lattices, and to demonstrate that the results in Theorem 3.2 cannot be strengthened.

Example 3.12** (A semidistributive lattice that is not completely semidistributive and is not a $\kappa$-lattice).**

Consider the lattice with elements [math], $1$, $X$ and $Y_{i}$ for $-\infty<i<\infty$, and relations that [math] is the minimum, $1$ is the maximum and $Y_{i}\leq Y_{j}$ for $i\leq j$. This lattice is complete and semidistributive. For each $Y_{i}$, we have $X\vee Y_{i}=1$. However, $X\vee\bigwedge_{i}Y_{i}=X\vee 0=X$. Also, $X$ is completely join irreducible with $X_{*}=0$, but there is no maximal element $\kappa(X)$ in the set $\{Y:X\wedge Y=0\}$.

Example 3.13** (A semidistributive $\kappa$-lattice that is not completely semidistributive and not well separated).**

Consider the lattice whose elements $L$ are called [math], $1$, $X_{i}$ and $Y_{i}$, for $i\in\mathbb{Z}$, with relations that [math] is the minimum, $1$ is the maximum, and $X_{i}\geq X_{j}$ and $Y_{i}\geq Y_{j}$ for $i\geq j$. This lattice is complete and semidistributive. However, it is not completely semidistributive for the same reason as in Example 3.12: any $X_{i}$ can play the role of $X$ from that example. Every element is both completely join-irreducible and completely meet-irreducible except [math] and $1$. Also, $0=\bigwedge L$ and $1=\bigvee L$. We have $\kappa(X_{i})=X_{i-1}$ and $\kappa(Y_{i})=Y_{i-1}$, and $\kappa^{d}$ is the inverse map, so $L$ is a $\kappa$-lattice. Also, $X_{0}\not\geq Y_{0}$, yet there is no completely join-irreducible element $J$ with $X_{0}\geq J$ and $\kappa(J)\geq Y_{0}$, so $L$ is not well separated.

Example 3.14** (A $\kappa$-lattice that is not semidistributive and not well separated).**

Consider the lattice $L$ with elements [math], $1$ and $X_{i}$, $Y_{i}$ and $Z_{i}$ for $-\infty<i<\infty$, and relations that [math] is the minimum, $1$ is the maximum, and $X_{i}\leq X_{j}$, $Y_{i}\leq Y_{j}$ and $Z_{i}\leq Z_{j}$ for $i\leq j$. As in Example 3.13, every element is both completely join-irreducible and completely meet-irreducible except [math] and $1$, and we have $\kappa(X_{i})=X_{i-1}$, $\kappa(Y_{i})=Y_{i-1}$ and $\kappa(Z_{i})=Z_{i-1}$. Thus $L$ is a $\kappa$-lattice. However, $X_{0}\wedge Y_{0}=X_{0}\wedge Z_{0}=0$ while $X_{0}\wedge(Y_{0}\vee Z_{0})=X_{0}\wedge 1=X_{0}$, so $L$ is not semidistributive. This lattice is not well separated for the same reason as Example 3.13.

Example 3.15** ($\kappa$-lattices not isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for a two-acyclic factorization system).**

Suppose $L$ is the lattice in Example 3.13 or the lattice in Example 3.14. Thus $L$ is a $\kappa$-lattice. By Theorem 1.3, if $L$ is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for a two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, then this system is isomorphic to $(\operatorname{JIrr}^{c}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$. However, ${\mathord{\to}}_{L}$ is not $\operatorname{Mult}({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$. Indeed, we have $\kappa(X_{1})=X_{0}\not\geq Y_{1}$ so $X_{1}\to Y_{1}$, but the only compositions $X_{1}\twoheadrightarrow P\hookrightarrow Q$ are those of the form $X_{1}\twoheadrightarrow X_{j}\hookrightarrow X_{k}$ for $1\geq j\geq k$, so we cannot factor $X_{1}\to Y_{1}$ into an ${\mathord{\twoheadrightarrow}}$ arrow and an ${\mathord{\hookrightarrow}}$ arrow.

We next give an example where meets and joins are cover-determined, but the lattice is not meet generated by its completely meet irreducible elements (because it has none). It is therefore also not a $\kappa$-lattice (well separated or otherwise). By Theorem 3.2, this example cannot be completely semidistributive; surprisingly, it is distributive!

Example 3.16** (A lattice where meets and joins are cover-determined, but $\operatorname{MIrr}^{c}(L)$ is empty).**

Let $\Omega$ be an infinite set. Let $L$ consist of the finite subsets of $\Omega$, as well as the set $\Omega$ itself, with $L$ ordered by containment. It is easy to check that $L$ is a complete lattice and meets and joins are cover-determined. However, $L$ has no meet irreducible elements, so it is not meet generated by them.

For completely semidistributive lattices, Theorem 3.2 tells us that having joins and meets cover-determined implies cover-separation, which in turn implies being meet and join generated by irreducibles. We now present two very similar examples, showing that these implications cannot be reversed. As Theorem 3.2 shows must be the case, these are $\kappa$-lattices but are not well separated.

Example 3.17** (A completely semidistributive $\kappa$-lattice that is meet and join generated by irreducibles but not cover-separated).**

We define a poset $L$ whose ground set is $5$ copies of the interval $[0,1]$ in the real numbers, with elements denoted by formal symbols $x^{++}$, $x^{+}$, $x^{\circ}$, $x^{-}$, and $x^{--}$ for each $x\in[0,1]$. For any $0\leq x<y\leq 1$, the poset $L$ induces the order depicted on the $10$ elements shown in Figure 1. The solid lines in the figure represent cover relations in $L$; the dashed lines are order relations in $L$ that are not covers. It is easy to check from Figure 1 that $L$ is a lattice. To see that it is actually a complete lattice, note that any set of the form $\{x^{++}:x\in A\}$ has a join (either $(\sup A)^{++}$ or $(\sup A)^{+}$, where $\sup A$ is the supremum), and similarly, sets consisting of elements $x^{+}$, $x^{\circ}$, $x^{-}$ or $x^{--}$ have joins. Thus we can compute the join of any subset of $L$ as the join of at most five elements, and similarly for meets.

Complete meet semidistributivity is verified in the following table. For any $u<v$, the table gives the unique maximal $p$ with $v\wedge p=u$, where $u$ is given in the row headings and $v$ in the column headings, assuming $x<y$. Complete join semidistributivity is dual.

[TABLE]

It is easy to check that it is meet and join generated by irreducibles; this, in combination with the fact that it is completely semidistributive, allows us to conclude by Theorem 3.2 that it is a $\kappa$-lattice.

However, this lattice is not cover-separated: There are no covers in the interval $[0^{\circ},1^{\circ}]$. It follows from Theorem 3.2 that it is not well separated.

Example 3.18** (A completely semidistributive $\kappa$-lattice which is cover-separated but where joins and meets are not cover-determined).**

Take Example 3.17 and remove the elements of the form $x^{\circ}$. As before, we can verify that this is a complete, completely semidistributive lattice, and it is now cover-separated. Theorem 3.2 therefore shows that it is a $\kappa$-lattice.

However, neither meets nor joins are cover-determined. For all $x$, we have $x^{-}\vee 0^{+}=x^{--}\vee 0^{+}=x^{+}$. So, for $0\leq x<z\leq 1$, we have $x^{-}\vee 0^{+}<z^{--}\vee 0^{+}$ but, for any cover $x^{-}<y^{--}\lessdot y^{-}<z^{--}$, we have $y^{-}\vee 0^{+}=y^{--}\vee 0^{+}$. Since meets and joins are not cover-determined, Theorem 3.2 shows that the lattice is not well separated.

We conclude with two examples that together show that semidistributivity is independent of all of the other conditions in Theorem 3.2. Example 3.20 is particularly frustrating, as we do not know what additional condition on the two-acyclic factorization system would rule out this lattice.

Example 3.19** (A completely semidistributive lattice which violates all other conditions in Theorem 3.2).**

Let $L$ be the interval $[0,1]$ in the real numbers with the standard total order. This is a complete and completely semidistributive (even distributive!) lattice, but it has no completely join-irreducible or completely meet-irreducible elements at all. Thus $L$ is neither meet nor join generated by irredicibles, and thus Theorem 3.2 shows that this lattice also violates the other conditions there.

Example 3.20** (A two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ such that the $\kappa$-lattice $\operatorname{Pairs}({\mathord{\to}})$ is not semidistributive.).**

We describe our lattice by giving a two-acyclic factorization system. We will show that $\operatorname{Pairs}({\mathord{\to}})$ is not semidistributive, but by Theorem 3.2, $\operatorname{Pairs}({\mathord{\to}})$ obeys all the other conditions in that theorem.

Let consist of elements $y$, $z$, and $x_{i}$ for $i\in\mathbb{Z}$. Define $\to$ by $x_{i}\to x_{j}$ for $i\geq j$, by $y\to x_{2i}$ and $z\to x_{2i+1}$ for $i\in\mathbb{Z}$, as well as $y\to y$ and $z\to z$. We define $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ to be $\operatorname{Fact}({\mathord{\to}})$. Explicitly,

• •

$x_{i}\twoheadrightarrow x_{j}$* if and only if $i\geq j$,*

• •

$x_{i}\hookrightarrow x_{j}$* if and only if $i\geq j$ and $i\equiv j\bmod 2$,*

• •

$y\hookrightarrow x_{2i}$* and $z\hookrightarrow x_{2i+1}$ for all $i\in\mathbb{Z}$, and*

• •

$y\hookrightarrow y$, $y\twoheadrightarrow y$, $z\hookrightarrow z$ and $z\twoheadrightarrow z$.

The relations ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ are shown in the left picture of Figure 2.

We see that if $p\to q$, then either $p\hookrightarrow q$ or $p\twoheadrightarrow q$, so $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})\supseteq{\mathord{\to}}$. By Proposition 2.3.3 we conclude that $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a factorization system. The union of the relations ${\mathord{\to}}$, ${\mathord{\twoheadrightarrow}}$, and ${\mathord{\hookrightarrow}}$ is acyclic, so in particular $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is two-acyclic.

For $j\in\mathbb{Z}\cup\{\pm\infty\}$, we put $X[j]=\{x_{i}:i\leq j\}$, $Y[j]=\{y\}\cup\{x_{i}:i\leq j\}$ and $Z[j]=\{z\}\cup\{x_{i}:i\leq j\}$. The closed sets are $X[i]$, $Y[2i]$ and $Z[2i+1]$, for $i\in\mathbb{Z}\cup\{\pm\infty\}$, as well as the whole set . The lattice of closed sets is shown as the right picture of Figure 2. We see that $X[0]\wedge Y[-\infty]=X[0]\wedge Z[-\infty]=X[-\infty]$, but that $X[0]\wedge(Y[-\infty]\vee Z[-\infty])=X[0]\wedge\Sha=X[0]$.

4. Factorization systems for intervals

Let $L$ be a finite semidistributive lattice. By the FTFSDL, it can be described as $\operatorname{Pairs}({\mathord{\to}})$ for a finite two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. An interval in $L$ is also a finite semidistributive lattice, so it is also describable by a finite two-acyclic factorization system. In this section, we show that this can be done very explicitly. More generally, we show that, if $L\cong\operatorname{Pairs}({\mathord{\to}})$ for a possibly infinite, two-acyclic factorization system $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, then every interval of $L$ can be described as $\operatorname{Pairs}({\mathord{\to}}^{\prime})$ for some two-acyclic factorization system $({}^{\prime},{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$.

We begin by describing lower intervals in $\operatorname{Pairs}({\mathord{\to}})$. We use notation such as ${\mathord{\to}}|_{X}$ for the restriction of relations to subsets $X\subseteq\Sha$.

Proposition 4.1**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system and let $(X,Y)\in\operatorname{Pairs}({\mathord{\to}})$. Then $(X,{\mathord{\to}}|_{X},{\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$ is a two-acyclic factorization system, where ${\mathord{\hookrightarrow}}^{\prime}$ is given by $\operatorname{Fact}({\mathord{\to}}|_{X})=({\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$. The map $(U,V)\mapsto(U,V\cap X)$ is an isomorphism from the interval \bigl{[}(\emptyset,\Sha),(X,Y)\bigr{]} in $\operatorname{Pairs}({\mathord{\to}})$ to the lattice $\operatorname{Pairs}({\mathord{\to}}|_{X})$. The closure operators associated to $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ and $(X,{\mathord{\to}}|_{X},{\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$ agree on subsets of $X$.

Proof.

Write $\operatorname{Fact}({\mathord{\to}}|_{X})=({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$. We have $({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})\supseteq({\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}|_{X})$ because $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a factorization system. We now check that ${\mathord{\twoheadrightarrow}}|_{X}\supseteq{\mathord{\twoheadrightarrow}}^{\prime}$. Suppose that $p$ and $q\in X$ with $p\not\twoheadrightarrow q$; we will check that $p\not\!{\mathord{\twoheadrightarrow}}^{\prime}q$. Then there is some $s\in\Sha$ with $p\not\to s$ and $q\to s$. Factor $q\to s$ as $q\twoheadrightarrow r\hookrightarrow s$; since $X$ is an ${\mathord{\twoheadrightarrow}}$-downset by Proposition 2.4, we have $r\in X$. If $p\to r$ then there is a factorization $p\twoheadrightarrow t\hookrightarrow r$ and thus $p\twoheadrightarrow t\hookrightarrow s$, contradicting the assumption that $p\not\to s$. We conclude that $p\not\to r$. Now $r\in X$ with $p\not\to r$ and $q\to r$, so $p\not\twoheadrightarrow^{\prime}q$.

To show that $(X,{\mathord{\to}}|_{X},{\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$ is a factorization system, it remains to show that $\operatorname{Mult}({\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})={\mathord{\to}}|_{X}$. By Proposition 2.3(3), we have $\operatorname{Mult}({\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})\subseteq{\mathord{\to}}|_{X}$. Suppose that $x$ and $z\in X$ and $x\to z$. Then $x\twoheadrightarrow y\hookrightarrow z$ for some $y\in\Sha$. Since $X$ is a downset for ${\mathord{\twoheadrightarrow}}$, we have $y\in X$. Using our above observation that ${\mathord{\hookrightarrow}}^{\prime}\supseteq{\mathord{\hookrightarrow}}|_{X}$, we have $x\twoheadrightarrow y\hookrightarrow^{\prime}z$, so $x$ is related to $z$ by $\operatorname{Mult}({\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$.

