A new tool to study the fixed point property of finite posets

TL;DR

This paper introduces a new topological tool for analyzing the fixed point property in finite posets, using a construction that induces an endofunctor in the homotopy category of finite T0-spaces.

Contribution

It presents a novel topological construction that provides a new method to study fixed point properties of finite posets, with applications demonstrated through examples.

Findings

The construction induces an endofunctor in the homotopy category of finite T0-spaces.

The tool helps analyze fixed point properties using topological methods.

Several applications illustrate the effectiveness of the new construction.

Abstract

We develop a novel tool to study the fixed point property of finite posets using a topological approach. Our tool is a construction which turns out to induce an endofunctor of the homotopy category of finite $T_0$--spaces. We study many properties of this construction and give several examples of application.

Equations

$$U_A^X=\bigcap _{a\in A}{U}_a^X\text{and}{F}_A^X=\bigcap _{a\in A}{F}_a^X$$

$${\mathcal{U}}_0(X)$$

$${\mathcal{F}}_0(X)$$

$$\mathcal{U}(X)$$

$$\mathcal{F}(X)$$

$$\text{mxl}([)\mathcal{C}(X)]=\{U_a\mid a\in \text{mxl}([)X]\}\text{and}\text{mnl}([)\mathcal{C}(X)]=\{F_a\mid a\in \text{mnl}([)X]\}.$$

$$F_A^X=\bigcap _{a\in A}{F}_a=\bigcap _{a\in A}{\left(U_a^{X^{\text{op}}}\right)}^{\text{op}}={\left(\bigcap _{a\in A}{U}_a^{X^{\text{op}}}\right)}^{\text{op}}={\left(U_A^{X^{\text{op}}}\right)}^{\text{op}}.$$

$$B^{♯}=\{a\in \text{mxl}([)X]\mid B\subseteq U_a\}\text{and}{B}^{♭}=\{a\in \text{mnl}([)X]\mid B\subseteq F_a\}.$$

$${\mathbf{C}}_{\mathcal{U}}(x)=\min \{C\in \mathcal{U}(X)\mid x\in C\}$$

$${\mathbf{C}}_{\mathcal{F}}(x)=\max \{C\in \mathcal{F}(X)\mid x\in C\}.$$

$$\mathcal{U}(f):\mathcal{U}(X)\to \mathcal{U}(Y)\text{by}\mathcal{U}(f)(C)=\min \{D\in \mathcal{U}(Y)\mid f(C)\subseteq D\},$$

$$\mathcal{F}(f):\mathcal{F}(X)\to \mathcal{F}(Y)\text{by}\mathcal{F}(f)(C)=\max \{D\in \mathcal{F}(Y)\mid f(C)\subseteq D\}.$$

$$\mathcal{C}(f):\mathcal{C}(X)\to \mathcal{C}(Y)\text{by}\mathcal{C}(f)(C)=\{\begin{array}{ll}\mathcal{U}(f)(C)& \text{if }C\in \mathcal{U}(X),\\ \mathcal{F}(f)(C)& \text{if }C\in \mathcal{F}(X).\end{array}$$

$$\mathcal{U}(f)\mathcal{U}(g)(\{2\})=\mathcal{U}(f)(\{0,1\})=U_3\ne \{2\}=\mathcal{U}(fg)(\{2\}).$$

$$fg(C)\subseteq f(\mathcal{U}(g)(C))\subseteq \mathcal{U}(f)(\mathcal{U}(g)(C)).$$

$$\bullet$$

$$f({\hat{U}}_{f(x)})\subseteq \{0,1,2,3,4,5\}\subseteq f(U_a\cup U_b)\subseteq U_{f(x)}.$$

$$\bullet$$

$$U_{a_1}$$

$$U_{a_2}$$

$$U_{a_3}$$

$$U_{b_1}$$

$$U_{b_j}$$

$$U_{b_n}$$

$$U_{c_j}$$

$$A$$

$$B$$

$$C$$

$$D$$

$$E$$

$$F$$

$$S$$

$$\#A=2n-3>n-k+2\ge \#B$$

$$D$$

$$\mathcal{U}(i)\mathcal{U}(r)(C)=\mathcal{U}(i)(r(C))=\mathcal{U}(ir)(C).$$

$$\mathcal{U}(i)\mathcal{U}(r)(C)=\mathcal{U}(i)(r(C))=\mathcal{U}(ir)(C)\le \mathcal{U}({\text{id}}_X)(C)=C$$

$$\text{mxl}([)\mathcal{U}(X)]\cap F_D^{\mathcal{U}(X)}=\text{mxl}([)\mathcal{U}(X)]\cap {\hat{F}}_C^{\mathcal{U}(X)}=\text{mxl}([)\mathcal{U}(X)]\cap F_C^{\mathcal{U}(X)}$$

Full text

Research partially supported by grant M044 (2016–2018) of SeCTyP, UNCuyo.

1. Introduction

In this article we develop a novel tool to study the fixed point property of finite $T_{0}$–spaces (or equivalently finite posets) which consists in analysing a new finite $T_{0}$–space which is constructed from the one whose fixed point property we want to study. More specifically, given a finite $T_{0}$–space $X$ we construct a new finite $T_{0}$–space $\mathscr{C}(X)$ whose elements are connected subspaces of $X$ and such that each continuous map $f\colon X\to X$ induces a continuous map $\mathscr{C}(f)\colon\mathscr{C}(X)\to\mathscr{C}(X)$. This assignment is not functorial but defines an endofunctor of the homotopy category of finite $T_{0}$–spaces.

We show with several examples that this tool is useful to study the fixed point property of finite $T_{0}$–spaces and we develop many properties of this construction so as to gain a better understanding of it. Among them, we prove that our construction preserves homotopy type and we study how it changes if we remove beat points of the given space.

2. Preliminaries

Finite topological spaces

Let $X$ be a finite topological space. For each $x\in X$ the minimal open set that contains $x$ is defined as the intersection of all the open subsets of $X$ that contain $x$ and is denoted by $U_{x}$ (or by $U_{x}^{X}$ if we need to make explicit the topological space in which it is considered). We denote by $F_{x}$ the intersection of all closed sets that contain $x$. We also denote $\widehat{U}_{x}=U_{x}-\{x\}$ and $\widehat{F}_{x}=F_{x}-\{x\}$. The set $\{U_{x}\mid x\in X\}$ is a basis for the topology of $X$. We can define a preorder in $X$ by $x_{1}\leq x_{2}$ if and only if $U_{x_{1}}\subseteq U_{x_{2}}$. Moreover, this defines a bijective correspondence between topologies and preorders in a given finite set, under which $T_{0}$–topologies correspond to partial orders [1]. This correspondence allows us to consider a finite $T_{0}$–space as a finite poset, and a finite poset as a finite $T_{0}$–space. In addition, a map between finite $T_{0}$–spaces is continuous if and only if it is order-preserving.

Let $X$ be a finite topological space with associated preorder $\leq$. We define $X^{\textup{op}}$ as the finite space with underlying set $X$ and associated preorder given by $x\leq^{\textup{op}}y$ if and only if $y\leq x$ for all $x,y\in X$.

Let $X$ and $Y$ be finite $T_{0}$–spaces and let $f,g\colon X\to Y$ be continuous maps. Then $f$ and $g$ are homotopic (denoted by $f\simeq g$) if and only if there exist $n\in\mathbb{N}$ and continuous maps $f_{0},f_{1},\ldots,f_{n}\colon X\to Y$ such that $f=f_{0}\leq f_{1}\geq\ldots\leq f_{n}=g$ [2, Corollary 1.2.6].

As a consequence, if $f$ and $g$ are such that for each $x\in X$, $f(x)$ and $g(x)$ are comparable, then $f\simeq g$ (since both $f$ and $g$ are comparable with the continuous map $h=\max\{f,g\}$).

Let $X$ be a finite $T_{0}$–space and let $x\in X$. We say that $x$ is a down beat point of $X$ if $\widehat{U}_{x}$ has a maximum element. We say that $x$ is an up beat point of $X$ is $\widehat{F}_{x}$ has a minimum element. If $x$ is a down beat point or an up beat point, we say that $x$ is a beat point of $X$. It is well known that if $X$ is a finite $T_{0}$–space and $x_{0}\in X$ is a beat point, then $X-\{x_{0}\}$ is a strong deformation retract of $X$ [2, Proposition 1.3.4].

bp–retracts

We recall the following definition from [3].

Definition 2.1** ([3, Definition 3.1]).**

Let $X$ be a finite $T_{0}$–space and let $A\subseteq X$. We say that $A$ is a dbp–retract (resp. ubp–retract) of $X$ if $A$ can be obtained from $X$ by successively removing down beat points (resp. up beat points), that is, if there exist $n\in\mathbb{N}_{0}$ and a sequence $X=X_{0}\supseteq X_{1}\supseteq\cdots\supseteq X_{n}=A$ of subspaces of $X$ such that, for all $i\in\{1,\ldots,n\}$, the space $X_{i}$ is obtained from $X_{i-1}$ by removing a single down beat point (resp. up beat point) of $X_{i-1}$.

Clearly, if $X$ is a finite $T_{0}$–space and $A$ is a dbp–retract of $X$ then the minimal elements of $X$ belong to $A$ since they can not be down beat points of any subspace of $X$. Similarly, if $X$ is a finite $T_{0}$–space and $A$ is an ubp–retract of $X$ then the maximal elements of $X$ belong to $A$.

Observe also that if $X$ is a finite $T_{0}$–space and $A\subseteq X$ then $A$ is an ubp–retract of $X$ if and only if $A^{\textup{op}}$ is a dbp–retract of $X^{\textup{op}}$.

The following results of [3] give characterizations of dbp–retracts which will be needed in this work. Clearly, similar results hold for ubp–retracts.

Theorem 2.2** ([3, Theorem 3.5]).**

Let $X$ be a finite $T_{0}$–space, let $A$ be a subspace of $X$ and let $i\colon A\to X$ be the inclusion map. Then, the following propositions are equivalent:

1. (1)

$A$* is a dbp–retract of $X$.* 2. (2)

There exists a continuous function $f\colon X\to X$ such that $f\leq\textup{id}_{X}$, $f^{2}=f$ and $f(X)=A$. 3. (3)

There exists a unique continuous function $f\colon X\to X$ such that $f\leq\textup{id}_{X}$, $f^{2}=f$ and $f(X)=A$. 4. (4)

There exists a retraction $r\colon X\to A$ of $i$ such that $ir\leq\textup{id}_{X}$. 5. (5)

There exists a unique retraction $r\colon X\to A$ of $i$ such that $ir\leq\textup{id}_{X}$.

Theorem 2.3** ([3, Theorem 3.7]).**

Let $X$ be a finite $T_{0}$–space and let $A$ be a subspace of $X$. Then, $A$ is a dbp–retract of $X$ if and only if $U^{X}_{x}\cap A$ has a maximum for every $x\in X$. Equivalently, $A$ is a dbp–retract of $X$ if and only if $U^{X}_{x}\cap A$ has a maximum for every $x\in X-A$.

From the proof of theorem 3.7 of [3] it follows that if $A$ is a dbp-retract of a finite $T_{0}$–space $X$ and $r\colon X\to A$ is the corresponding retraction, then $r(x)=\max(U^{X}_{x}\cap A)$ for all $x\in X$.

Fixed point property

Let $X$ be a topological space. We say that a continuous map $f\colon X\to X$ has a fixed point if there exists $x\in X$ such that $f(x)=x$. We say that $X$ has the fixed point property if all continuous maps $f\colon X\to X$ have a fixed point. It is easy to see that if a topological space fails to be $T_{0}$ or connected, then it does not have the fixed point propery. Thus, we will only consider connected finite $T_{0}$–spaces in this article.

It is known that if $X$ is a finite $T_{0}$–space, $f\colon X\to X$ is a continuous map and there exists $x\in X$ such that $f(x)$ and $x$ are comparable, then $f$ has a fixed point. From this fact we deduce the following result.

Proposition 2.4**.**

Let $X$ be a finite $T_{0}$–space and let $f\colon X\to X$ be a continuous map. If there exists $x\in X$ such that $\widehat{U}_{x}$ (respectively $\widehat{F}_{x}$) has the fixed point property, and $f(\widehat{U}_{x})\subseteq U_{x}$ (respectively $f(\widehat{F}_{x})\subseteq F_{x}$), then $f$ has a fixed point.

Proof.

Suppose that $f(\widehat{U}_{x})\subseteq U_{x}$. If there exists $y\in\widehat{U}_{x}$ such that $f(y)=x$, then $f$ has a fixed point. Otherwise, $f(\widehat{U}_{x})\subseteq\widehat{U}_{x}$ and hence $f$ has a fixed point. ∎

We recall the following result.

Proposition 2.5**.**

[2*, Corollary 10.1.4]**

Let $X$ and $Y$ be homotopy equivalent finite $T_{0}$–spaces. If $X$ has the fixed point property, then $Y$ also has the fixed point property.*

It follows from the previous proposition that if $X$ is a finite $T_{0}$–space with the fixed point property and $x$ is a beat point of $X$, then $X-\{x\}$ also has the fixed point property. As a further consequence, if $X$ is a contractible finite $T_{0}$–space, then it has the fixed point property. Therefore, if a finite $T_{0}$–space has a maximum or minimum element, then it is contractible and hence it has the fixed point property.

Let $X$ be a finite $T_{0}$–space and let $n\in\mathbb{N}$ such that $n\geq 2$. We say that $X$ is a $2n$–crown if $X$ is homeomorphic to the finite $T_{0}$–space given by the following Hasse diagram

\bullet$$1$$\bullet$$2$$\cdots$$\bullet$${n-1}$$\bullet$$n$$\bullet$$n+1$$\bullet$$n+2$$\cdots$$\bullet$$2n-1$$\bullet$$2n

For each element $a$ in a $2n$–crown $X$, we have that $X-\{a\}$ is contractible. From this fact we obtain the following proposition.