We now check the two-acyclicity of $(X,{\mathord{\to}}|_{X},{\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$: The relation ${\mathord{\twoheadrightarrow}}|_{X}$ inherits acyclicity from ${\mathord{\twoheadrightarrow}}$. Suppose $p\twoheadrightarrow q\hookrightarrow^{\prime}p$. Since $\operatorname{Mult}({\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})={\mathord{\to}}|_{X}$, we have $q\to p$, so there exists $z\in\Sha$ such that $q\twoheadrightarrow z\hookrightarrow p$. Thus $p\twoheadrightarrow z\hookrightarrow p$, so $z=p$ by the brick condition on $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. Thus $q\twoheadrightarrow p\twoheadrightarrow q$, so that $p=q$ by the acyclicity of $\twoheadrightarrow$.

To complete the proof of two-acyclicity  suppose $p\hookrightarrow^{\prime}q\hookrightarrow^{\prime}p$. Again, since $q\hookrightarrow^{\prime}p$, there exists $z\in\Sha$ such that $q\twoheadrightarrow z\hookrightarrow p$. Since $X$ is a downset for ${\mathord{\twoheadrightarrow}}$, we have $z\in X$. Thus by the observation that ${\mathord{\hookrightarrow}}^{\prime}\supseteq{\mathord{\hookrightarrow}}|_{X}$, we have $q\twoheadrightarrow z\hookrightarrow^{\prime}p$, so $q\twoheadrightarrow z\hookrightarrow^{\prime}q$. By the brick condition on $(X,{\mathord{\to}}|_{X},{\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$, established above, we have $z=q$, so $q\hookrightarrow p$. Similarly, since $p\hookrightarrow^{\prime}q$, there exists $w\in X$ with $p\twoheadrightarrow w\hookrightarrow q$, and thus $p\twoheadrightarrow w\hookrightarrow^{\prime}q$, so $p\twoheadrightarrow w\hookrightarrow^{\prime}p$, and thus $w=p$, so that $p\hookrightarrow q$. By the acyclicity of ${\mathord{\hookrightarrow}}$, we see that $p=q$.

Writing $\triangle$ for orthogonality operations with respect to ${\mathord{\to}}|_{X}$, we now establish that ${}^{\triangle}(U^{\triangle})={}^{\perp}(U^{\perp})$ for any $U\subseteq X$. Unpacking definitions yields

[TABLE]

It is clear that $U^{\perp}\supseteq(U^{\perp}\cap X)\cup Y$, and thus ${}^{\perp}(U^{\perp})\subseteq{}^{\triangle}(U^{\triangle})$.

For the reverse containment, suppose $p\not\in{}^{\perp}(U^{\perp})$. Since $U\subseteq X$, we have $U^{\perp}\supseteq X^{\perp}=Y$ and thus ${}^{\perp}(U^{\perp})\subseteq Y^{\perp}=X$. If $p\not\in X$, then $p\not\in{}^{\triangle}(U^{\triangle})$, so we may as well assume $p\in X$. Since $p\not\in{}^{\perp}(U^{\perp})$, there exists $r\in U^{\perp}$ with $p\to r$. Factor $p\to r$ as $p\twoheadrightarrow q\hookrightarrow r$. Since $X$ is an ${\mathord{\twoheadrightarrow}}$-downset and $p\in X$, we have $q\in X$. If $q\not\in U^{\perp}$, then there exists $u\in U$ with $u\rightarrow q$. Since $q\hookrightarrow r$, this implies $u\rightarrow r$, contradicting $r\in U^{\perp}$. Thus $q\in U^{\perp}$. Since also $q\in X$, we have $q\in U^{\triangle}$. Since $p\to q$, we see that $p\not\in{}^{\triangle}(U^{\triangle})$, as desired.

Since the closure operations coincide on subsets of $X$, the lattices of closed sets coincide. Thus $(U,U^{\perp})\mapsto(U,U^{\triangle})$ is an isomorphism from \bigl{[}(\emptyset,\Sha),(X,Y)\bigr{]} to $\operatorname{Pairs}(\to^{\prime})$. The equality $U^{\triangle}=U^{\perp}\cap X$ gives the formulation of this result in the statement of the Theorem. ∎

We explicitly state the dual result for upper intervals of $\operatorname{Pairs}({\mathord{\to}})$.

Proposition 4.2**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system and let $(X,Y)\in\operatorname{Pairs}({\mathord{\to}})$. Then $(Y,{\mathord{\to}}|_{Y},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}|_{Y})$ is a two-acyclic factorization system, where ${\mathord{\twoheadrightarrow}}^{\prime}$ is given by $\operatorname{Fact}({\mathord{\to}}|_{Y})=({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}|_{Y})$. The map $(U,V)\mapsto(U\cap Y,V)$ is an isomorphism from the interval \bigl{[}(X,Y),(\Sha,\emptyset)\bigr{]} in $\operatorname{Pairs}({\mathord{\to}})$ to the lattice $\operatorname{Pairs}({\mathord{\to}}|_{Y})$.

Concatenating Propositions 4.1 and 4.2, we obtain:

Theorem 4.3**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system and let \bigl{[}(X_{1},Y_{1}),(X_{2},Y_{2})\bigr{]} be an interval in $\operatorname{Pairs}({\mathord{\to}})$. Let ${\mathord{\to}}^{\prime}$ be the restriction of ${\mathord{\to}}$ to $X_{2}\cap Y_{1}$ and let $({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})=\operatorname{Fact}({\mathord{\to}}^{\prime})$. Then $(X_{2}\cap Y_{1},{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$ is a two-acyclic factorization system. The map $(U,V)\mapsto(U\cap Y_{1},V\cap X_{2})$ is an isomorphism from \bigl{[}(X_{1},Y_{1}),(X_{2},Y_{2})\bigr{]} to $\operatorname{Pairs}(\to^{\prime})$.

Proof.

By Proposition 4.1, $(X_{2},{\mathord{\to}}|_{X_{2}},{\mathord{\twoheadrightarrow}}|_{X_{2}},{\mathord{\hookrightarrow}}^{\prime\prime})$ is a two-acyclic factorization system, where ${\mathord{\hookrightarrow}}^{\prime\prime}$ is defined by $\operatorname{Fact}({\mathord{\to}}|_{X_{2}})=({\mathord{\twoheadrightarrow}}|_{X_{2}},{\mathord{\hookrightarrow}}^{\prime\prime})$. Furthermore, $(X_{1},X_{2}\cap Y_{1})$ is in $\operatorname{Pairs}({\mathord{\to}}|_{X_{2}})$. The relation $\to^{\prime}$ is the restriction of ${\mathord{\to}}|_{X_{2}}$ to $X_{2}\cap Y_{1}$, so Proposition 4.2 states that $(X_{2}\cap Y_{1},{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$ is a two-acyclic factorization system. By Proposition 4.1, the interval \bigl{[}(X_{1},Y_{1}),(X_{2},Y_{2})\bigr{]} in $\operatorname{Pairs}({\mathord{\to}})$ is isomorphic to the interval \bigl{[}(X_{1},X_{2}\cap Y_{1}),(X_{2},\emptyset)\bigr{]} in $\operatorname{Pairs}({\mathord{\to}}|_{X_{2}})$. By Proposition 4.2, the latter is isomorphic to $\operatorname{Pairs}(\to^{\prime})$. The isomorphisms given in the theorems are $(U,V)\mapsto(U,V\cap X_{2})\mapsto(U\cap Y_{1},V\cap X_{2})$. ∎

We conclude with a corollary of Theorem 4.3 which will be useful in the next section.

Lemma 4.4**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system and suppose $(X_{1},Y_{1})\leq(X_{2},Y_{2})$ in $\operatorname{Pairs}({\mathord{\to}})$. Then $(X_{2},Y_{2})$ covers $(X_{1},Y_{1})$ if and only if $X_{2}\cap Y_{1}$ is a singleton.

Proof.

By definition, $(X_{2},Y_{2})$ covers $(X_{1},Y_{1})$ if and only if \bigl{[}(X_{1},Y_{1}),(X_{2},Y_{2})\bigr{]} has precisely two elements. Equivalently (by Theorem 1.3), the lattice $\operatorname{Pairs}({\mathord{\to}}^{\prime})$ of Theorem 4.3 has $|\operatorname{JIrr}^{c}(\operatorname{Pairs}({\mathord{\to}}^{\prime}))|=1$. In other words, ${|X_{2}\cap Y_{1}|=1}$. ∎

Remark 4.5**.**

See [10, Theorem 3.3(b)] for a similar result about covers in the lattice of torsion classes.

5. Covers and canonical join representations

In this section, we describe cover relations in the lattice of maximal orthogonal pairs of a two-acyclic factorization system. We then use the description of covers to describe canonical join representations in the finite case. Throughout the section, $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ will be a two-acyclic factorization system. All references to orthogonality operations $\perp$ refer to this system, and references to closures and closed sets refer to the closure operator $X\mapsto\overline{X}={}^{\perp}(X^{\perp})$.

5.1. Covers

Given $c\in X$, define $\operatorname{Del}(X,c)=X\setminus\{x\in X:x\twoheadrightarrow c\}$. This notation suggests “deleting $c$ from $X$”, as $\operatorname{Del}(X,c)$ is the largest ${\mathord{\twoheadrightarrow}}$-downset contained in $X$ but not containing $c$. Given a closed set $X$, define $\operatorname{Cov}(X)$ to be the set of elements $c\in X$ such that $\operatorname{Del}(X,c)$ is closed and $\overline{\operatorname{Del}(X,c)\cup\{c\}}=X$. We will prove the following theorem.

Theorem 5.1**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system and let $X$ be a closed set. Then $\operatorname{Del}(X,c)\lessdot X$ in the lattice of closed sets for all $c\in\operatorname{Cov}(X)$, and every element covered by $X$ is $\operatorname{Del}(X,c)$ for a unique $c\in\operatorname{Cov}(X)$.

Proposition 2.11 lets us restate Theorem 5.1 in terms of maximal orthogonal pairs: Given $(X,Y)\in\operatorname{Pairs}({\mathord{\to}})$, the elements covered by $(X,Y)$ are precisely the pairs $(\operatorname{Del}(X,c),\operatorname{Del}(X,c)^{\perp})$ for $c\in\operatorname{Cov}(X)$, with exactly one cover for each $c\in\operatorname{Cov}(X)$.

We will also prove some alternative descriptions of $\operatorname{Cov}(X)$. The first is general and explicit, but unilluminating. The second is general but comes at the cost of computing $\operatorname{Fact}({\mathord{\to}}|_{X})$. The third applies only to special cases, including the finite case, but provides some insight into the general case.

Proposition 5.2**.**

$\operatorname{Cov}(X)$* consists of those elements $c$ of $X$ which have the following properties:*

• •

$c$* is ${\mathord{\hookrightarrow}}$-maximal in $X$,*

• •

for all $y\in\Sha$, if there exists $x\in X$ such that $x\hookrightarrow y$, then there exists $x^{\prime}\in\operatorname{Del}(X,c)\cup\{c\}$ such that $x^{\prime}\hookrightarrow y$.

Theorem 5.3**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system and let $X$ be a closed set. Write $\operatorname{Fact}({\mathord{\to}}|_{X})=({\mathord{\twoheadrightarrow}}|_{X},{\mathord{\hookrightarrow}}^{\prime})$ as in Proposition 4.1. Then $\operatorname{Cov}(X)$ is the set of ${\mathord{\hookrightarrow}}^{\prime}$-maximal elements of $X$.

Theorem 5.4**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system and let $X$ be a closed set. Let $C(X)$ be the set of ${\mathord{\hookrightarrow}}$-maximal elements of $X$. Then every element of $\operatorname{Cov}(X)$ is an ${\mathord{\twoheadrightarrow}}$-maximal element of $C(X)$. If is finite, or more generally if $X$ is finite, then $\operatorname{Cov}(X)$ is precisely the set of ${\mathord{\twoheadrightarrow}}$-maximal elements of $C(X)$.

We now proceed to prove Theorem 5.1. Our proof uses the following lemma.

Lemma 5.5**.**

If $(X,Y)\in\operatorname{Pairs}({\mathord{\to}})$ and $c$ is ${\mathord{\hookrightarrow}}$-maximal in $X$, then

[TABLE]

Consequently $\operatorname{Del}(X,c)$ is closed.

Proof.

The first equality says that $x\twoheadrightarrow c$ if and only if $x\to c$. If $x\twoheadrightarrow c$, then $x\twoheadrightarrow c\hookrightarrow c$, so $x\to c$. If $x\to c$ then $x\twoheadrightarrow z\hookrightarrow c$ for some $z\in\Sha$. Proposition 2.4 says that $X$ is a downset for ${\mathord{\twoheadrightarrow}}$, so $z\in X$. Since $c$ is ${\mathord{\hookrightarrow}}$-maximal in $X$, we have $z=c$, so $x\twoheadrightarrow c$ as desired. For the second equality, note that ${}^{\perp}(Y\cup\{c\})=\left({}^{\perp}Y\right)\cap\left({}^{\perp}\{c\}\right)=X\cap\left({}^{\perp}\{c\}\right)=X\setminus\{x\to c\}$. The set $\operatorname{Del}(X,c)$ is closed because any set of the form ${}^{\perp}S$ is closed (Proposition 2.12). ∎

Proof of Theorem 5.1.

For $c\in\operatorname{Cov}(X)$, we have that $\operatorname{Del}(X,c)$ is a closed set, and any strictly larger closed set contains $c$, so contains the closure of $\operatorname{Del}(X,c)\cup\{c\}$, which is $X$. This shows $\operatorname{Del}(X,c)\lessdot X$.

Suppose now that $(X^{\prime},Y^{\prime})\lessdot(X,X^{\perp})$ in the lattice of maximal orthogonal pairs. From Lemma 4.4, we have $X\cap Y^{\prime}=\{c\}$ for some $c\in X$. We will establish that $c\in\operatorname{Cov}(X)$ and $X^{\prime}=\operatorname{Del}(X,c)$.

If there exists $c^{\prime}\in X\setminus\{c\}$ with $c^{\prime}\hookrightarrow c$, then since $Y^{\prime}$ is an ${\mathord{\hookrightarrow}}$-upset, we have $c^{\prime}\in Y^{\prime}$. Thus because $X\cap Y^{\prime}$ is the singleton $\{c\}$, we see that $c$ is ${\mathord{\hookrightarrow}}$-maximal in $X$. Thus by Lemma 5.5, $\operatorname{Del}(X,c)$ is closed.

Since $X^{\prime}$ is a downset for ${\mathord{\twoheadrightarrow}}$ and $c\not\in X^{\prime}$, we must have $X^{\prime}\subseteq\operatorname{Del}(X,c)$. It therefore follows from the fact that $X^{\prime}\lessdot X$ that $X^{\prime}=\operatorname{Del}(X,c)$. Since we now know that $\operatorname{Del}(X,c)\lessdot X$, the closure of $\operatorname{Del}(X,c)\cup\{c\}$ must be $X$. We have proved that $c\in\operatorname{Cov}(X)$.