Proposition 2.6**.**

Let $n\in\mathbb{N}$ such that $n\geq 2$ and let $X$ be a $2n$–crown. Let $f\colon X\to X$ be a continuous map. If $f$ is not bijective then $f$ has a fixed point.

3. The $\mathscr{C}$ construction

Let $S$ be a set. The power set of $X$ will be denoted by $\mathcal{P}(S)$ and the set $\mathcal{P}(S)-\{\varnothing\}$ will be denoted by $\mathcal{P}_{\neq\varnothing}(S)$.

Definition 3.1**.**

Let $X$ be a finite $T_{0}$–space. Let $A$ be a non-empty subset of $X$. We define

[TABLE]

When there is no risk of confusion, the superindices $X$ of the previous definition will be omitted and thus $U^{X}_{A}$ and $F^{X}_{A}$ will be denoted by $U_{A}$ and $F_{A}$ respectively.

Note that if $X$ is a finite $T_{0}$–space and $A$ is a non-empty subset of $X$, then $U_{A}$ is an open subset of $X$ and $F_{A}$ is a closed subset of $X$.

From now on, if $X$ is a topological space, $\mathcal{C}(X)$ will denote the set of connected components of $X$.

Definition 3.2**.**

Let $X$ be a finite $T_{0}$–space. We define

[TABLE]

We also define

[TABLE]

Note that $\mathscr{U}(X)$ and $\mathscr{F}(X)$ are finite sets.

Remark 3.3*.*

Let $X$ be a finite $T_{0}$–space.

(1) Clearly, the elements of $\mathscr{F}(X)$ are closed subsets of $X$. In addition, since finite topological spaces are locally connected it follows that the elements of $\mathscr{U}(X)$ are open subsets of $X$.

(2) For all $a\in\textup{mxl}([)X]$, $U_{a}\in\mathscr{U}(X)$ since $U_{a}$ is connected. In a similar way, for all $b\in\textup{mnl}([)X]$, $F_{b}\in\mathscr{F}(X)$.

The following proposition states that the sets $\mathscr{U}(X)$ and $\mathscr{F}(X)$ of the previous definition are disjoint, except in a trivial case.

Proposition 3.4**.**

Let $X$ be a finite $T_{0}$–space.

1. (a)

If $X$ is connected and $\mathscr{U}(X)\cap\mathscr{F}(X)\neq\varnothing$, then $X$ has a maximum element and a minimum element. 2. (b)

If $X$ has a maximum element then $\mathscr{U}(X)=\{X\}$. 3. (c)

If $X$ has a minimum element then $\mathscr{F}(X)=\{X\}$. 4. (d)

If $X$ is connected and $\mathscr{U}(X)\cap\mathscr{F}(X)\neq\varnothing$, then $\mathscr{U}(X)=\mathscr{F}(X)=\{X\}$.

Proof.

(a) Let $C\in\mathscr{U}(X)\cap\mathscr{F}(X)$. Then $C$ is non-empty, open and closed in $X$ by remark 3.3. Since $X$ is connected, $C=X$.

Since $C\in\mathscr{U}(X)$, there exists $U\in\mathscr{U}_{0}(X)$ such that $C$ is a connected component of $U$. Hence $U=X$. Since $X$ is finite and $X=U$ has at most one maximal element of $X$, then $X$ has a unique maximal element, that is, $X$ has a maximum element.

In a similar way, it follows that $X$ has a minimum element.

(b) If $a=\max X$, then $\mathscr{U}_{0}(X)=\{U_{a}\}=\{X\}$ and hence $\mathscr{U}(X)=\{X\}$.

(c) Analogous to the proof of (b).

(d) It follows immediately from (a), (b) and (c). ∎

Observe that in the previous proposition, the hypothesis that $X$ is connected is necessary for parts (a) and (d). Indeed, consider the discrete space of two points $X=\{0,1\}$. Clearly $F_{0}=\{0\}=U_{0}$. Hence $\{0\}\in\mathscr{U}(X)\cap\mathscr{F}(X)$, but $X$ has neither a maximum element nor a minimum element. Therefore item ($a$) does not hold. In addition, it is easy to verify that $\mathscr{U}(X)=\mathscr{F}(X)=\{\{0\},\{1\}\}$, and thus item ($d$) does not hold.

Definition 3.5**.**

Let $X$ be a finite $T_{0}$–space. We define the relation $\leq$ in $\mathscr{U}(X)\cup\mathscr{F}(X)$ as follows.

Given $C_{1},C_{2}\in\mathscr{U}(X)\cup\mathscr{F}(X)$, we will say that $C_{1}\leq C_{2}$ if and only if one of the following conditions holds:

• •

$C_{1},C_{2}\in\mathscr{U}(X)$ and $C_{1}\subseteq C_{2}$,

• •

$C_{1},C_{2}\in\mathscr{F}(X)$ and $C_{1}\supseteq C_{2}$,

• •

$C_{1}\in\mathscr{F}(X)$, $C_{2}\in\mathscr{U}(X)$ and $C_{1}\cap C_{2}\neq\varnothing$.

It is not difficult to prove that the previously defined relation $\leq$ is an order relation in $\mathscr{U}(X)\cup\mathscr{F}(X)$.

Definition 3.6**.**

Let $X$ be a finite $T_{0}$–space.

• •

We define $\mathscr{U}(X)$ as the finite $T_{0}$–space associated to the poset $(\mathscr{U}(X),\subseteq)$.

• •

We define $\mathscr{F}(X)$ as the finite $T_{0}$–space associated to the poset $(\mathscr{F}(X),\supseteq)$.

• •

We define $\mathscr{C}(X)$ as the finite $T_{0}$–space associated to the poset $(\mathscr{U}(X)\cup\mathscr{F}(X),\leq)$, where $\leq$ is the order defined in 3.5.

Observe that $\mathscr{U}(X)$ and $\mathscr{F}(X)$ are subspaces of $\mathscr{C}(X)$.

The following is a simple example of the $\mathscr{C}$ construction.

Example 3.7**.**

Let $X$ be the finite $T_{0}$–space given by the following Hasse diagram.

$\bullet$0$\bullet$1$\bullet$2$\bullet$3$\bullet$4

Then the Hasse diagram of $\mathscr{C}(X)$ is

2030 140 101342 3 403 41

The following proposition describes the maximal and minimal elements of the poset $\mathscr{C}(X)$ for any given finite $T_{0}$–space $X$.

Proposition 3.8**.**

Let $X$ be a finite $T_{0}$–space. Then

[TABLE]

Proof.

Let $a\in\textup{mxl}([)X]$. Since $U_{a}$ is connected, $U_{a}\in\mathscr{C}(X)$. Let $C\in\mathscr{U}(X)$ such that $C\geq U_{a}$. There exists a non-empty subset $A\subseteq\textup{mxl}([)X]$ such that $C$ is a connected component of $U_{A}$. Then $U_{a}\subseteq C\subseteq U_{A}$. It follows that $a\leq x$ for all $x\in A$, and thus $a=x$ for all $x\in A$. Hence $A=\{a\}$ and $U_{a}=U_{A}=C$. Therefore, $U_{a}\in\textup{mxl}([)\mathscr{C}(X)]$.