The uniqueness of $c$ such that $X^{\prime}=\operatorname{Del}(X,c)$ follows because $c$ is the unique ${\mathord{\twoheadrightarrow}}$-minimal element of $\{x\in X:x\twoheadrightarrow c\}$ by the acyclicity of ${\mathord{\twoheadrightarrow}}$. ∎

We now prove Proposition 5.2 and Theorems 5.3 and  5.4.

Proof of Proposition 5.2.

Suppose $c$ satisfies the conditions of the proposition. Since $c$ is ${\mathord{\hookrightarrow}}$-maximal in $X$, by Lemma 5.5, $\operatorname{Del}(X,c)$ is closed.

Let $y\in\Sha$. Then $y\not\in X^{\perp}$ if and only if there exists some $x\in X$ with $x\to y$. By factoring $x\to y$ as $x\twoheadrightarrow z\hookrightarrow y$, and noting that $X$ is a downset for $\twoheadrightarrow$, we see that the existence of $x\in X$ with $x\to y$ is equivalent to the existence of $x\in X$ with $x\hookrightarrow y$. By the hypothesis, this is equivalent to the existence of $x^{\prime}\in\operatorname{Del}(X,c)\cup\{c\}$ with $x^{\prime}\hookrightarrow y$, which is equivalent to $y\not\in(\operatorname{Del}(X,c)\cup\{c\})^{\perp}$. This shows $X$ and $\operatorname{Del}(X,c)\cup\{c\}$ have the same right perpendiculars, so their closures agree. We have now shown that $c\in\operatorname{Cov}(X)$.

Conversely, suppose that $c\in\operatorname{Cov}(X)$. The argument in the previous paragraph also proves that $c$ satisfies the second condition of the proposition. The fact that $c$ in ${\mathord{\hookrightarrow}}$-maximal in $X$ was established in the proof of Theorem 5.1. ∎

Theorem 5.3 follows from the following proposition, in light of Proposition 4.1.

Proposition 5.6**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system. Then $\operatorname{Cov}(\Sha)$ is the set of ${\mathord{\hookrightarrow}}$-maximal elements of .

Proof.

We showed in the proof of Theorem 5.1 that each element of $\operatorname{Cov}(X)$ is ${\mathord{\hookrightarrow}}$-maximal in $X$. Conversely, if $c$ is ${\mathord{\hookrightarrow}}$-maximal in , then since ${}^{\perp}=\emptyset$, Lemma 5.5 says that $\Sha\setminus\{x\in\Sha:x\twoheadrightarrow c\}$ is ${}^{\perp}\{c\}$. Since $c$ is ${\mathord{\hookrightarrow}}$-maximal, $\{c\}=F(x)$, so Proposition 2.14 says in particular that $(\Sha\setminus\{x\in\Sha:x\twoheadrightarrow c\},\{c\})$ is a maximal orthogonal pair. Since $\{c\}$ is a singleton, this pair is covered by $(\Sha,\emptyset)$. ∎

Proof of Theorem 5.4.

The proof of Theorem 5.1 established that each element $c$ of $\operatorname{Cov}(X)$ is in $C(x)$. If $c^{\prime}\neq c$ is also in $C(X)$ and $c^{\prime}\twoheadrightarrow c$, then $X\supsetneq\operatorname{Del}(X,c^{\prime})\supsetneq\operatorname{Del}(X,c)$, because ${\mathord{\twoheadrightarrow}}$ is a partial order. By Lemma 5.5, these are all closed sets, contradicting the fact that $\operatorname{Del}(X,c)\lessdot X$. Thus $c$ is ${\mathord{\twoheadrightarrow}}$-maximal in $C(X)$.

Now suppose $X$ is finite. We will show that every ${\mathord{\twoheadrightarrow}}$-maximal element $c$ of $C(X)$ is in $\operatorname{Cov}(X)$. If $X$ does not cover $\operatorname{Del}(X,c)$, then $X$ must cover some $X^{\prime}$ with $X^{\prime}\supsetneq\operatorname{Del}(X,c)$. By Theorem 5.1, the set $X^{\prime}$ is $\operatorname{Del}(X,c^{\prime})$ for some $c^{\prime}\in\operatorname{Cov}(X)$. In particular, $c^{\prime}\in C(X)$. But then the containment $\operatorname{Del}(X,c^{\prime})\supsetneq\operatorname{Del}(X,c)$ implies that $c^{\prime}\twoheadrightarrow c$ with $c^{\prime}\neq c$, contradicting the fact that $c$ is ${\mathord{\twoheadrightarrow}}$-maximal in $C(x)$. We conclude that $X$ covers $\operatorname{Del}(X,c)$, so $c\in\operatorname{Cov}(X)$ by Theorem 5.1. ∎

We pause to point out a lemma that will be useful later.

Lemma 5.7**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system and let $X$ be a closed set. If $c\in\operatorname{Cov}(X)$, then $\operatorname{Del}(X,c)\wedge T(c)=T_{*}(c)$ and $\operatorname{Del}(X,c)\vee T(c)=X$.

Proof.

The first assertion follows from the definitions. The second follows from the definitions and the fact that $\operatorname{Del}(X,c)\lessdot X$. ∎

5.2. Canonical join representations

For $S$ and $S^{\prime}$ subsets of $L$, we write $S\leq\!\!\!\leq S^{\prime}$ if, for every $s\in S$, there exists $s^{\prime}\in S^{\prime}$ with $s\leq s^{\prime}$. The relation $\leq\!\!\!\leq$ is a preorder, and becomes a partial order when restricted to antichains.

We say that $S$ is the canonical join representation of $x\in L$ if

1. (1)

$x=\bigvee S$ 2. (2)

$S\leq\!\!\!\leq S^{\prime}$ for any set $S^{\prime}$ such that $x=\bigvee S^{\prime}$ and 3. (3)

$S$ is an antichain.

Remark 5.8**.**

We make several remarks on this definition: Since $\leq\!\!\!\leq$ is a partial order on antichains, $x$ has at most one canonical join representation. If $L$ is finite and $S$ obeys the first two conditions to be the canonical join representation of $x$, then the set of maximal elements of $S$ is the canonical representation of $x$. Finally, if $S$ is the canonical join representation of $x$ and $S^{\prime}\subsetneq S$, then $\bigvee S^{\prime}<x$.

The following well-known fact can be found, for example in [1, Theorem 3-1.4] or [11, Theorem 2.24].

Theorem 5.9**.**

Suppose $L$ is a finite lattice. Then $L$ is join semidistributive if and only if every element of $L$ has a canonical join representation.

Thus, if $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a finite two-acyclic factorization system, we deduce that each element of $\operatorname{Pairs}({\mathord{\to}})$ has a canonical join representation. We will now give an explicit description of this representation. Recall the notation $T(x)$ for $\{x^{\prime}:x\twoheadrightarrow x^{\prime}\}$.

Theorem 5.10**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a finite two-acyclic factorization system. If $X$ is a closed set, then the canonical join representation of $X$ is $\{T(c):c\in\operatorname{Cov}(X)\}$.

Proof.

Let $S=\{T(c):c\in\operatorname{Cov}(X)\}$. We first verify that $\bigvee S=X$. For each $c\in\operatorname{Cov}(X)$, we have $T(c)\subseteq X$, so $\bigvee S\subseteq X$. Suppose for the sake of contradiction that this containment were strict. Then there would be some $X^{\prime}$, covered by $X$, with $\bigvee S\subseteq X^{\prime}\subsetneq X$. (This is the one use of finiteness in this proof.) By Theorem 5.1, $X^{\prime}=\operatorname{Del}(X,c)$ for some $c\in\operatorname{Cov}(X)$ and then $T(c)\not\subseteq X^{\prime}$, a contradiction.

We next show that, if $X=\bigvee S^{\prime}$, then $S\leq\!\!\!\leq S^{\prime}$. We must show that, for all $c\in\operatorname{Cov}(X)$, there is some $X^{\prime}\in S^{\prime}$ with $T(c)\subseteq X^{\prime}$. We have $T(c)\subseteq X^{\prime}$ if and only if $c\in X^{\prime}$ so suppose, for the sake of contradiction that $c$ is not in any $X^{\prime}\in S^{\prime}$. Since $c\not\in X^{\prime}$, we have $X^{\prime}\cap\{x\in\Sha:x\twoheadrightarrow c\}=\emptyset$, and we deduce that the closed set $\operatorname{Del}(X,c)$ contains $X^{\prime}$ for all $X^{\prime}\in S^{\prime}$. But then $\bigvee S^{\prime}\subseteq\operatorname{Del}(X,c)\subsetneq X$, a contradiction.

Theorem 5.4 implies in particular that $\operatorname{Cov}(X)$ is an antichain in the partial order ${\mathord{\twoheadrightarrow}}$. Therefore also $S$ is an antichain. ∎

By Proposition 4.1, the conclusion of Theorem  5.10 also holds when the hypothesis that is finite is replaced with the weaker hypothesis that $X$ is finite.

For ease of exposition, let $L$ be a finite lattice. We say that $S\subseteq L$ joins canonically if $S$ is the canonical join representation of an element (which is necessarily $\bigvee S$). When $L$ is join semidistributive, the collection of all $S\subset L$ such that $S$ joins canonically is a simplicial complex [23, Proposition 2.2]. This simplicial complex is called the canonical join complex. Its vertex set is $\operatorname{JIrr}(L)$.

A simplicial complex is called flag if all of its minimal non-faces are edges, or equivalently, if it is the clique complex of its $1$-skeleton. The following theorem of Barnard [4, Theorem 1.1] characterizes finite semidistributive lattices among all finite join semidistributive lattices in terms of their canonical join complexes.

Theorem 5.11**.**

A finite join semidistributive lattice is semidistributive if and only if its canonical join complex is flag.

In particular, to understand the canonical join complex of a finite semidistributive lattice $L$, it suffices to know which $2$-element subsets of $\operatorname{JIrr}(L)$ join canonically. This turns out to be intimately connected with FTFSDL.

Theorem 5.12**.**

Let $L$ be a finite semidistributive lattice, identified with $\operatorname{Pairs}({\mathord{\to}}_{L})$ for ${\mathord{\to}}_{L}$ as in Theorem 1.2. The faces of the canonical join complex of $L$ are the subsets $S$ of $\operatorname{JIrr}(L)$ such that $x\not\to_{L}y$ for all distinct $x,y\in S$.

Proof.

In light of Theorem 5.11, we need only prove that for $i,j\in\operatorname{JIrr}(L)$, the set $\{i,j\}$ joins canonically if and only neither $i\to_{L}j$ nor $j\to_{L}i$. If $i\to_{L}j$ then $i\vee j_{*}=i\vee j$ by Lemma 2.26. Since $\{i,j_{*}\}\leq\!\!\!\leq\{i,j\}$, the set $\{i,j\}$ does not join canonically. Conversely, suppose $\{i,j\}$ does not join canonically and write $x=i\vee j$. If $x=\bigvee S$ is the canonical join representation of $x$, then $S\leq\!\!\!\leq\{i,j\}$. Since $x=\bigvee S$ is irredundant, $S\not\supseteq\{i,j\}$, so without loss of generality, $j\not\in S$. By the definition of $\leq\!\!\!\leq$, and since $j\not\in S$, every $s\in S$ has either $s\leq i$ or $s\leq j_{*}$. Thus $S\leq\!\!\!\leq\{i,j_{*}\}$, so that $x=\bigvee S\leq i\vee j_{*}\leq i\vee j=x$. We see that $i\vee j_{*}=i\vee j$, so that $i\to_{L}j$. ∎

Remark 5.13**.**

A priori, it is not unreasonable to hope to find a Fundamental Theorem of Finite Join-Semidistributive Lattices along the same lines as FTFSDL. However, in view of Theorem 5.11, we expect that such a hypothetical FTFJDL would not simply involve binary relations between the elements of , but rather relations between $k$-tuples of elements for $k>2$.

6. Quotient Lattices

In this section, we prove Theorem 1.4 and Corollary 1.5, which describe all lattice congruences and quotients of a finite semidistributive lattice. We prove these results as special cases of results on lattices of maximal orthogonal pairs for not-necessarily-finite two-acyclic factorization systems.

To begin the section, we give background and terminology necessary to understand the results and their generalizations. Further background and proofs, from a point of view compatible with the present treatment, can be found in [24, Section 9-5] and [10, Section 2].

A lattice homomorphism is a map $\eta:L_{1}\to L_{2}$ with $\eta(x\wedge y)=\eta(x)\wedge\eta(y)$ and $\eta(x\vee y)=\eta(x)\vee\eta(y)$ for all $x,y\in L_{1}$. Equivalently, $\eta(\bigvee X)=\bigvee(\eta X)$ for all finite, non-empty subsets $X\subseteq L_{1}$. A complete lattice homomorphism has $\eta(\bigvee X)=\bigvee(\eta X)$ for infinite subsets $X$ as well.

A congruence on a lattice $L$ is an equivalence relation $\Theta$ on $L$ that respects the meet and join operations in the following sense: If $x_{1}\equiv y_{1}$ and $x_{2}\equiv y_{2}$ mod $\Theta$, then $x_{1}\wedge x_{2}\equiv y_{1}\wedge y_{2}$ and $x_{1}\vee x_{2}\equiv y_{1}\vee y_{2}$ mod $\Theta$. Equivalently, if $I$ is a finite, non-empty indexing set and $(x_{i}:i\in I)$ and $(y_{i}:i\in I)$ are elements of $L$ such that $x_{i}\equiv y_{i}\bmod\Theta$ for all $i\in I$, then $\bigwedge_{i\in I}x_{i}\equiv\bigwedge_{i\in I}y_{i}$ and $\bigvee_{i\in I}x_{i}\equiv\bigvee_{i\in I}y_{i}$ mod $\Theta$. If the same condition holds for all infinite indexing sets, then $\Theta$ is a complete congruence. The quotient $L/\Theta$ is the set of congruence classes, with the obvious lattice structure. The map sending an element to its congruence class is a surjective lattice homomorphism from $L$ to $L/\Theta$. On the other hand, given a (complete) surjective lattice homomorphism $\eta:L_{1}\to L_{2}$, the fibers of $\eta$ are the congruence classes of a (complete) congruence $\Theta$ on $L_{1}$, and $\eta$ induces an isomorphism from $L_{1}/\Theta$ to $L_{2}$. Every congruence class of a complete lattice congruence is an interval in $L$.

The set of all equivalence relations on a set $X$ forms a lattice (the lattice of set partitions of $X$), where $\Theta_{1}$ is smaller than $\Theta_{2}$ if $\Theta_{1}$ refines $\Theta_{2}$ (in other words, if $x\equiv y\bmod\Theta_{1}$ implies $x\equiv y\bmod\Theta_{2}$). The set $\operatorname{Con}(L)$ of all congruences on $L$ is a sublattice of the lattice of set partitions of $L$. While the lattice of set partitions is not even semidistributive (for $|X|>2$), the lattice $\operatorname{Con}(L)$ is distributive for any lattice $L$ [13, Theorem 149].