Now suppose that $C\in\textup{mxl}([)\mathscr{C}(X)]$. Suppose, in addition, that $C\in\mathscr{F}(X)$ and let $x\in C$. Since $X$ is finite, there exists $b\in\textup{mxl}([)X]$ such that $x\leq b$. Hence $b\in C$ since $C$ is closed. Thus, $C\cap U_{b}\neq\varnothing$ and then $C<U_{b}$ which contradicts the maximality of $C$.

Thus $C\in\mathscr{U}(X)$. Then there exists a non-empty subset $A\subseteq\textup{mxl}([)X]$ such that $C$ is a connected component of $U_{A}$. Let $a\in A$. Then $C\subseteq U_{a}$, that is, $C\leq U_{a}$. Hence $C=U_{a}$.

The second identity can be proved in a similar way. ∎

The following proposition shows that the construction $\mathscr{F}$ can be interpreted as a dual version of $\mathscr{U}$.

Proposition 3.9**.**

Let $X$ be a finite $T_{0}$–space.

1. (a)

For all $A\subseteq\textup{mnl}([)X]$, $F_{A}^{X}=\left(U_{A}^{X^{\textup{op}}}\right)^{\textup{op}}$. 2. (b)

$\mathscr{F}(X)=\mathscr{U}(X^{\textup{op}})$* as sets.* 3. (c)

$\mathscr{C}(X^{\textup{op}})=\left(\mathscr{C}(X)\right)^{\textup{op}}$. In particular, $\mathscr{F}(X)=(\mathscr{U}(X^{\textup{op}}))^{\textup{op}}$ as posets with the corresponding partial orders induced from $\mathscr{C}(X)$ and $\mathscr{C}(X^{\textup{op}})$ respectively.

Proof.

(a) Let $A\subseteq\textup{mnl}([)X]$. Note that for all $a\in A$, $F_{a}=\left(U_{a}^{X^{\textup{op}}}\right)^{\textup{op}}$. Hence

[TABLE]

(b) Follows from (a).

($c$) It follows from ($b$) that $\mathscr{U}(X^{\textup{op}})\cup\mathscr{F}(X^{\textup{op}})=\mathscr{F}(X)\cup\mathscr{U}(X)$. Then the underlying sets of $\mathscr{C}(X^{\textup{op}})$ and $\left(\mathscr{C}(X)\right)^{\textup{op}}$ coincide. Let $\leq$ denote the partial order of $\mathscr{C}(X)$ and let $\preceq$ denote the partial order of $\mathscr{C}(X^{\textup{op}})$. Let $C_{1},C_{2}\in\mathscr{C}(X^{\textup{op}})$.

• •

If $C_{1},C_{2}\in\mathscr{U}(X^{\textup{op}})=\mathscr{F}(X)$, then $C_{1}\preceq C_{2}$ if and only if $C_{1}\subseteq C_{2}$ if and only if $C_{2}\leq C_{1}$ if and only if $C_{1}\leq^{\textup{op}}C_{2}$.

• •

If $C_{1},C_{2}\in\mathscr{F}(X^{\textup{op}})=\mathscr{U}(X)$, then $C_{1}\preceq C_{2}$ if and only if $C_{1}\supseteq C_{2}$ if and only if $C_{2}\leq C_{1}$ if and only if $C_{1}\leq^{\textup{op}}C_{2}$.

• •

If $C_{1}\in\mathscr{F}(X^{\textup{op}})$ and $C_{2}\in\mathscr{U}(X^{\textup{op}})$, then $C_{1}\preceq C_{2}$ if and only if $C_{1}\cap C_{2}\neq\varnothing$ if and only if $C_{2}\leq C_{1}$ if and only if $C_{1}\leq^{\textup{op}}C_{2}$.

Hence, the partial orders $\preceq$ and $\leq^{\textup{op}}$ coincide. ∎

Definition 3.10**.**

Let $X$ be a finite $T_{0}$–space and let $B\subseteq X$ be a subset. We define

[TABLE]

Remark 3.11*.*

Let $X$ be a finite $T_{0}$–space and let $B\subseteq X$ be a subset.

If $B^{\sharp}\neq\varnothing$ then $B\subseteq U_{B^{\sharp}}$. If, in addition, $B$ is a connected subspace of $X$ then there exists a unique connected component $C_{0}$ of $U_{B^{\sharp}}$ such that $B\subseteq C_{0}$.

Similarly, if $B^{\flat}\neq\varnothing$ then $B\subseteq F_{B^{\flat}}$, and if, in addition, $B$ is a connected subspace of $X$ then there exists a unique connected component $C_{0}$ of $F_{B^{\flat}}$ such that $B\subseteq C_{0}$.

Lemma 3.12**.**

Let $X$ be a finite $T_{0}$–space and let $B\subseteq X$ be a connected subspace. Then:

1. (a)

$B^{\sharp}\neq\varnothing$* if and only if $\{C\in\mathscr{U}(X)\mid B\subseteq C\}\neq\varnothing$.* 2. (b)

$B^{\flat}\neq\varnothing$* if and only if $\{C\in\mathscr{F}(X)\mid B\subseteq C\}\neq\varnothing$.*

Proof.

We will prove only item (a). The proof of item (b) is similar.

($\Rightarrow$) Let $a_{0}\in B^{\sharp}$. Then $U_{a_{0}}\in\{C\in\mathscr{U}(X)\mid B\subseteq C\}$.

($\Leftarrow$) Let $C_{0}\in\mathscr{U}(X)$ such that $B\subseteq C_{0}$ and let $A\subseteq\textup{mxl}([)X]$ be a non-empty subset such that $C_{0}$ is a connected component of $U_{A}$. Let $a_{0}\in A$. Then $B\subseteq U_{A}\subseteq U_{a_{0}}$. Hence $a_{0}\in B^{\sharp}$. ∎

The following proposition states a key property of the $\mathscr{C}$ construction.

Proposition 3.13**.**

Let $X$ be a finite $T_{0}$–space and let $B\subseteq X$ be a connected non-empty subspace.

1. (a)

If $\{C\in\mathscr{U}(X)\mid B\subseteq C\}$ is non-empty, then it has a minimum element, which is the connected component of $U_{B^{\sharp}}$ that contains $B$. 2. (b)

If $\{C\in\mathscr{F}(X)\mid B\subseteq C\}$ is non-empty, then it has a maximum element, which is the connected component of $F_{B^{\flat}}$ that contains $B$.

Proof.

By the previous lemma, $B^{\sharp}\neq\varnothing$. Let $C_{0}$ be the unique connected component of $U_{B^{\sharp}}$ such that $B\subseteq C_{0}$.