In particular, when $L$ is finite, we can understand $\operatorname{Con}(L)$ by way of the FTFDL. Given a congruence $\Theta$ on a finite lattice $L$ and an edge $a\lessdot b$ in the Hasse diagram of $L$, we say that $\Theta$ contracts $a\lessdot b$ if $a\equiv b$ mod $\Theta$. There is a unique finest congruence contracting $a\lessdot b$ (the meet in $\operatorname{Con}(L)$ of all congruences contracting $b\lessdot b$), and we write $\operatorname{con}(a,b)$ for this congruence. The following proposition is well known; see, for example, [24, Proposition 9.5.14].

Proposition 6.1**.**

Let $L$ be a finite lattice and $\Theta$ a congruence on $L$. The following are equivalent:

1. (i)

$\Theta$* is a join-irreducible element in the lattice $\operatorname{Con}(L)$.* 2. (ii)

$\Theta$* is of the form $\operatorname{con}(a,b)$ for some cover $a\lessdot b$ of $L$.* 3. (iii)

$\Theta$* is of the form $\operatorname{con}(j_{*},j)$ for some join-irreducible element $j$.* 4. (iv)

$\Theta$* is of the form $\operatorname{con}(m,m^{*})$ for some meet-irreducible element $m$.*

Since $\operatorname{Con}(L)$ is a distributive lattice, the FTFDL states that it is isomorphic to the lattice of downsets of $\operatorname{JIrr}(\operatorname{Con}(L))$. Proposition 6.1 states that $j\mapsto\operatorname{con}(j_{*},j)$ gives a surjection from $\operatorname{JIrr}(L)$ to $\operatorname{JIrr}(\operatorname{Con}(L))$. We define the forcing preorder on $\operatorname{JIrr}(L)$ by saying that $j$ forces $j^{\prime}$ if and only if $\operatorname{con}(j_{*},j)\geq\operatorname{con}(j^{\prime}_{*},j^{\prime})$ or, in other words, if every congruence contracting $j$ also contracts $j^{\prime}$. Forcing is a preorder on $\operatorname{JIrr}(L)$, meaning it is reflexive and transitive, but perhaps not antisymmetric. Thus $\operatorname{Con}(L)$ is isomorphic to the poset of downsets for the forcing preorder on $\operatorname{JIrr}(L)$.

On a finite lattice, congruences are determined by the set of covers that they contract. The same is not true for infinite complete lattices. For example, consider the $\mathbb{N}\cup\{\infty\}$ with the obvious partial order. There are two complete congruences that contract all covers, one with a single congruence class and one with two classes $\mathbb{N}$ and $\{\infty\}$. We will consider the congruences that are determined by the covers they contract.

Given a congruence $\Theta$ on a lattice $L$, if $x\leq y$ and $x\equiv y\bmod\Theta$, then for all $u,v$ with $x\leq u\lessdot v\leq y$, we have $u\equiv v\bmod\Theta$. The congruence $\Theta$ is cover-determined if the converse holds as well: If $x\leq y$ and $u\equiv v\bmod\Theta$ for all $u,v$ with $x\leq u\lessdot v\leq y$, then $x\equiv y\bmod\Theta$. This notion was mentioned earlier in Remark 3.1.

We have given the lattice-theoretic background to Theorems 1.4 and its generalization. We now need two more definitions related to two-acyclic factorization systems.

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a two-acyclic factorization system. Given $x$, $y\in\Sha$, we write $x\rightsquigarrow y$ and say $x$ directly forces $y$ if and only if either

1. (i)

$x$ is ${\mathord{\twoheadrightarrow}}$-minimal in $F(y)=\{x^{\prime}\in\Sha:x^{\prime}\hookrightarrow y\}$, or 2. (ii)

$x$ is ${\mathord{\hookrightarrow}}$-maximal in $T(y)=\{x^{\prime}\in\Sha:y\twoheadrightarrow x^{\prime}\}$.

Recall that a $\mathord{\rightsquigarrow}$-upset means a subset of such that, if $x\rightsquigarrow y$ and $y\in\Che$, then $x\in\Che$.

Let $x\to z$, so $\{y\in\Sha:x\twoheadrightarrow y\hookrightarrow z\}$ is nonempty. We define an ${\mathord{\twoheadrightarrow}}$-minimal element of $\{y\in\Sha:x\twoheadrightarrow y\hookrightarrow z\}$ to be an image of $x\to z$ and define an ${\mathord{\hookrightarrow}}$-maximal element to be a co-image of $x\to z$. We will say that a factorization system satisfies the image condition if, for every $x$ and $z\in\Sha$ with $x\to z$, there is at least one image and at least one co-image of $x\to z$.

We will prove the following generalization of Theorem 1.4.

Theorem 6.2**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system that obeys the image condition and $\Che\subseteq\Sha$ is a $\mathord{\rightsquigarrow}$-upset. Then the restriction of $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ to is a two-acyclic factorization system that obeys the image condition. The map $(X,Y)\mapsto(X\cap\Che,Y\cap\Che)$ is a surjective complete lattice homomorphism between the corresponding lattices of maximal orthogonal pairs. The complete congruence associated to $(X,Y)\mapsto(X\cap\Che,Y\cap\Che)$ is cover-determined. Every cover-determined congruence on $\operatorname{Pairs}({\mathord{\to}})$ arises from a $\mathord{\rightsquigarrow}$-upset in this way.

We note that it is not obvious that $(X\cap\Che,Y\cap\Che)$ is a maximal orthogonal pair in $\operatorname{Pairs}({\mathord{\to}}|)$; this is checked as Proposition 6.5.

Since every finite factorization system satisfies the image condition and every congruence on a finite lattice is complete and cover-determined, Theorem 1.4 is an immediate corollary of Theorem 6.2. We now proceed to prove Theorem 6.2. Following that proof, we will prove Corollary 1.5 as an immediate corollary of an infinite version (Corollary 6.14).

Conventions: For the rest of the section, we will assume that $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system that obeys the image condition. We will also assume that $\Che\subset\Sha$ is a $\mathord{\rightsquigarrow}$-upset. The symbols ${\mathord{\to}}^{\prime}$, ${\mathord{\twoheadrightarrow}}^{\prime}$, and ${\mathord{\hookrightarrow}}^{\prime}$ will stand for the restrictions of ${\mathord{\to}}$, ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ to .

Throughout the proof, the image condition will almost always be used through the following lemma:

Lemma 6.3**.**

If $y$ is an image of $x\to z$, then $y\rightsquigarrow z$. If $y$ is a co-image of $x\to z$ then $y\rightsquigarrow x$.

Proof.

We prove the first claim; the second is dual. Suppose that $y\not\rightsquigarrow z$. By hypothesis, $y\hookrightarrow z$, so $y$ must not be ${\mathord{\twoheadrightarrow}}$-minimal in $\{y^{\prime}:y^{\prime}\hookrightarrow z\}$. Thus there exists $y^{\prime}\neq y$ with $y\twoheadrightarrow y^{\prime}\hookrightarrow z$. But then $x\twoheadrightarrow y^{\prime}\hookrightarrow z$, contradicting the fact that $y$ is an image. ∎

We prove Theorem 6.2 as a series of propositions.

Proposition 6.4**.**

$(\Che,{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$* is a two-acyclic factorization system that obeys the image condition.*

Proof.

Two-acyclicity is obvious, and it is clear that $\operatorname{Mult}({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})\subseteq{\mathord{\to}}^{\prime}$ and $\operatorname{Fact}({\mathord{\to}}^{\prime})\supseteq({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$. We now prove the latter containments are equalities.

In order to show that $\operatorname{Mult}({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})\supseteq{\mathord{\to}}^{\prime}$, we suppose $x,z\in\Che$ have $x\to z$ and show that there exists $y\in\Che$ with $x\twoheadrightarrow y\hookrightarrow z$. The image condition says there is an image $y$ of $x\to z$. By Lemma 6.3, $y\rightsquigarrow z$, and since is a $\mathord{\rightsquigarrow}$-upset, we have $y\in\Che$, as desired. (We pause to note that $y$ is also an image of $x\to z$ in .)

To show that $\operatorname{Fact}({\mathord{\to}}^{\prime})\subseteq({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$, let $\operatorname{Fact}({\mathord{\to}}^{\prime})=({\mathord{\twoheadrightarrow}}^{\prime\prime},{\mathord{\hookrightarrow}}^{\prime\prime})$. We show that ${\mathord{\twoheadrightarrow}}^{\prime\prime}\subseteq{\mathord{\twoheadrightarrow}}^{\prime}$; the proof that ${\mathord{\hookrightarrow}}^{\prime\prime}\subseteq{\mathord{\hookrightarrow}}^{\prime}$ is dual. We first suppose that $x_{1}\twoheadrightarrow^{\prime\prime}x_{2}$ for some $x_{1}$ and $x_{2}\in\Che$ and show that $x_{1}\twoheadrightarrow x_{2}$. In other words, the hypothesis is that $\{y\in\Che:x_{1}\to y\}\supseteq\{y\in\Che:x_{2}\to y\}$, and the desired conclusion is that $\{z\in\Sha:x_{1}\to z\}\supseteq\{z\in\Sha:x_{2}\to z\}$. Suppose $x_{2}\to z$ and let $y$ be a co-image of $x_{2}\to z$. By Lemma 6.3, $y\rightsquigarrow x_{2}$, and thus $y\in\Che$ because is a $\mathord{\rightsquigarrow}$-upset. Since $y$ is a co-image of $x_{2}\to z$, in particular, $x_{2}\twoheadrightarrow y\hookrightarrow z$. By the hypothesis, $x_{1}\twoheadrightarrow y\hookrightarrow z$, so that $x_{1}\to z$ as desired. (Again, we note that $y$ is also a co-image of $x_{2}\to z$ in .)

We have shown that $(\Che,{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$ is a two-acyclic factorization system. In the process, we have shown how to find both an image and a co-image for an arbitrary ${\mathord{\to}}^{\prime}$-arrow, so $(\Che,{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$ obeys the image condition. ∎

Proposition 6.5**.**

If $(X,Y)\in\operatorname{Pairs}({\mathord{\to}})$, then $(X\cap\Che,Y\cap\Che)\in\operatorname{Pairs}({\mathord{\to}}^{\prime})$.

Proof.

We write $X^{\prime}$ for $X\cap\Che$ and $Y^{\prime}$ for $Y\cap\Che$. We use the symbol $\triangle$ for $\perp$ with respect to $(\Che,{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$. Since $(X,Y)$ is a maximal orthogonal pair for $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, we have $Y^{\prime}\subseteq(X^{\prime})^{\triangle}$ and $X^{\prime}\subseteq{}^{\triangle}(Y^{\prime})$. We will argue that $Y^{\prime}\supseteq(X^{\prime})^{\triangle}$; the argument that $X^{\prime}\supseteq{}^{\triangle}(Y^{\prime})$ is dual.

To show that $Y^{\prime}\supseteq(X^{\prime})^{\triangle}$, we must show that, if $z\in\Che$ and $z\not\in Y$, then there exists $y\in X^{\prime}$ with $y\to z$. Since $z\not\in Y$, by definition, there is some $x\in X$ with $x\to z$. Let $y$ be an image of $x\to z$, so that $y\rightsquigarrow z$ by Lemma 6.3. But is a $\mathord{\rightsquigarrow}$-upset, so this implies that $y\in\Che$. Also, $x\twoheadrightarrow y$, so $y\in X$ by Proposition 2.4, and thus $y\in X^{\prime}$. Since $y\hookrightarrow z$, we have $y\to z$, as desired. ∎

Proposition 6.6**.**

The map $(X,Y)\mapsto(X\cap\Che,Y\cap\Che)$ is a surjective complete lattice homomorphism from $\operatorname{Pairs}({\mathord{\to}})$ to $\operatorname{Pairs}({\mathord{\to}}^{\prime})$.

Proof.

Identifying $\operatorname{Pairs}({\mathord{\to}})$ with the lattice of closed sets, the meet is intersection. The assertion that the map respects arbitrary meets is $\bigcap_{i}(X_{i}\cap\Che)=\left(\bigcap X_{i}\right)\cap\Che$, which is obvious. The proof that the map respects meets is dual (identifying $\operatorname{Pairs}({\mathord{\to}}^{\prime})$ with the dual notion of the lattice of closed sets by considering the second entries in pairs).

We now show that this map is surjective. Let $(X^{\prime},Y^{\prime})$ be a maximal orthogonal pair for ${\mathord{\to}}^{\prime}$. Let $X={}^{\perp}Y^{\prime}$ (using this $\perp$ operator in the sense of $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$). Then $X$ is closed (in the sense of ${\mathord{\to}}$), and unfolding the definitions immediately shows that $X\cap\Che=X^{\prime}$. Let $Y=X^{\perp}$. Then $(X,Y)$ is a maximal orthogonal pair, so $(X\cap\Che,Y\cap\Che)$ is a maximal orthogonal pair by Proposition 6.5. Since $X\cap\Che=X^{\prime}$ and $(X^{\prime},Y^{\prime})$ is a maximal orthogonal pair, we also have $Y\cap\Che=Y^{\prime}$. ∎

The following is [10, Proposition 2.6].

Proposition 6.7**.**

A complete lattice congruence $\Theta$ on a complete lattice $L$ is cover-determined if and only if the quotient $L/\Theta$ is cover-separated.

Theorem 3.2 says that $\operatorname{Pairs}({\mathord{\to}}^{\prime})$ is cover-separated, and thus Proposition 6.7 implies the following proposition.

Proposition 6.8**.**

The complete congruence associated to $(X,Y)\mapsto(X\cap\Che,Y\cap\Che)$ is cover-determined.

To prove the rest of Theorem 6.2, we need two simple, known lemmas. The second is essentially the easy direction of [8, Lemma 3.2], but appears in that source without proof.

Lemma 6.9**.**

Let $L$ be a $\kappa$-lattice, let $\eta:L\to L^{\prime}$ be a lattice homomorphism, and let $j$ be a completely join irreducible element of $L$. Then $\eta(j)=\eta(j_{*})$ if and only if $\eta(\kappa(j))=\eta(\kappa(j)^{*})$.

Proof.

For brevity, put $m=\kappa(j)$. We have $j\vee m=m^{*}$ and $j_{*}\vee m=m$, so if $\eta(j)=\eta(j_{*})$ then $\eta(m)=\eta(m^{*})$. We have $m\wedge j=j_{*}$, so if $\eta(m)=\eta(m^{*})$ then $\eta(j)=\eta(j_{*})$. ∎

Lemma 6.10**.**

Suppose $L$ is a lattice and $\Theta$ is a lattice congruence on $L$. Suppose we have elements $a$, $b$, $c$, $d$ of $L$ with $a\leq b$ and $a\equiv b\bmod\Theta$. Suppose that either $a\leq c\leq d\leq b\vee c$ or else $a\wedge d\leq c\leq d\leq b$. Then $c\equiv d\bmod\Theta$.

Proof.

We consider the case $a\leq c\leq d\leq b\vee c$; the other case is similar. Then $a\equiv b\bmod\Theta$, so $c=a\vee c\equiv b\vee c\bmod\Theta$. Since $c\leq d\leq b\vee c$, we have $c=c\wedge d\equiv(b\vee c)\wedge d=d\bmod\Theta$. ∎

Additional conditions: We continue to assume that $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system that obeys the image condition. In addition, for the rest of the section, we will assume that $\Theta$ is a complete lattice congruence on $\operatorname{Pairs}({\mathord{\to}})$. Furthermore, we define

[TABLE]

Lemma 6.9 and Proposition 2.17 combine to imply that

[TABLE]

We will want both descriptions of Θ.

Proposition 6.11**.**

Θ* is a $\mathord{\rightsquigarrow}$-upset.*

Proof.

We prove the equivalent assertion that $\Sha\setminus{}_{\Theta}$ is a $\mathord{\rightsquigarrow}$-downset. Suppose $x\not\in{}_{\Theta}$, meaning that $(T_{*}(x),T_{*}(x)^{\perp})\equiv(T(x),T(x)^{\perp})$, and suppose $x\rightsquigarrow y$.

We suppose that $x$ is ${\mathord{\twoheadrightarrow}}$-minimal in $F(y)$; the case where $x$ is ${\mathord{\hookrightarrow}}$-maximal in $T(y)$ is similar. We will apply the first case of Lemma 6.10 with ${a=(T_{*}(x),T_{*}(x)^{\perp})}$, $b=(T(x),T(x)^{\perp})$, $c=({}^{\perp}F(y),F(y))$ and $d=({}^{\perp}F^{*}(y),F^{*}(y))$. To that end, we must confirm that $(T_{*}(x),T_{*}(x)^{\perp})\leq({}^{\perp}F(y),F(y))$ and $({}^{\perp}F^{*}(y),F^{*}(y))\leq(T(x),T(x)^{\perp})\vee({}^{\perp}F(y),F(y))$.

To show that $(T_{*}(x),T_{*}(x)^{\perp})\leq({}^{\perp}F(y),F(y))$, we must show that $T_{*}(x)\subseteq{}^{\perp}F(y)$. Suppose for the sake of contradiction that there exists $x^{\prime}\in T_{*}(x)$ with $x^{\prime}\not\in{}^{\perp}F(y)$. That is, there exist $x$ and $y$ in with $x\twoheadrightarrow x^{\prime}\neq x$ and $x^{\prime}\to y^{\prime}\hookrightarrow y$. Factor the arrow $x^{\prime}\to y^{\prime}$ as $x^{\prime}\twoheadrightarrow z\hookrightarrow y^{\prime}$, so that $x\twoheadrightarrow z\hookrightarrow y$. Since $x\neq x^{\prime}$, also $x\neq z$, contradicting the ${\mathord{\twoheadrightarrow}}$-minimality of $x$ in $F(y)$.

To show that $({}^{\perp}F^{*}(y),F^{*}(y))\leq(T(x),T(x)^{\perp})\vee({}^{\perp}F(y),F(y))$, we must show that $F^{*}(y)\supseteq T(x)^{\perp}\cap F(y)$. The only element of $F(y)$ not in $F^{*}(y)$ is $y$, so we simply must show that $y\not\in T(x)^{\perp}$. But $x\in F(y)$, so in particular $x\to y$ and thus $y\not\in T(x)^{\perp}$.

Now Lemma 6.10 says that $({}^{\perp}F(y),F(y))\equiv({}^{\perp}F^{*}(y),F^{*}(y))\bmod\Theta$, so that $y\in{}_{\Theta}$, using the second description of Θ. ∎

In light of Proposition 6.11, Proposition 6.4 says the restriction of $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ to Θ is a two-acyclic factorization system satisfying the image condition, and Proposition 6.6 says the map $(X,Y)\mapsto(X\cap{}_{\Theta},Y\cap{}_{\Theta})$ is a complete surjective lattice homomorphism from $\operatorname{Pairs}({\mathord{\to}})$ to the lattice of maximal orthogonal pairs for the restriction.

Proposition 6.12**.**

If $\Theta$ is cover-determined, then $\Theta$ is the congruence associated to $(X,Y)\mapsto{(X\cap{}_{\Theta},Y\cap{}_{\Theta})}$.

The following lemma will be useful in proving Proposition 6.12. The lemma depends neither on the image condition nor on the completeness of $\Theta$. It follows immediately from Lemma 5.7 and the definition of a congruence.

Lemma 6.13**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system and $\Theta$ is a congruence on $\operatorname{Pairs}({\mathord{\to}})$. If $(X,Y)\in\operatorname{Pairs}({\mathord{\to}})$ and $c\in\operatorname{Cov}(X)$, then $(\operatorname{Del}(X,c),\operatorname{Del}(X,c)^{\perp})\equiv(X,Y)$ if and only if $(T_{*}(c),T_{*}(c)^{\perp})\equiv(T(c),T(c)^{\perp})$ mod $\Theta$.

Proof of Proposition 6.12.

Proposition 6.8 says that the complete congruence on $\operatorname{Pairs}$ associated to $(X,Y)\mapsto{(X\cap{}_{\Theta},Y\cap{}_{\Theta})}$ is cover-determined. Since $\Theta$ is also cover-determined, we only need to show that if $(X_{1},Y_{1})\lessdot(X_{2},Y_{2})$ in $\operatorname{Pairs}({\mathord{\to}})$, then $(X_{1},Y_{1})\equiv(X_{2},Y_{2})\bmod\Theta$ if and only if $X_{1}\cap{}_{\Theta}=X_{2}\cap{}_{\Theta}$.

Suppose $(X_{1},Y_{1})\lessdot(X_{2},Y_{2})$ in $\operatorname{Pairs}({\mathord{\to}})$. Theorem 5.1 says that there exists $c$ such that $X_{1}=\operatorname{Del}(x_{2},c)$. Since $X_{1}\cap{}_{\Theta}$ is a downset for the restriction of ${\mathord{\twoheadrightarrow}}$ to Θ, we have $X_{1}\cap{}_{\Theta}=X_{2}\cap{}_{\Theta}$ if and only if $c\not\in{}_{\Theta}$. By definition, $c\not\in{}_{\Theta}$ if and only if $(T_{*}(c),T_{*}(c)^{\perp})\equiv(T(c),T(c)^{\perp})\bmod\Theta$. By Lemma 6.13, this is if and only if $(X_{1},Y_{1})\equiv(X_{2},Y_{2})\bmod\Theta$. ∎

We have completed the proof of Theorem 6.2. We now proceed to generalize and prove Corollary 1.5.

Given a complete lattice $L$, let $\operatorname{Con}^{\operatorname{cc}}(L)$ be the set of complete cover-determined congruences on $L$. Then $\operatorname{Con}^{\operatorname{cc}}(L)$ is a complete lattice under refinement order. More specifically, [10, Proposition 2.7] says that it is a complete meet-sublattice of the lattice of complete congruences on $L$, which is a complete meet-sublattice of $\operatorname{Con}(L)$. It is thus a complete meet-semilattice, and since it also has a maximal element, it is a complete lattice by a standard argument.

Given a cover relation $a\lessdot b$ in $L$, we use the notation $\operatorname{con}^{\operatorname{cc}}(a,b)$ for the finest complete cover-determined congruence having $a\equiv b$. This is the meet, in $\operatorname{Con}^{\operatorname{cc}}(L)$, of all complete cover-determined congruences with $a\equiv b$; this meet has $a\equiv b$ because $\operatorname{Con}^{\operatorname{cc}}(L)$ is a meet-sublattice of the lattice of partitions of $L$. A cover-determined complete congruence $\Theta$ is the join, in $\operatorname{Con}^{\operatorname{cc}}(L)$, of the congruences $\operatorname{con}^{\operatorname{cc}}(a,b)$ for all covers $a\lessdot b$ contracted by $\Theta$.

When $L$ is $\operatorname{Pairs}({\mathord{\to}})$ for a two-acyclic factorization system, Lemma 6.13 implies that a cover-determined congruence $\Theta$ is the join of the congruences $\operatorname{con}^{\operatorname{cc}}(a,b)$ for all covers $a\lessdot b$ contracted by $\Theta$ such that $b$ is $(T(x),T(x)^{\perp})$ and $a$ is $(T_{*}(x),T_{*}(x)^{\perp})$. We are led to define the cover-determined forcing preorder on $\operatorname{JIrr}^{c}(\operatorname{Pairs}({\mathord{\to}}))$: We say that $(T(x),T(x)^{\perp})$ forces $(T(y),T(y)^{\perp})$ if and only if every cover-determined congruence contracting the cover $(T_{*}(x),T_{*}(x)^{\perp})\lessdot(T(x),T(x)^{\perp})$ also contracts the cover $(T_{*}(y),T_{*}(y)^{\perp})\lessdot(T(y),T(y)^{\perp})$.

We are now prepared to state and prove the generalization of Corollary 1.5.

Corollary 6.14**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system that obeys the image condition. The map $x\mapsto(T(x),T(x)^{\perp})$ is an isomorphism from the transitive closure of $\mathord{\rightsquigarrow}$ on to the cover-determined forcing preorder on $\operatorname{JIrr}^{c}(\operatorname{Pairs}({\mathord{\to}}))$. This map induces an isomorphism from the poset of $\mathord{\rightsquigarrow}$-downsets under containment to the lattice $\operatorname{Con}^{\operatorname{cc}}(\operatorname{Pairs}({\mathord{\to}}))$.

Proof.

By Proposition 2.14, the map $x\mapsto(T(x),T(x)^{\perp})$ is a bijection from to $\operatorname{JIrr}(\operatorname{Pairs}({\mathord{\to}}))$. Theorem 6.2 implies that for $x,y\in\Sha$, $(T(x),T(x)^{\perp})$ forces $(T(y),T(y)^{\perp})$ in the cover-determined forcing preorder if and only if there is a directed path in the relation $\mathord{\rightsquigarrow}$ from $x$ to $y$. Theorem 6.2 also implies that cover-determined complete congruences are completely specified by the set of elements $x\in\Sha$ such that $(T_{*}(x),T_{*}(x)^{\perp})\lessdot(T(x),T(x)^{\perp})$ is contracted, and that the sets arising are precisely the $\mathord{\rightsquigarrow}$-downsets. The refinement order on cover-determined complete congruences is containment order on these $\mathord{\rightsquigarrow}$-downsets. ∎

7. Relation to other types of finite lattices

7.1. Distributive lattices

A lattice $L$ is distributive if $x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z)$ and $x\vee(y\wedge z)=(x\vee y)\wedge(x\vee z)$ for all $x$, $y$ and $z\in L$. This condition is clearly stronger than semidistributivity, so every distributive lattice is semidistributive and thus, every finite distributive lattice is of the form $\operatorname{Pairs}({\mathord{\to}})$ where $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system.

We now state three results that summarize the relationship between the Fundamental Theorem of Finite Distributive Lattices (FTFDL—Theorem 1.1) and the Fundamental Theorem of Finite Semidistributive Lattices (FTFSDL—Theorem 1.2). These statements are easily proved, and we omit the proofs. Theorem 7.2 is stated in a way that allows convenient comparison with Theorem 2.31.

Proposition 7.1**.**

Suppose $P$ is a set and ${\mathord{\to}}$ is a partial order on $P$. Then $(P,{\mathord{\to}},{\mathord{\to}},{\mathord{\to}})$ is a two-acyclic factorization system. The map $(X,Y)\mapsto X$ is an isomorphism from $\operatorname{Pairs}({\mathord{\to}})$ to $\operatorname{Downsets}(P)$, with inverse $X\mapsto(X,P\setminus X)$.

Theorem 7.2** (FTFDL, rephrased).**

A finite poset $L$ is a distributive lattice if and only if it is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for some partial order ${\mathord{\to}}$ on a finite set $P$. In this case, $(P,{\mathord{\to}})$ is isomorphic to $(\operatorname{JIrr}(L),\geq)$, where $\geq$ is the partial order induced from $L$. The map $x\mapsto(\{j\in\operatorname{JIrr}(L):j\leq x\},\ \{j\in\operatorname{JIrr}(L):j\not\leq x\}))$ is an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\to}})$, with inverse $(X,Y)\mapsto\bigvee X$.

Proposition 7.3**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a finite two-acyclic factorization system. Then the following are equivalent:

$\operatorname{Pairs}({\mathord{\to}})$* is distributive.* 2. 2.

${\mathord{\to}}$* is a partial order.* 3. 3.

${\mathord{\twoheadrightarrow}}={\mathord{\hookrightarrow}}$. 4. 4.

${\mathord{\to}}={\mathord{\twoheadrightarrow}}$. 5. 5.

${\mathord{\to}}={\mathord{\hookrightarrow}}$.

Using Proposition 7.3, we see that if $L$ is finite and distributive then $x\mathord{\rightsquigarrow}y$ if and only if $x=y$. From there, by Corollary 1.5, we recover the well known fact that $\operatorname{JIrr}(\operatorname{Con}(L))$ is an antichain.

7.2. Congruence uniform lattices

We recall our discussion of lattice congruences from Section 6. As we described, the set of congruences on a finite lattice $L$ forms a distributive lattice $\operatorname{Con}(L)$. Since $\operatorname{Con}(L)$ is distributive, it is isomorphic to the poset of downsets of $\operatorname{JIrr}(\operatorname{Con}(L))$. Proposition 6.1 described surjective maps to $\operatorname{JIrr}(\operatorname{Con}(L))$ from $\operatorname{JIrr}(L)$, from $\operatorname{MIrr}(L)$, and from the set of covers of $L$. When $L$ is semidistributive, Lemma 6.13 implies that surjection from $\mathrm{Covers}(L)$ to $\operatorname{JIrr}(\operatorname{Con}(L))$ factors (as a composition of surjections) through $\operatorname{JIrr}(L)$. Dually, the map factors through $\operatorname{MIrr}(L)$. Also when $L$ is semidistributive, $\kappa$ gives a bijection between $\operatorname{JIrr}(L)$ and $\operatorname{MIrr}(L)$. These maps are shown in the diagram below, and they all commute:

[TABLE]

A finite lattice $L$ is conguence uniform if the map $j\mapsto\operatorname{con}(j_{*},j)$ from $\operatorname{JIrr}(L)$ to $\operatorname{JIrr}(\operatorname{Con}(L))$ and the map $m\mapsto\operatorname{con}(m,m^{*})$ from $\operatorname{MIrr}(L)$ to $\operatorname{JIrr}(\operatorname{Con}(L))$ are both bijections. In this case, $L$ is semidistributive (see [8, Lemma 4.2] and [8, Theorem 5.1]), with $\kappa$ and $\kappa^{d}$ being the unique maps that make the diagram above commute. Thus a finite lattice $L$ is conguence uniform if and only if it is semidistributive and one (and therefore both) of the maps $j\mapsto\operatorname{con}(j_{*},j)$ and $m\mapsto\operatorname{con}(m,m^{*})$ is a bijection. Since the forcing preorder on $\operatorname{JIrr}(L)$ is the pullback of the partial order on $\operatorname{JIrr}(\operatorname{Con}(L))$, we can alternatively say that $L$ is congruence uniform if it is semidistributive and the forcing preorder on $\operatorname{JIrr}(L)$ is a partial order. Thus we have the following corollary of the FTFSDL and Corollary 1.5.