Let $C\in\mathscr{U}(X)$ such that $B\subseteq C$. There exists a non-empty subset $A\subseteq\textup{mxl}([)X]$ such that $C$ is a connected component of $U_{A}$. Hence $B\subseteq U_{a}$ for all $a\in A$. Then, $A\subseteq B^{\sharp}$ and thus $U_{B^{\sharp}}\subseteq U_{A}$. Hence $C_{0}\subseteq U_{A}$.

Since $C$ is a connected component of $U_{A}$ and $\varnothing\neq B\subseteq C_{0}\cap C$, it follows that $C_{0}\subseteq C$, that is, $C_{0}\leq C$. Hence $C_{0}=\min\{C\in\mathscr{U}(X)\mid B\subseteq C\}$.

The second part can be proved in a similar way. ∎

The following corollary follows immediately from the previous proposition.

Corollary 3.14**.**

Let $X$ be a finite $T_{0}$–space.

1. (a)

Let $B\in\mathscr{U}(X)$. Then $B$ is a connected component of $U_{B^{\sharp}}$. 2. (b)

Let $B\in\mathscr{F}(X)$. Then $B$ is a connected component of $F_{B^{\flat}}$.

Proposition 3.13 allows us to give the following definition.

Definition 3.15**.**

Let $X$ be a finite $T_{0}$–space and let $x\in X$. We define

[TABLE]

and

[TABLE]

Proposition 3.16**.**

Let $X$ be a finite $T_{0}$–space.

1. (a)

The set $\mathscr{U}(X)$ is a basis for a topology in $X$ which is coarser than the topology of $X$. 2. (b)

Let $\mathcal{T}$ be the topology generated by $\mathscr{U}(X)$. The Kolmogorov quotient of $(X,\mathcal{T})$ is homeomorphic to a subspace of the finite $T_{0}$–space associated to the poset $\mathscr{U}(X)$.

Proof.

($a$) Since $U_{a}\in\mathscr{U}(X)$ for all $a\in\textup{mxl}([)X]$ it follows that $\displaystyle\bigcup_{C\in\mathscr{U}(X)}C=X$. Now let $C_{1},C_{2}\in\mathscr{U}(X)$ and let $x\in C_{1}\cap C_{2}$. Then $x\in{\bf C}_{\mathscr{U}}(x)\subseteq C_{1}\cap C_{2}$ by 3.13. Thus, $\mathscr{U}(X)$ is a basis for a topology in $X$ which, by 3.3, is clearly coarser than the topology of $X$.

($b$) Let $K$ be the Kolmogorov quotient of $(X,\mathcal{T})$ and let $q\colon X\to K$ be the quotient map. Let $f\colon X\to\mathscr{U}(X)$ be defined by $f(x)={\bf C}_{\mathscr{U}}(x)$. We will prove that $f$ is order-preserving. Let $a,b\in X$ be such that $a\leq b$ then $a\in{\bf C}_{\mathscr{U}}(b)$ since ${\bf C}_{\mathscr{U}}(b)$ is an open subset of $X$ which contains $b$. Thus ${\bf C}_{\mathscr{U}}(a)\subseteq{\bf C}_{\mathscr{U}}(b)$. Hence, $f$ is order-preserving, and therefore, it is a continuous map.

Now, let $c,d\in X$ be such that $q(c)=q(d)$. Thus, $\{C\in U(X)\mid q(c)\in C\}=\{C\in U(X)\mid q(d)\in C\}$ since $\mathscr{U}(X)$ is a basis for the topology $\mathcal{T}$. Hence, $f(c)={\bf C}_{\mathscr{U}}(c)={\bf C}_{\mathscr{U}}(d)=f(d)$. Therefore, there exists a continuous map $\overline{f}\colon K\to\mathscr{U}(X)$ such that $\overline{f}q=f$.

We will prove now that if $k,l\in K$ are such that $\overline{f}(k)\leq\overline{f}(l)$ then $k\leq l$. Let $k,l\in K$ be such that $\overline{f}(k)\leq\overline{f}(l)$. Let $x,y\in X$ be such that $k=q(x)$ and $l=q(y)$. Then $f(x)=\overline{f}(k)\leq\overline{f}(l)=f(y)$. Hence ${\bf C}_{\mathscr{U}}(x)\subseteq{\bf C}_{\mathscr{U}}(y)$. Then, for all $C\in\mathscr{U}(X)$ such that $y\in C$ we obtain that $x\in{\bf C}_{\mathscr{U}}(x)\subseteq{\bf C}_{\mathscr{U}}(y)\subseteq C$, and thus $x\in C$. Therefore $x\leq y$. Hence, $k=q(x)\leq q(y)=l$. ∎

Definition 3.17**.**

Let $X$ and $Y$ be finite $T_{0}$–spaces and let $f\colon X\to Y$ be a continuous map. We define

[TABLE]

and

[TABLE]

If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$, we define

[TABLE]

We will prove now that $\mathscr{U}(f)$ is well defined. Let $C\in\mathscr{U}(X)$. Then there exists $a\in\textup{mxl}([)X]$ such that $C\subseteq U_{a}$. Let $b\in\textup{mxl}([)Y]$ such that $f(a)\leq b$. Then $f(C)\subseteq U_{f(a)}\subseteq U_{b}$. Hence $f(C)^{\sharp}\neq\varnothing$. Then $\{D\in\mathscr{U}(Y)\mid f(C)\subseteq D\}$ is non-empty by 3.12, and thus it has a minimum element by 3.13.

In a similar way it can be proved that $\mathscr{F}(f)$ is well defined and thus $\mathscr{C}(f)$ is also well defined.

Note that the map $\mathscr{C}(f)$ might not be well defined without the assumption $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$. Note also that if $\mathscr{U}(X)\cap\mathscr{F}(X)\neq\varnothing$ then either $X$ is not connected or it has minimum and maximum elements by 3.4.

Remark 3.18*.*

Let $X$ and $Y$ be finite $T_{0}$–spaces and let $f\colon X\to Y$ be a continuous map.

Observe that, if $C\in\mathscr{U}(X)$, then, by 3.13, $\mathscr{U}(f)(C)$ is the connected component of $U_{f(C)^{\sharp}}$ that contains $f(C)$.

In a similar way, if $C\in\mathscr{F}(X)$, then $\mathscr{F}(f)(C)$ is the connected component of $F_{f(C)^{\flat}}$ that contains $f(C)$.

We will show in the following proposition that the maps defined in 3.17 are continuous.

Proposition 3.19**.**

Let $X$ and $Y$ be finite $T_{0}$–spaces and let $f\colon X\to Y$ be a continuous map. Then the maps $\mathscr{U}(f)$ and $\mathscr{F}(f)$ are continuous. If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ then the map $\mathscr{C}(f)\colon\mathscr{C}(X)\to\mathscr{C}(Y)$ is continuous.

Proof.

Let $C_{1},C_{2}\in\mathscr{C}(X)$ such that $C_{1}\leq C_{2}$.

• •

If $C_{1},C_{2}\in\mathscr{U}(X)$, then $C_{1}\subseteq C_{2}$. Hence $f(C_{1})\subseteq f(C_{2})\subseteq\mathscr{U}(f)(C_{2})$. Then $\mathscr{U}(f)(C_{1})\subseteq\mathscr{U}(f)(C_{2})$ by definition of $\mathscr{U}(f)$. Thus, $\mathscr{U}(f)(C_{1})\leq\mathscr{U}(f)(C_{2})$.

• •

If $C_{1},C_{2}\in\mathscr{F}(X)$, then $C_{2}\subseteq C_{1}$ and $f(C_{2})\subseteq f(C_{1})\subseteq\mathscr{F}(f)(C_{1})$. Hence $\mathscr{F}(f)(C_{2})\subseteq\mathscr{F}(f)(C_{1})$ by definition of $\mathscr{F}(f)$. Thus, $\mathscr{F}(f)(C_{1})\leq\mathscr{F}(f)(C_{2})$.

• •

If $C_{1}\in\mathscr{F}(X)$ and $C_{2}\in\mathscr{U}(X)$, then $C_{1}\cap C_{2}\neq\varnothing$. Hence $\varnothing\neq f(C_{1})\cap f(C_{2})\subseteq\mathscr{C}(f)(C_{1})\cap\mathscr{C}(f)(C_{2})$. Thus, $\mathscr{C}(f)(C_{1})\leq\mathscr{C}(f)(C_{2})$.

The result follows. ∎

Proposition 3.20**.**

Let $X$ and $Y$ be finite $T_{0}$–spaces and let $f\colon X\to Y$ be a homeomorphism. Then

1. (a)

for all $C\in\mathscr{U}(X)$, $\mathscr{U}(f)(C)=f(C)$, 2. (b)

for all $C\in\mathscr{F}(X)$, $\mathscr{F}(f)(C)=f(C)$, 3. (c)

$\mathscr{U}(f)$* and $\mathscr{F}(f)$ are homeomorphisms.*

Proof.

($a$) Since $f$ is a homeomorphism, $f(\textup{mxl}([)X])=\textup{mxl}([)Y]$ and $f(U_{A})=U_{f(A)}$ for each non-empty subset $A\subseteq\textup{mxl}([)X]$. Let $C\in\mathscr{U}(X)$. Then $C$ is a connected component of $U_{C^{\sharp}}$ and hence $f(C)$ is a connected component of $U_{f(C^{\sharp})}$. Thus $\mathscr{U}(f)(C)=f(C)$.

($b$) Analogous to ($a$).

($c$) Let $C\in\mathscr{U}(X)$. Then applying ($a$) we obtain that $\mathscr{U}(f^{-1})\mathscr{U}(f)(C)=\mathscr{U}(f^{-1})(f(C))=f^{-1}(f(C))=C$. Thus $\mathscr{U}(f^{-1})\mathscr{U}(f)=\textup{id}_{\mathscr{U}(X)}$. Similarly $\mathscr{U}(f)\mathscr{U}(f^{-1})=\textup{id}_{\mathscr{U}(Y)}$.

The proof for $\mathscr{F}(f)$ follows similarly. ∎

In the following example we will see that the constructions $\mathscr{U}$, $\mathscr{F}$ and $\mathscr{C}$ are not functorial.

Example 3.21**.**

Let $X$ be the finite $T_{0}$–space whose underlying set is $\{0,1,2,3,4\}$ and whose topology is generated by the basis $\{\{0\},\{0,1\},\{2\},\{0,1,2,3\},\{0,1,2,4\}\}$. The Hase diagrams of $X$ and $\mathscr{U}(X)$ are shown below.

$\bullet$0$\bullet$1$\bullet$2$\bullet$3$\bullet$4$X$

31 2041 20102$\mathscr{U}(X)$

Let $f\colon X\to X$ be the map defined by

[TABLE]

Clearly $f$ is a continuous map. Let $g\colon X\to X$ be the constant map with value 0. Then $fg$ is the constant map with value 2. It follows that $\mathscr{U}(g)$ is the constant map with value $\{0,1\}$ and that $\mathscr{U}(fg)$ is the constant map with value $\{2\}$. Then

[TABLE]

Hence, $\mathscr{U}(fg)\neq\mathscr{U}(f)\mathscr{U}(g)$.

This implies that $\mathscr{C}(fg)\neq\mathscr{C}(f)\mathscr{C}(g)$, and the fact that $\mathscr{F}$ is not functorial can be deduced from this example considering opposite spaces.

Proposition 3.22**.**

Let $X$, $Y$ and $Z$ be finite $T_{0}$–spaces and let $g\colon X\to Y$ and $f\colon Y\to Z$ be continuous maps. Then $\mathscr{U}(fg)\leq\mathscr{U}(f)\mathscr{U}(g)$ and $\mathscr{F}(fg)\geq\mathscr{F}(f)\mathscr{F}(g)$. If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{U}(Y)\cap\mathscr{F}(Y)=\varnothing$ then $\mathscr{C}(fg)\simeq\mathscr{C}(f)\mathscr{C}(g)$.

Proof.

Let $C\in\mathscr{U}(X)$. Then

[TABLE]

Hence, $\mathscr{U}(fg)(C)\subseteq\mathscr{U}(f)(\mathscr{U}(g)(C))$. Therefore, $\mathscr{U}(fg)\leq\mathscr{U}(f)\mathscr{U}(g)$. In a similar way it can be proved that $\mathscr{F}(fg)\geq\mathscr{F}(f)\mathscr{F}(g)$.

If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{U}(Y)\cap\mathscr{F}(Y)=\varnothing$ then the maps $\mathscr{C}(fg)$ and $\mathscr{C}(f)\mathscr{C}(g)$ are defined. Since $\mathscr{U}(fg)\leq\mathscr{U}(f)\mathscr{U}(g)$ and $\mathscr{F}(fg)\geq\mathscr{F}(f)\mathscr{F}(g)$ we obtain that, for all $C\in\mathscr{C}(X)$, the elements $\mathscr{C}(fg)(C)$ and $\mathscr{C}(f)\mathscr{C}(g)(C)$ are comparable. Thus, the last assertion follows. ∎

Proposition 3.23**.**

Let $X$ and $Y$ be finite $T_{0}$–spaces and let $f,g\colon X\to Y$ be continuous maps.

1. (a)

If $f\leq g$, then $\mathscr{U}(f)\leq\mathscr{U}(g)$ and $\mathscr{F}(f)\leq\mathscr{F}(g)$. 2. (b)

If $f\leq g$ and $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$, then $\mathscr{C}(f)\leq\mathscr{C}(g)$. 3. (c)

If $f\simeq g$, then $\mathscr{U}(f)\simeq\mathscr{U}(g)$ and $\mathscr{F}(f)\simeq\mathscr{F}(g)$. 4. (d)

If $f\simeq g$ and $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$, then $\mathscr{C}(f)\simeq\mathscr{C}(g)$.

Proof.

We will prove ($a$). Item ($b$) follows from ($a$), while items ($c$) and ($d$) follow from ($a$) and ($b$).