Corollary 7.4**.**

Suppose $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a finite two-acyclic factorization system. Then $\operatorname{Pairs}({\mathord{\to}})$ is congruence uniform if and only if $\mathord{\rightsquigarrow}$ is acyclic.

As mentioned above, every finite congruence uniform lattice is also semidistributive. Thus the concatenation of Corollary 7.4 and Theorem 1.2 can be thought of as a Fundamental Theorem of Finite Congruence Uniform Lattices.

Another characterization of congruence uniformity, due to Day, uses the notion of doubling that we now recall. Let $\mathbf{2}$ be the two-element lattice with elements $1<2$. Let $L$ be a finite lattice and let $I$ be an interval in $L$. Let $L[I]$ be the set $(L\setminus I)\cup(I\times\mathbf{2})$ and let $\pi:L[I]\to L$ be the obvious projection. The doubling of $I$ in $L$ is $L[I]$ equipped with the partial order that $x\leq y$ if

1. (i)

$\pi(x)\leq\pi(y)$ and 2. (ii)

if $x=(a,i)$ and $y=(b,j)\in I\times\mathbf{2}$, then $i\leq j$.

The following is [8, Corollary 5.4].

Theorem 7.5**.**

A finite lattice $L$ is congruence uniform if and only if there is a sequence of lattices $L_{0},L_{1},\ldots,L_{k}=L$ such that $L_{0}$ has exactly one element and for all $j=1,\ldots,k$, there is an interval $I_{j-1}$ in $L_{j-1}$ such that $L_{j}\cong L_{j-1}[I_{j-1}]$.

In light of Corollary 7.4 and Theorem 7.5, it is natural to ask how doubling happens in a two-acyclic factorization system. This question is answered by the following theorem, whose proof we omit.

Theorem 7.6**.**

Suppose that $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a finite two-acyclic factorization system and suppose that $(X_{1},Y_{1})\leq(X_{2},Y_{2})$ in $\operatorname{Pairs}({\mathord{\to}})$. Then the doubling $\operatorname{Pairs}({\mathord{\to}})[(X_{1},Y_{1}),(X_{2},Y_{2})]$ is a semidistributive lattice isomorphic to $\operatorname{Pairs}({\mathord{\to}}^{\prime})$ for a two-acyclic factorization system $(\Sha\cup\{a\},{\mathord{\to}}^{\prime},{\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$ for some $a\not\in\Sha$. The relations ${\mathord{\to}}^{\prime}$, ${\mathord{\twoheadrightarrow}}^{\prime}$, and ${\mathord{\hookrightarrow}}^{\prime}$ agree with ${\mathord{\to}}$, ${\mathord{\twoheadrightarrow}}$, and ${\mathord{\hookrightarrow}}$ on . In addition, there are the reflexive relations $a\to^{\prime}a$, $a\twoheadrightarrow^{\prime}a$, and $a\hookrightarrow^{\prime}a$ and the following relations:

1. (i)

$x\to^{\prime}a$* for all $x\in\Sha\setminus X_{2}$ and $a\to^{\prime}y$ for all $y\in\Sha\setminus Y_{1}$.* 2. (ii)

$a\twoheadrightarrow^{\prime}y$* for all $y\in X_{1}$ and $x\twoheadrightarrow^{\prime}a$ for all $x\in\Sha\setminus X_{2}$ such that $x\twoheadrightarrow z$ for all $z\in X_{1}$.* 3. (iii)

$x\hookrightarrow^{\prime}a$* for all $x\in Y_{2}$ and $a\hookrightarrow^{\prime}y$ for all $y\in\Sha\setminus Y_{1}$ such that $z\hookrightarrow y$ for all $z\in Y_{2}$.*

Part of Theorem 7.6 is the following proposition, which is well known. (See, for example, [9, Theorem B].)

Proposition 7.7**.**

If $L$ is a finite semidistributive lattice and $I$ is an interval in $L$, then $L[I]$ is a semidistributive lattice.

7.3. General finite lattices

One may wonder whether there is a Fundamental Theorem of Finite Lattices, giving a canonical way of realizing an arbitrary lattice as something like $\operatorname{Pairs}({\mathord{\to}})$. Such a result is given by a theorem of Markowsky [16]. We restate Markowsky’s result in a manner that allows easy comparison to FTFSDL. Whereas FTFSDL realizes a finite semidistributive lattice as $\operatorname{Pairs}({\mathord{\to}})$ for a relation ${\mathord{\to}}$ on a single set , Markowsky’s theorem realizes an arbitrary finite lattice as $\operatorname{Pairs}({\mathord{\to}})$ for a relation ${\mathord{\to}}$ between two sets and . ( and are the Cyrillic letters “el” and “er”, which we hope are mnemonic for “left” and “right”.) Given a relation ${\mathord{\to}}$ between sets and , for any subset $X$ of , put $X^{\perp}=\{y\in\Er:\ x\not\to y\ \forall x\in X\}$ and, for any subset $Y$ of , put ${}^{\perp}Y=\{x\in\El:\ x\not\to y\ \forall y\in Y\}$. We define $(X,Y)$ to be a maximal orthogonal pair if $Y=X^{\perp}$ and $X={}^{\perp}Y$ and we write $\operatorname{Pairs}({\mathord{\to}})$ for the set of maximal orthogonal pairs, ordered by containment on the first element. We partially order $\operatorname{Pairs}({\mathord{\to}})$ by containment in the first entries of pairs (or equivalently reverse containment in the second entries).

We write $\operatorname{Fact}({\mathord{\to}})$ for the pair $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ of preorders ${\mathord{\twoheadrightarrow}}$ and ${\mathord{\hookrightarrow}}$ on and respectively defined as follows: $x_{1}\twoheadrightarrow x_{2}$ if $x_{2}\to y$ implies $x_{1}\to y$; and $y_{1}\hookrightarrow y_{2}$ if $x\to y_{1}$ implies $x\to y_{2}$. (If $\El=\Er$, then this coincides with the earlier definition of $\operatorname{Fact}({\mathord{\to}})$ in connection with FTFSDL.) Given $x\in\El$, a right companion of $x$ is an element $y\in\Er$ such that $x\to y$ and such that if $x\twoheadrightarrow x^{\prime}\to y$ then $x=x^{\prime}$. Similarly, for $y\in\Er$, a left companion of $y$ is an element $x\in\El$ such that $x\to y$ and such that if $x\to y^{\prime}\hookrightarrow y$ then $y=y^{\prime}$. We say that ${\mathord{\to}}$ is companionable if every $x\in\El$ has a right companion and every $y\in\Er$ has a left companion.

Theorem 7.8** (FTFL [16]).**

A finite poset $L$ is a lattice if and only if it is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for a companionable relation ${\mathord{\to}}$ between finite sets and . In this case, $(\El,\Er,{\mathord{\to}})$ is isomorphic to $(\operatorname{JIrr}(L),\operatorname{MIrr}(L),{\mathord{\to}}_{L})$, where $j\to_{L}m$ if and only if $j\not\leq m$. Writing $\operatorname{Fact}({\mathord{\to}}_{L})=({\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$, we see that ${\mathord{\twoheadrightarrow}}_{L}$ is the partial order induced on $\operatorname{JIrr}(L)$ as a subset of $L$, while ${\mathord{\hookrightarrow}}_{L}$ is the partial order induced on $\operatorname{MIrr}(L)$ as a subset of $L$. The map

[TABLE]

is an isomorphism from $L$ to $\operatorname{Pairs}({\mathord{\to}}_{L})$, with inverse $(X,Y)\mapsto\bigvee X=\bigwedge Y$.

Theorem 7.8 shows that a finite lattice can always be written as $\operatorname{Pairs}({\mathord{\to}})$ by taking $\El=\operatorname{JIrr}(L)$, $\Er=\operatorname{MIrr}(L)$ and $j\to m$ if $j\not\leq m$, whether or not that lattice is semidistributive. What is special in the semidistributive case is, first, that $\operatorname{JIrr}(L)$ and $\operatorname{MIrr}(L)$ can be identified with a single set such that each element is its own left and right companion and, second, that we can recognize those $(\El,\Er,{\mathord{\to}})$ which come from semidistributive lattices by axioms reminiscent of abelian categories. In Section 7.4, we will see a different case in which and can be identified.

The following proof spells out how Theorem 7.8 is a restatement of Markowsky’s result [16].

Proof of Theorem 7.8.

By Proposition 2.27, $\operatorname{JIrr}(L)$ and $\operatorname{MIrr}(L)$ are join- and meet-generating subsets of $L$, so they can play the roles of $X$ and $Y$ in [16, Theorem 5]. By [16, Theorem 5(a)], the given map from $L$ to $\operatorname{Pairs}({\mathord{\to}}_{L})$ is an isomorphism.

Now, let $(\El,\Er,{\mathord{\to}})$ be two sets and a relation such that $L\cong\operatorname{Pairs}({\mathord{\to}})$. By [16, Theorem 9], we have $(\El,\Er,{\mathord{\to}})\cong(\operatorname{JIrr}(L),\operatorname{MIrr}(L),\not\leq)$ if and only if

1. (1)

For all $x\in\operatorname{JIrr}(L)$, if $\Delta\subseteq\operatorname{JIrr}(L)$ is such that $\{y\in\Er:x\to y\}={\{y\in\Er:\exists x^{\prime}\in\Delta\ \mbox{with}\ x^{\prime}\to y\}}$, then $x\in\Delta$ and 2. (2)

For all $y\in\operatorname{JIrr}(L)$, if $\Gamma\subseteq\operatorname{MIrr}(L)$ is such that $\{x\in\El:x\to y\}={\{x\in\El:\exists y^{\prime}\in\Gamma\ \mbox{with}\ x\to y^{\prime}\}}$, then $y\in\Gamma$.

We will show that the first condition is equivalent to saying that every $x\in\El$ has a right companion; analogously, the second condition is is equivalent to saying that every $y\in\Er$ has a left companion.

So, suppose that $x$ has a right companion $y$ and suppose that $\Delta\subseteq\operatorname{JIrr}(L)$ is such that $\{y\in\Er:x\to y\}=\{y\in\Er:\exists x^{\prime}\in\Delta\ \mbox{with}\ x^{\prime}\to y\}$. By the definition of a right companion, $y$ is contained in the left hand side, so there is some $x^{\prime}\in\Delta$ with $x^{\prime}\to y$. If $x\not\twoheadrightarrow x^{\prime}$, there is some $y^{\prime}$ with $x^{\prime}\to y^{\prime}$ and $x\not\to y^{\prime}$, in which case $y^{\prime}$ is in the right hand side but not the left, a contradiction. So $x\twoheadrightarrow x^{\prime}$ and $x^{\prime}\to y$; then the definition of a right companion shows that $x=x^{\prime}$, so we have shown that $x\in\Delta$.

Conversely, suppose that $x$ does not have a right companion. If $\Delta$ is the set ${\{x^{\prime}:x\twoheadrightarrow x^{\prime},\ x\neq x^{\prime}\}}$, then $\{y\in\Er:x\to y\}=\{y\in\Er:\exists x^{\prime}\in\Delta\ \mbox{with}\ x^{\prime}\to y\}$ but $x\not\in\Delta$. ∎

7.4. Extremal lattices

Let $L$ be a finite lattice. Let $x_{0}<x_{1}<\cdots<x_{n}$ be a chain of elements in $L$. Then, for each $1\leq i\leq n$, there must be some join-irreducible element $j$ with $j\leq x_{i}$ and $j\not\leq x_{i-1}$, so $|J(L)|\geq n$. Similarly, $|M(L)|\geq n$. A lattice is called extremal if it has a chain of length $N$ where $|J(L)|=|M(L)|=N$. An acyclic reflexive relation is a reflexive relation that, when interpreted as a directed graph, is acyclic except for $1$-cycles.

We will want the following notation: given a bijection $\mu$ from $\operatorname{JIrr}(L)$ to $\operatorname{MIrr}(L)$, define a relation $\to^{\mu}$ on $\operatorname{JIrr}(L)$ by $i\to^{\mu}j$ if and only if $i\to_{L}\mu(j)$. (Recall that, for $j\in\operatorname{JIrr}(L)$ and $m\in\operatorname{MIrr}(L)$, we write $j\to_{L}m$ if $j\not\leq m$.) The following is a restatement of a characterization of extremal lattices due to Markowsky [17].

Theorem 7.9** (FTFEL).**

A finite poset $L$ is an extremal lattice if and only if it is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$ for an acyclic reflexive relation ${\mathord{\to}}$ on a finite set . In this case, $(\Sha,{\mathord{\to}})$ is unique up to isomorphism. More precisely, in the case where $L$ is an extremal lattice, there is a unique bijection $\mu$ from $\operatorname{JIrr}(L)$ to $\operatorname{MIrr}(L)$ for which $\to^{\mu}$ is a reflexive relation, and $(\Sha,{\mathord{\to}})$ is isomorphic to $(\operatorname{JIrr}(L),\to^{\mu})$.

Proof.

We prove the first two assertions of the theorem by combining [17, Theorem 13] with other results of Markowsky that appear here as Theorem 7.8. In our notation, [17, Theorem 13] says a finite lattice $L$ is extremal if and only if there is a bijection $\mu$ from $\operatorname{JIrr}(L)$ to $\operatorname{MIrr}(L)$ such that ${\mathord{\to}}^{\mu}$ is an acyclic reflexive relation.