Let $C\in\mathscr{U}(X)$. If $x\in C$ then $f(x)\leq g(x)\in g(C)\subseteq\mathscr{U}(g)(C)$, and since $\mathscr{U}(g)(C)$ is an open subset of $Y$ it follows that $f(x)\in\mathscr{U}(g)(C)$. Therefore, $f(C)\subseteq\mathscr{U}(g)(C)$. Hence $\mathscr{U}(f)(C)\leq\mathscr{U}(g)(C)$.

Now let $C\in\mathscr{F}(X)$. If $x\in C$ then $g(x)\geq f(x)\in f(C)\subseteq\mathscr{F}(f)(C)$, and since $\mathscr{F}(f)(C)$ is a closed subset of $Y$ it follows that $g(x)\in\mathscr{F}(f)(C)$. Therefore, $g(C)\subseteq\mathscr{F}(f)(C)$. Hence $\mathscr{F}(f)(C)\leq\mathscr{F}(g)(C)$. ∎

Corollary 3.24**.**

Let $X$ and $Y$ be finite $T_{0}$–spaces and let $f\colon X\to Y$ be a homotopy equivalence. Then $\mathscr{U}(f)$ and $\mathscr{F}(f)$ are homotopy equivalences. If, in addition, $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{U}(Y)\cap\mathscr{F}(Y)=\varnothing$ then $\mathscr{C}(f)$ is a homotopy equivalence.

4. Fixed points

The following proposition and its corollaries relate the $\mathscr{C}$ construction to the fixed point property.

Proposition 4.1**.**

Let $X$ be a finite $T_{0}$–space and let $f\colon X\to X$ be a continuous map. Let $C\in\mathscr{C}(X)$ be a subspace of $X$ with the fixed point property.

1. (a)

If $C\in\mathscr{U}(X)$ and $\mathscr{U}(f)(C)\leq C$ then $f$ has a fixed point. 2. (b)

If $C\in\mathscr{F}(X)$ and $\mathscr{F}(f)(C)\geq C$ then $f$ has a fixed point. 3. (c)

If $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$ and $\mathscr{C}(f)(C)=C$, then $f$ has a fixed point.

Proof.

Under the assumptions of item ($a$) we obtain that $f(C)\subseteq\mathscr{U}(f)(C)\subseteq C$. Hence the map $f$ has a fixed point. Item ($b$) can be proved in a similar way and item ($c$) follows immediately from ($a$) and ($b$). ∎

Corollary 4.2**.**

Let $X$ be a finite $T_{0}$–space.

• •

If the space $\mathscr{U}(X)$ (resp. $\mathscr{F}(X)$) has the fixed point property and every $C\in\mathscr{U}(X)$ (resp. every $C\in\mathscr{F}(X)$) has the fixed point property then $X$ has the fixed point property.

• •

If $\mathscr{U}(X)\cap\mathscr{F}(X)=\varnothing$, the space $\mathscr{C}(X)$ has the fixed point property and every $C\in\mathscr{C}(X)$ has the fixed point property then $X$ has the fixed point property.

Proof.

We will prove the result in the case that $\mathscr{U}(X)$ has the fixed property. Let $f\colon X\to X$ be a continuous map. Then the map $\mathscr{U}(f)\colon\mathscr{U}(X)\to\mathscr{U}(X)$ has a fixed point. Thus $f$ has a fixed point by the previous proposition. ∎

Corollary 4.3**.**

Let $X$ be a finite $T_{0}$–space and let $f\colon X\to X$ be a continuous map.

1. (a)

If there exist $a\in\textup{mxl}([)X]$ and $C\in\mathscr{U}(X)$ such that $U_{a}\geq C$ and $\mathscr{U}(f)(C)=U_{a}$, then $f$ has a fixed point. 2. (b)

If there exist $a\in\textup{mnl}([)X]$ and $C\in\mathscr{F}(X)$ such that $F_{a}\leq C$ and $\mathscr{F}(f)(C)=F_{a}$, then $f$ has a fixed point.

Proof.

From the hypotheses of item ($a$) we obtain that $\mathscr{U}(f)(U_{a})\geq\mathscr{U}(f)(C)=U_{a}$. And since $U_{a}\in\textup{mxl}([)\mathscr{C}(X)]$, $\mathscr{U}(f)(U_{a})=U_{a}$. Then $f$ has a fixed point by proposition 4.1.

The second item follows in a similar way. ∎

Proposition 4.4**.**

Let $X$ be a finite $T_{0}$–space and let $f\colon X\to X$ be a continuous map. If $f$ has a fixed point, then the maps $\mathscr{U}(f)$ and $\mathscr{F}(f)$ have fixed points.

Proof.

Let $x\in X$ be such that $f(x)=x$. Let $C_{0}=\min\{C\in\mathscr{U}(X)\mid x\in C\}$. Note that this minimum element exists by proposition 3.13. It follows that $x=f(x)\in f(C_{0})\subseteq\mathscr{U}(f)(C_{0})$. Then $C_{0}\leq\mathscr{U}(f)(C_{0})$. Hence, $\mathscr{U}(f)$ has a fixed point.

In a similar way it can be proved that the map $\mathscr{F}(f)$ has a fixed point. ∎

The following example shows that the converse of the previous proposition does not hold.

Example 4.5**.**

Let $X$ be the finite $T_{0}$–space whose underlying set is $\{0,1,2,3,4,5\}$ and whose topology is generated by the basis $\{\{0\},\{1\},\{0,1,2\},\{0,1,3\},\{0,1,2,3,4\},\{0,1,2,3,5\}\}$. The Hase diagrams of $X$ and $\mathscr{U}(X)$ are shown below.

$\bullet$0$\bullet$1$\bullet$2$\bullet$3$\bullet$4$\bullet$5$X$

42 30 152 30 12 30 1$\mathscr{U}(X)$

Let $f\colon X\to X$ be the continuous map defined by

[TABLE]

Clearly $\mathscr{U}(X)$ has the fixed point property and hence the map $\mathscr{U}(f)$ has a fixed point. In a similar way, $\mathscr{F}(f)$ has a fixed point. However, the map $f$ does not have fixed points.

5. Examples

In this section we give several examples of application which show that the tools of section 3 are useful to study the fixed point property of finite posets.

First we give the following lemmas.

Lemma 5.1**.**

Let $X$ the finite $T_{0}$–space given by the following Hasse diagram.

[TABLE]

If $f\colon X\to X$ is a continuous map without fixed points, then $f(\{3,4,5\})=\{3,4,5\}$.

Proof.

We will prove first that $f(\{3,4,5\})\subseteq X-\textup{mxl}([)X]$. Let $x\in\{3,4,5\}$ and suppose that $f(x)\in\textup{mxl}([)X]$. Then $f(F_{x})=\{f(x)\}$ and hence there exist $a,b\in\textup{mxl}([)X]$ such that $a\neq b$ and $f(a)=f(b)=f(x)$. Observe that $\{0,1,2,3,4,5\}\subseteq U_{a}\cup U_{b}$. Hence

[TABLE]

Since $\widehat{U}_{f(x)}$ is contractible, $f$ has a fixed point by 2.4 which entails a contradiction.