Theorem 7.8 says that in any case, $L$ is isomorphic to $\operatorname{Pairs}({\mathord{\to}}_{L})$. If there exists a bijection $\mu$ from $\operatorname{JIrr}(L)$ to $\operatorname{MIrr}(L)$, then there is an isomorphism from $\operatorname{Pairs}({\mathord{\to}}_{L})$ to $\operatorname{Pairs}({\mathord{\to}}^{\mu})$ sending $(X,Y)$ to $(X,\mu^{-1}(Y))$. We conclude that if $L$ is extremal, then there is a bijection $\mu$ from $\operatorname{JIrr}(L)$ to $\operatorname{MIrr}(L)$ such that ${\mathord{\to}}^{\mu}$ is an acyclic reflexive relation, and $L$ is isomorphic to $\operatorname{Pairs}({\mathord{\to}}^{\mu})$.

Conversely, if there exists an acyclic reflexive relation ${\mathord{\to}}$ on a set such that $L$ is isomorphic to $\operatorname{Pairs}({\mathord{\to}})$, then $(\Sha,\Sha,{\mathord{\to}})$ is a companionable relation; each $x\in\Sha$ is its own left companion and its own right companion. Thus Theorem 7.8 (FTFL) says that $(\Sha,\Sha,{\mathord{\to}})$ is isomorphic to $(\operatorname{JIrr}(L),\operatorname{MIrr}(L),{\mathord{\to}}_{L})$. This isomorphism features bijections between and $\operatorname{JIrr}(L)$ and between and $\operatorname{MIrr}(L)$. Composing these bijections, we obtain a bijection $\mu$ from $\operatorname{JIrr}(L)$ to $\operatorname{MIrr}(L)$ such that ${\mathord{\to}}^{\mu}$ coincides with ${\mathord{\to}}$. Again appealing to [17, Theorem 13], we see that $L$ is extremal. Furthermore, we have established the uniqueness of $(\Sha,{\mathord{\to}})$ up to isomorphism.

The final assertion of the theorem is that the bijection $\mu$ is the unique bijection that makes ${\mathord{\to}}^{\mu}$ reflexive. This more precise statement of uniqueness is not in Markowsky’s work, but can be seen more generally as follows: Let $A$ and $B$ be two finite sets with a relation ${\mathord{\to}}$ between them, and let $\mu$ and $\nu$ be bijections between $A$ and $B$. Define a relation ${\mathord{\to}}^{\mu}$ on $A$ with $x\to^{\mu}y$ if and only if $x\to\mu(y)$. If ${\mathord{\to}}^{\mu}$ is reflexive and ${\mathord{\to}}^{\nu}$ is an acyclic reflexive relation, then we claim that $\mu=\nu$.

Suppose to the contrary that $\mu\neq\nu$. Then we can find some $r\geq 2$ and some $a_{1}$, $a_{2}$, …, $a_{r}\in A$ with $\mu(a_{j})=\nu(a_{j+1})$, with indices periodic modulo $r$. Since $\mu$ is reflexive, we have $a_{j}\to\mu(a_{j})$, and thus we have $a_{1}\to^{\nu}a_{2}\to^{\nu}a_{3}\to^{\nu}\cdots\to^{\nu}a_{r}\to^{\nu}a_{1}$, contradicting that ${\mathord{\to}}^{\nu}$ was assumed to be an acyclic reflexive relation. ∎

If $L$ is a finite extremal lattice, then the relation $(\Sha,{\mathord{\to}})$ of Theorem 7.9 is obtained from the relation $(\operatorname{JIrr}(L),\operatorname{MIrr}(L),{\mathord{\to}}_{L})$ of Theorem 7.8 by identifying $\operatorname{JIrr}(L)$ with $\operatorname{MIrr}(L)$ by the bijection $\mu$. If, instead, $L$ is a semidistributive lattice then $(\Sha,{\mathord{\to}})$ is obtained analogously from $(\operatorname{JIrr}(L),\operatorname{MIrr}(L),{\mathord{\to}}_{L})$ by identifying $\operatorname{JIrr}(L)$ and $\operatorname{MIrr}(L)$ by the bijection $\kappa$. In this latter case, $(\Sha,{\mathord{\to}})$ need not be acyclic.

The condition of being extremal neither implies nor is implied by being semidistributive: We present examples of lattices that satisfy each condition without satisfying the other, followed by two propositions that describe how the conditions can be satisfied simultaneously. In the examples, if any sort of arrow is drawn between two vertices, we mean that the ${\mathord{\to}}$ relation holds between them, and we have decorated this arrow to indicate whether the ${\mathord{\twoheadrightarrow}}$ or ${\mathord{\hookrightarrow}}$ relation also holds where $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}}):=\operatorname{Fact}({\mathord{\to}})$. Conveniently, if $x\twoheadrightarrow y$ or $x\hookrightarrow y$, then $x\to y$, so we can always draw diagrams in this way. All relations are assumed to be reflexive, although this is not indicated pictorially.

Example 7.10**.**

The quadruple $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ shown on the left of Figure 3 obeys the axioms of a two-acyclic factorization system, but ${\mathord{\to}}$ has $3$-cycles. Thus the corresponding lattice of closed sets, shown on the right of Figure 3, is semidistributive but not extremal.

Example 7.11**.**

The quadruple $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ shown on the left of Figure 4 is an acyclic reflexive relation. However, the middle ${\mathord{\to}}$-arrow cannot be factored into an ${\mathord{\twoheadrightarrow}}$-arrow and an ${\mathord{\hookrightarrow}}$-arrow, so the corresponding lattice of closed sets, also pictured, is extremal without being semidistributive.

Proposition 7.12**.**

Let ${\mathord{\to}}$ be an acyclic reflexive relation on a finite set and let $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})=\operatorname{Fact}({\mathord{\to}})$. Then $\operatorname{Pairs}({\mathord{\to}})$ is semidistributive if and only if ${\mathord{\to}}=\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, in which case $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system.

Proof.

Write $L$ for $\operatorname{Pairs}({\mathord{\to}})$. By Theorem 7.9, we can take to be $\operatorname{JIrr}(L)$ and ${\mathord{\to}}$ to be derived from the bijection $\mu$ in Theorem 7.9.

Suppose $L$ is semidistributive so that, by Theorem 1.2, $L$ is isomorphic to $\operatorname{Pairs}({\mathord{\to}}_{L})$ for the finite two-acyclic factorization system $(\operatorname{JIrr}(L),{\mathord{\to}}_{L},{\mathord{\twoheadrightarrow}}_{L},{\mathord{\hookrightarrow}}_{L})$. Since in particular $i\not\leq\kappa(i)$ for all $i\in\operatorname{JIrr}(L)$, we see that $\kappa$ is the bijection $\mu$ from Theorem 7.9. Thus ${\mathord{\to}}={\mathord{\to}}^{\kappa}={\mathord{\to}}_{L}$ has the desired properties.

Conversely, suppose that ${\mathord{\to}}=\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. By definition, $({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})=\operatorname{Fact}({\mathord{\to}})$, so $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a factorization system. Since ${\mathord{\twoheadrightarrow}}\subseteq{\mathord{\to}}$ and ${\mathord{\hookrightarrow}}\subseteq{\mathord{\to}}$, the two-acyclicity of $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ follows from the acyclicity of ${\mathord{\to}}$. ∎

Proposition 7.13**.**

Let $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ be a finite two-acyclic factorization system. Then $\operatorname{Pairs}({\mathord{\to}})$ is extremal if and only if ${\mathord{\to}}$ is an acyclic reflexive relation.

Proof.

By Theorem 7.9, if ${\mathord{\to}}$ is an acyclic reflexive relation, then $\operatorname{Pairs}({\mathord{\to}})$ is extremal. Conversely, if $\operatorname{Pairs}({\mathord{\to}})$ is extremal, then it is isomorphic to $\operatorname{Pairs}({\mathord{\to}}^{\prime})$ for an acyclic reflexive relation ${\mathord{\to}}^{\prime}$ on a set ′. By Proposition 7.12, $({}^{\prime},{\mathord{\to}}^{\prime},\operatorname{Mult}({\mathord{\to}}^{\prime}))$ is a two-acyclic factorization system. By Theorem 1.2, $({}^{\prime},{\mathord{\to}}^{\prime},\operatorname{Mult}({\mathord{\to}}^{\prime}))$ and $(\Sha,{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ are isomorphic, so ${\mathord{\to}}$ is an acyclic reflexive relation. ∎

It was shown in [26] that if a lattice is extremal and semidistributive, then it is also left modular (and therefore trim, since by definition a lattice is trim if it is extremal and left modular). A central topic in [26] is the representation of lattices which are trim but not necessarily semidistributive as maximal orthogonal pairs for a suitable relation.

8. Motivating examples

In this section, we connect the FTFSDL and its generalizations to the two main examples that motivated it, namely posets of regions of hyperplane arrangements and lattices of torsion classes of finite-dimensional algebras. Posets of regions, and their quotients, have motivated much of the interest in the combinatorial study of semidistributivity, while lattices of torsion classes provided clues leading to the definition of a two-acyclic factorization system.

Our discussion here does two things: In Section 8.1, we apply the FTFSDL and a known characterization of semidistributivity of the poset of regions to construct two-acyclic factorization systems for a class of posets of regions that includes the simplicial case. In Section 8.2, given a finite-dimensional algebra $A$, we construct a two-acyclic factorization system whose lattice of maximal orthogonal pairs is isomorphic to the lattice $\operatorname{tors}(A)$ of torsion classes of $A$. As a consequence, $\operatorname{tors}(A)$ is a well separated $\kappa$-lattice. When $\operatorname{tors}(A)$ is finite, this recovers the finite case of [10, Theorem 1.3], namely that $\operatorname{tors}(A)$ is semidistributive. We point out that [10, Theorem 1.3] states that $\operatorname{tors}(A)$ is completely semidistributive even when it is infinite. We do not have a combinatorial hypothesis which would let us prove this result, since a well separated $\kappa$-lattice need not be semidistributive (see Example 3.20).

8.1. Posets of regions

Background on posets of regions can be found in [24]. Here we give the basic definitions. A (real, central) hyperplane arrangement $\mathcal{A}$ is a collection of linear hyperplanes in $\mathbb{R}^{n}$. We assume throughout the adjectives real and central, but will not repeat them. The regions of $\mathcal{A}$ are the closures of the connected components of the complement $\mathbb{R}^{n}\setminus(\bigcup_{H\in\mathcal{A}}H)$. Fixing one region $B$ to be the base region, each region $R$ is specified by its separating set $S(R)$, the set of hyperplanes in $\mathcal{A}$ that separate $R$ from $B$. The poset of regions $\operatorname{Pos}(\mathcal{A},B)$ of $\mathcal{A}$ with respect to $B$ is the set of regions, partially ordered by containment of their separating sets.

The regions of a finite hyperplane arrangement are full-dimensional polyhedral cones. A facet $F$ of $R$ is a lower facet of $R$ if the hyperplane defining it is in the separating set of $R$. Otherwise, $F$ is an upper facet. A region $R$ of $\mathcal{A}$ is tight with respect to $B$ if for every pair of lower facets of $R$, the intersection of the two facets is a codimension-$2$ face, and if for every pair of upper facets of $R$, the intersection of the two facets is a codimension-$2$ face. The arrangement $\mathcal{A}$ is tight with respect to $B$ if each of its regions is tight with respect to $B$. The arrangement is simplicial if each of its maximal cones has exactly $n$ facets. It is immediate that a simplicial arrangement is tight with respect to any choice of $B$.

The following is [24, Theorem 9-3.8] combined with [24, Corollary 9-3.9].

Theorem 8.1**.**

The poset of regions $\operatorname{Pos}(\mathcal{A},B)$ is a semidistributive lattice if and only if $\mathcal{A}$ is tight with respect to $B$. If particular, if $\mathcal{A}$ is simplicial, $\operatorname{Pos}(\mathcal{A},B)$ is a semidistributive lattice for any choice of $B$.

A rank-two subarrangement of $\mathcal{A}$ is a subset $\mathcal{A}^{\prime}$ of $\mathcal{A}$ with $|\mathcal{A}^{\prime}|\geq 2$ such that there exists a codimension-$2$ subspace $U$ having $\mathcal{A}^{\prime}=\{H\in\mathcal{A}:U\subseteq H\}$. Given a rank-two subarrangement $\mathcal{A}^{\prime}$ there is a unique $\mathcal{A}^{\prime}$-region $B^{\prime}$ containing the $\mathcal{A}$-region $B$. The basic hyperplanes of $\mathcal{A}^{\prime}$ are the two hyperplanes that define the facets of $B^{\prime}$.

Given distinct hyperplanes $H$ and $H^{\prime}$, there is a unique rank-two subarrangement $\mathcal{A}^{\prime}$ containing $H$ and $H^{\prime}$. We say $H^{\prime}$ cuts $H$ if $H^{\prime}$ is basic in $\mathcal{A}^{\prime}$ and $H$ is not basic in $\mathcal{A}^{\prime}$. For each hyperplane $H\in\mathcal{A}$, consider the subset $H\setminus(\bigcup_{H^{\prime}}H^{\prime}\cap H)$, where the union is taken over all $H^{\prime}\in\mathcal{A}$ such that $H^{\prime}$ cuts $H$. The closures of the connected components of this subset are called the shards in $H$. The set of shards of $\mathcal{A}$ is the union of the sets of shards of all of the hyperplanes of $\mathcal{A}$. The decomposition of the hyperplanes of $\mathcal{A}$ into shards depends strongly on the choice of $B$, and to emphasize that, we sometimes refer to shards of $\mathcal{A}$ with respect to $B$.

Given a shard $\Sigma$ of $\mathcal{A}$ contained in a hyperplane $H$, an upper region of $\Sigma$ is a region $R$ whose intersection with $\Sigma$ is $(n-1)$-dimensional and which has $H\in S(R)$. A lower region of $\Sigma$ is a region $R$ having an $(n-1)$-dimensional intersection with $\Sigma$, with $H\not\in S(R)$. [24, Proposition 9-7.8] asserts that, when $\mathcal{A}$ is tight with respect to $B$, the collection of upper regions of $\Sigma$ contains a unique minimal region (in the sense of $\operatorname{Pos}(\mathcal{A},B)$). Furthermore, this region is join-irreducible in $\operatorname{Pos}(\mathcal{A},B)$. We write $J(\Sigma)$ for this element. Every join-irreducible element is $J(\Sigma)$ for a unique shard $\Sigma$. Dually, there is a unique maximal region $M(\Sigma)$ among lower regions of $\Sigma$, this region is meet-irreducible in $\operatorname{Pos}(\mathcal{A},B)$, and every meet-irreducible element arises in this way.

Let $\Sha(\mathcal{A},B)$ stand for the set of shards of $\mathcal{A}$ with respect to $B$. Define relations ${\mathord{\to}}$, ${\mathord{\twoheadrightarrow}}$, and ${\mathord{\hookrightarrow}}$ on $\Sha(\mathcal{A},B)$ as follows. For $\Sigma,\Sigma^{\prime}\in\Sha(\mathcal{A},B)$, set $\Sigma\to\Sigma^{\prime}$ if and only if $J(\Sigma)\not\leq M(\Sigma^{\prime})$, set $\Sigma\twoheadrightarrow\Sigma^{\prime}$ if and only if $J(\Sigma)\geq J(\Sigma^{\prime})$, and set $\Sigma\hookrightarrow\Sigma^{\prime}$ if and only if $M(\Sigma)\geq M(\Sigma^{\prime})$.

Theorem 8.2**.**

If $\mathcal{A}$ is tight with respect to $B$, then $(\Sha(\mathcal{A},B),{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system. The map

[TABLE]

is an isomorphism from $\operatorname{Pos}(\mathcal{A},B)$ to $\operatorname{Pairs}({\mathord{\to}})$, with inverse

[TABLE]

Proof.

Theorem 8.1 says that $\operatorname{Pos}(\mathcal{A},B)$ is semidistributive, and thus is a $\kappa$-lattice by Theorem 2.28. Given a join-irreducible element $J(\Sigma)$ of $\operatorname{Pos}(\mathcal{A},B)$, write $J_{*}(\Sigma)$ for the unique element covered by $J(\Sigma)$. The unique element $M^{*}(\Sigma)$ covering $M(\Sigma)$ is an upper region of $\Sigma$, so $J(\Sigma)\leq M^{*}(\Sigma)$. Thus $M(\Sigma)$ is a maximal element of the set ${\{R\in\operatorname{Pos}(\mathcal{A},B):J(\Sigma)\wedge R=J_{*}(\Sigma)\}}$. We conclude that $\kappa(J(\Sigma))=M(\Sigma)$. Now Theorem 1.2 (FTFSDL) says that $(\Sha(\mathcal{A},B),{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$ is a two-acyclic factorization system and gives the desired isomorphism. ∎

8.2. Representation theory of finite-dimensional algebras

Let $k$ be a field and let $A$ be a finite-dimensional algebra over $k$. A full subcategory of the category of finite-dimensional $A$-modules is called a torsion class if it is closed under extensions and quotients. Because the intersection of an arbitrary collection of torsion classes is a torsion class, the torsion classes form a complete lattice $\operatorname{tors}(A)$. This lattice is known to be completely semidistributive [10, Theorem 3.1(a)]. To each torsion class $\mathcal{T}$ there is an associated torsion-free class $\mathcal{F}={\{M:\operatorname{Hom}(L,M)=0\textrm{ for all }L\in\mathcal{T}\}}$. Dually, the torsion class corresponding to a torsion-free class $\mathcal{F}$ can be obtained by $\mathcal{T}=\{L:\operatorname{Hom}(L,M)=0\textrm{ for all }M\in\mathcal{F}\}$. Inclusion of torsion classes corresponds to reverse inclusion of torsion-free classes. Given such a torsion pair $(\mathcal{T},\mathcal{F})$, for any module $M$, there is a maximal submodule of $M$ which lies in $\mathcal{T}$. This is denoted $t_{\mathcal{T}}M$, or $tM$ if the intended torsion class is clear. The quotient $M/t_{\mathcal{T}}M$ is the maximal quotient of $M$ which is contained in $\mathcal{F}$. A good reference for basic facts about torsion classes is [3, Section VI.1].

We now explain how to construct a set and relations ${\mathord{\twoheadrightarrow}}$, ${\mathord{\hookrightarrow}}$, ${\mathord{\to}}$ in terms of the representation theory of $A$, so that we recover the lattice of torsion classes of $A$ as $\operatorname{Pairs}({\mathord{\to}})$. This construction was one of the inspirations for this project.

An $A$-module is called a brick if its endomorphism ring is a division algebra (i.e., if all its non-zero endomorphisms are invertible). We write $\operatorname{Bricks}(A)$ for the collection of isomorphism classes of bricks.

For $[L]$, $[M]\in\operatorname{Bricks}(A)$, we define $[L]\to[M]$ iff $\operatorname{Hom}(L,M)\neq 0$. We define $[L]\twoheadrightarrow[M]$ iff $M$ is filtered by quotients of $L$ and define $[L]\hookrightarrow[M]$ iff $L$ is filtered by submodules of $M$. Note that a surjective morphism of modules from $L$ to $M$ implies $[L]\twoheadrightarrow[M]$ but the converse does not hold, and similarly for injective maps and ${\mathord{\hookrightarrow}}$.

Example 8.3**.**

Let $A$ be the path algebra of the Kronecker quiver with two arrows $a$ and $b$. We write a representation $V$ of $A$ as $V_{1}\rightrightarrows V_{2}$. Let $M$ be the representation $k\rightrightarrows k$ where $a=b=1$ and let $N$ be the representation $k^{2}\rightrightarrows k$ where $a=\left[\begin{smallmatrix}1\\ 0\end{smallmatrix}\right]$ and $b=\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right]$. We clearly cannot have a surjection $M\to N$ as $N$ has larger dimension. We cannot even have a surjection $M^{\oplus r}\to N$, as every map $M\to N$ has image lying in $(k\left[\begin{smallmatrix}1\\ 1\end{smallmatrix}\right]\rightrightarrows k)\ \subset N$. Note that this submodule of $N$ is isomorphic to $M$, and that the quotient of $N$ by this submodule is the simple module $k\rightrightarrows 0$, which is a quotient of $M$. Thus, $N$ is filtered by quotients of $M$ and $[M]\twoheadrightarrow[N]$.

We will need one useful fact about bricks. This is a special case of [2, Lemma 1.7(1)]. We give the short proof from [2].

Lemma 8.4**.**

If $S$ is a brick, and $M$ is in the torsion class consisting of modules filtered by quotients of $S$, then any non-zero morphism from $M$ to $S$ is surjective.

Proof.

There is a filtration $0<M_{1}<\dots<M_{r}=M$ with each $M_{i}/M_{i-1}$ a quotient of $S$. Let $i$ be maximal such that $f|_{M_{i}}=0$. Then $f$ descends to a map from $M_{i+1}/M_{i}$ to $S$. Composing this with the quotient map from $S$ to $M_{i+1}/M_{i}$, we obtain a non-zero endomorphism of $S$. Since $S$ is a brick, this endomorphism must be invertible, so the map from $M_{i+1}/M_{i}$ to $S$ must be surjective, and thus the map from $M$ to $S$ must be surjective.∎

The dual statement also holds for torsion-free classes.

We also need the following simple lemma which says that a torsion class is determined by the bricks it contains.

Lemma 8.5**.**

Let $\mathcal{T}$ be a torsion class. Then every module in $\mathcal{T}$ is filtered by bricks in $\mathcal{T}$.

We can thus recover $\mathcal{T}$ from the set of bricks in $\mathcal{T}$.

Proof.

Let $M$ be a minimal-dimensional counter-example. Then $M$ is not a brick, so it has some non-invertible non-zero endomorphism. Let $I$ be the image of this endomorphism. It is both a submodule of $M$ and a quotient of $M$. Since it is a quotient of $M$, $I\in\mathcal{T}$. Also, $M/I\in\mathcal{T}$. By the hypothesis that $M$ was the minimal counterexample, both $I$ and $M/I$ are filtered by bricks in $\mathcal{T}$. But then so is $M$. ∎

Theorem 8.6**.**

$(\operatorname{Bricks}(A),{\mathord{\to}},{\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$* is a two-acyclic factorization system. Furthermore, $(\mathcal{T},\mathcal{F})\mapsto(\mathcal{T}\cap\operatorname{Bricks}(A),\mathcal{F}\cap\operatorname{Bricks}(A))$ is an isomorphism from $\operatorname{tors}(A)$ to $\operatorname{Pairs}({\mathord{\to}})$. The inverse map takes $(X,Y)$ to $(\mathcal{T},\mathcal{F})$, where $\mathcal{T}$ consists of the modules filtered by quotients of bricks from $X$, and $\mathcal{F}$ consists of the modules filtered by submodules of modules from $Y$.*

Proof.

We first prove that $\operatorname{Fact}({\mathord{\to}})=({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$. Write $({\mathord{\twoheadrightarrow}}^{\prime},{\mathord{\hookrightarrow}}^{\prime})$ for $\operatorname{Fact}({\mathord{\to}})$.

Suppose $[L]\twoheadrightarrow[M]$. Specifically, suppose that $M$ has a filtration $0=M_{0}\subset M_{1}\subset\cdots\subset M_{r}=M$ with surjections from $L$ to $M_{j}/M_{j-1}$. If $[N]$ in $\operatorname{Bricks}(A)$ has $[M]\to[N]$, let $\psi:M\to N$ be a nonzero map. Let $j$ be minimal such that $\psi$ does not restrict to zero on $M_{j}$, so $\psi$ descends to a nonzero map from $M_{j}/M_{j-1}$ to $N$. The composition $L\to M_{j}/M_{j-1}\to N$ is then nonzero, so $[L]\to[N]$. We see that $[L]\twoheadrightarrow^{\prime}[M]$.

Conversely, suppose that $[L]\twoheadrightarrow^{\prime}[M]$. We will show that $M$ is filtered by quotients of $L$. Let $\mathcal{T}$ be the torsion class consisting of modules filtered by quotients of $L$. Let $tM$ be the torsion part of $M$ with respect to $\mathcal{T}$. If $tM=M$, then we are done, since this means that $M\in\mathcal{T}$, and thus $M$ is filtered by quotients of $L$. Otherwise, consider $M/tM$. It is in the torsion-free class corresponding to $\mathcal{T}$, so it admits no morphism from any module in $\mathcal{T}$, and in particular, it admits no morphism from $L$. If $M/tM$ were a brick, we would have found a contradiction, because we would have $[M]\to[M/tM]$ but $[L]\not\to[M/tM]$, contradicting our assumption that $[L]\twoheadrightarrow[M]$. In general, though, we can take a module $N$ which is minimal-dimensional among modules which are both quotients and submodules of $M/tM$. This $N$ must be a brick, because if it has a non-invertible endomorphism, its image would be a smaller quotient and submodule of $M/tM$. Now $\operatorname{Hom}(L,N)=0$, but $\operatorname{Hom}(M,N)\neq 0$, contrary to our assumption that $[L]\twoheadrightarrow^{\prime}[M]$. We conclude that $[L]\twoheadrightarrow[M]$.

The argument that $[L]\hookrightarrow[M]$ if and only if $[L]\hookrightarrow^{\prime}[M]$ is dual, and we see that $\operatorname{Fact}({\mathord{\to}})=({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$.

We next prove that $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})={\mathord{\to}}$. Since $\operatorname{Fact}({\mathord{\to}})=({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})$, Proposition 2.3.3 says that $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})\subseteq{\mathord{\to}}$. Conversely, if $\operatorname{Hom}(L,M)\neq 0$, we can define $N$ to be the image of such a map. This $N$ need not be a brick, but there is a minimal-dimensional module $N^{\prime}$ which is both a submodule and a quotient module of $N$. Then $[L]\twoheadrightarrow[N^{\prime}]\hookrightarrow[M]$. We have showed that $\operatorname{Mult}({\mathord{\twoheadrightarrow}},{\mathord{\hookrightarrow}})={\mathord{\to}}$.

Suppose that we have two non-isomorphic bricks $S$ and $T$ with $[S]\twoheadrightarrow[T]$. We will show that $\operatorname{Hom}(T,S)=0$. Assume otherwise. Lemma 8.4 says that any non-zero map from $T$ to $S$ must be surjective. But there is also a non-zero map from $S$ to $T$, and composing these we get a noninvertible endomorphism of $T$, contrary to the hypothesis that $T$ is a brick.

It follows that $S$ is not in the torsion class consisting of modules filtered by quotients of $T$, so we do not have $[T]\twoheadrightarrow[S]$. This establishes the order condition for ${\mathord{\twoheadrightarrow}}$. It also follows that $T$ is not in the torsion-free class of modules filtered by submodules of $S$, so we do not have $[T]\hookrightarrow[S]$. This establishes the brick condition. The proof of the order condition for ${\mathord{\hookrightarrow}}$ is dual to the order condition for ${\mathord{\twoheadrightarrow}}$.

Finally, we verify the isomorphism. Let $(\mathcal{T},\mathcal{F})$ be a torsion pair for $\mathrm{mod}A$. Define $X=\operatorname{Bricks}(A)\cap\mathcal{T}$ and $Y=\operatorname{Bricks}(A)\cap\mathcal{F}$. As recalled above, $\mathcal{T}$ consists of those modules which have no non-zero morphisms to any module in $\mathcal{F}$. Since by the dual of Lemma 8.5 any module in $\mathcal{F}$ is filtered by submoduless of bricks in $\mathcal{F}$, we can also describe $\mathcal{T}$ as consisting of those modules which have no non-zero morphisms to any module in $Y$. We therefore have that $X={}^{\perp}Y$. Dually, $Y=X^{\perp}$.

Conversely, suppose $(X,Y)\in\operatorname{Pairs}({\mathord{\to}})$. Define $\mathcal{T}$ to consist of modules filtered by quotients of modules from $X$, and $\mathcal{F}$ to consist of modules filtered by submodules of modules from $Y$. Clearly, there are no morphisms from any module in $\mathcal{T}$ to any module in $\mathcal{F}$. We now prove by induction on the dimension of $M$ that if $M$ is in $\mathcal{T}^{\perp}$, then $M\in\mathcal{F}$. Let $B$ be minimal dimensional among modules which are both submodules and quotient modules of $M$. $B$ is necessarily a brick. Since $B$ is a submodule of $M$, we must have $B\in\mathcal{T}^{\perp}$, so $B\in Y\subset\mathcal{F}$. Let $K$ be the kernel of a surjection from $M$ to $B$. Now $K\in\mathcal{T}^{\perp}$, so by the induction hypothesis, $K\in\mathcal{F}$. So $M$, which is the extension of $B$ and $K$, is also in $\mathcal{F}$. The dual argument shows that ${}^{\perp}\mathcal{F}=\mathcal{T}$. ∎

Combining Theorem 8.6 with Theorem 1.3 or with Theorem 1.2, we obtain the following corollaries.

Corollary 8.7**.**

The poset $\operatorname{tors}(A)$ is a well separated $\kappa$-lattice.

Corollary 8.8**.**

If $\operatorname{tors}(A)$ is finite, then it is a semidistributive lattice.

Corollary 8.8 recovers [12, Theorem 4.5]. This is a special case of [10, Theorem 1.3], which says that $\operatorname{tors}(A)$ is completely semidistributive, without the hypothesis that $\operatorname{tors}(A)$ is finite. However, Corollary 8.7 accomplishes something different from [10, Theorem 1.3]: Recall that Examples 3.19 and 3.20 show that there are no implications between being a well separated $\kappa$-lattice and being a completely semidistributive lattice.

Acknowledgements

H.T. would like to thank Laurent Demonet and Osamu Iyama for their hospitality at Nagoya University and helpful conversations.