Since $f(\{3,4,5\})\subseteq X-\textup{mxl}([)X]$, it follows that $f(X-\textup{mxl}([)X])\subseteq X-\textup{mxl}([)X]$. And since $f$ does not have fixed points, $f$ is bijective in this subspace by 2.6. In particular $f(\{3,4,5\})=\{3,4,5\}$. ∎

Lemma 5.2**.**

Let $X$ the finite $T_{0}$–space given by the following Hasse diagram.

[TABLE]

Let $f\colon X\to X$ a continuous map without fixed points. Then $f(\{3,4,5\})=\{3,4,5\}$.

Proof.

Let $x\in\{0,1\}$. Since $f$ does not have fixed points, $f(x)$ is not comparable with $x$. Hence $f(x)\in\{0,1,2\}-\{x\}$. In addition, note that $\widehat{F}_{2}\subseteq F_{x}$. Then, if $f(x)=2$, it follows that $f(\widehat{F}_{2})\subseteq f(F_{x})\subseteq F_{2}$, with $\widehat{F}_{2}$ contractible. Hence by 2.4, $f$ has a fixed point, which entails a contradiction. Then $f(0)=1$ and $f(1)=0$. It follows that $f(F_{0,1})\subseteq F_{0,1}$. In addition, as $f$ does not have fixed points, the restriction $f|\colon F_{0,1}\to F_{0,1}$ is bijective by 2.6. In particular, $f(\{3,4,5\})=\{3,4,5\}$. ∎

Now we are ready to give the examples.

Example 5.3**.**

We will prove that the space $P^{3323}$ of [5, Fig.1] has the fixed point property. Its Hasse diagram is given in figure 1(a). The Hasse diagrams of $\mathscr{U}(P^{3323})$ and $\mathscr{F}(P^{3323})$ are shown in figure 2 and the shaded elements of these diagrams are contractible subspaces of $P^{3323}$.

Suppose that there exists a continuous map $f\colon P^{3323}\to P^{3323}$ without fixed points.

Let $C_{1}=\{0,1,2,3,4,5\}\in\mathscr{U}(X)$. Since the poset $\mathscr{U}(P^{3323})$ has a minimum element, the map $\mathscr{U}(f)\colon\mathscr{U}(P^{3323})\to\mathscr{U}(P^{3323})$ has a fixed point, which must be $C_{1}$ by 4.1. Thus, $f(C_{1})\subseteq\mathscr{U}(f)(C_{1})=C_{1}$.

Let $C_{2}=\{3,6,7,8,9,10\}\in\mathscr{F}(P^{3323})$. Proceeding as in the previous paragraph we obtain that $C_{2}$ is a fixed point of $\mathscr{F}(f)$. Thus, $f(C_{2})\subseteq\mathscr{F}(f)(C_{2})=C_{2}$.

Hence, $f(C_{1}\cap C_{2})\subseteq C_{1}\cap C_{2}$ and hence $3$ is a fixed point of $f$, which entails a contradiction.

Therefore, the space $P^{3323}$ has the fixed point property.

Example 5.4**.**

We will prove that the space $P^{343}_{1}$ given in [5, Fig.1] has the fixed point property. Its Hasse diagram is given in figure 1(b). The Hasse diagram of $\mathscr{C}(P^{343}_{1})$ is shown in figure 3(a).

We will prove first that $\mathscr{C}(P^{343}_{1})$ is homotopy equivalent to $P^{3323}$. Note that the minimal elements of $\mathscr{U}(P^{343}_{1})$ are down beat points of $\mathscr{C}(P^{343}_{1})$ and that the maximal elements of $\mathscr{F}(P^{343}_{1})$ are up beat points of $\mathscr{C}(P^{343}_{1})$. Then $\mathscr{C}(P^{343}_{1})$ is homotopy equivalent to the finite $T_{0}$–space that is given by the Hasse diagram of figure 4(a). Now observe that the element labeled $a$ in this figure is a down beat point, and by removing it we obtain the space given in figure 4(b), which is homeomorphic to $P^{3323}$. Therefore $\mathscr{C}(P^{343}_{1})$ is homotopy equivalent to $P^{3323}$.

Now, let $f\colon P^{343}_{1}\to P^{343}_{1}$ be a continuous map. In example 5.3 we proved that $P^{3323}$ has the fixed point property. Hence, by 2.5, $\mathscr{C}(P^{343}_{1})$ has the fixed point property. Thus, the map $\mathscr{C}(f)$ has a fixed point. And since all the elements of $\mathscr{C}(P^{343}_{1})$ are contractible subspaces of $P^{343}_{1}$, applying proposition 4.1, we obtain that $f$ has a fixed point. Therefore $P^{343}_{1}$ has the fixed point property.

Example 5.5**.**

We will prove that the space $P^{343}_{2}$ given in [5, Fig.1] has the fixed point property. Its Hasse diagram is given in figure 1(c). The Hasse diagram of $\mathscr{C}(P^{343}_{2})$ is shown in figure 3(b). Let $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ be the sets indicated in that figure.

Suppose that $f\colon P^{343}_{2}\to P^{343}_{2}$ is a continuous map without fixed points. Since all the elements of $\mathscr{U}(P^{343}_{2})$ are contractible subspaces of $P^{343}_{2}$, then the map $\mathscr{U}(f)$ does not have fixed points by 4.1. By lemma 5.2, $\mathscr{U}(f)(\{A,B,C\})\subseteq\{A,B,C\}$. In a similar way, $\mathscr{F}(f)(\{A^{\prime},B^{\prime},C^{\prime}\})\subseteq\{A^{\prime},B^{\prime},C^{\prime}\}$. Consider the subspace $M=\{A,B,C,A^{\prime},B^{\prime},C^{\prime}\}\subseteq\mathscr{C}(P^{343}_{2})$. It follows that $\mathscr{C}(f)(M)\subseteq M$. Since $M$ is contractible, the map $\mathscr{C}(f)$ has fixed points and then, from 4.1 we obtain that $f$ has fixed points, which entails a contradiction.

Therefore, $P^{343}_{2}$ has the fixed point property.

Example 5.6**.**

Let $n,k\in\mathbb{N}$ such that $n\geq 4$ and $2\leq k\leq n-1$. Let $X_{n,k}$ the finite $T_{0}$–space whose underlying set is $\{a_{1},a_{2},a_{3},b_{1},b_{2},\ldots,b_{n},c_{1},c_{2},\ldots,c_{n}\}$ and whose minimal open sets are

[TABLE]

The Hasse diagram of $X_{n,k}$ is

\bullet$$c_{1}$$\bullet$$c_{2}$$\cdots$$\cdots$$\bullet$$c_{k}$$\bullet$$c_{n-2}$$\bullet$$c_{n-1}$$\bullet$$c_{n}$$\bullet$$b_{1}$$\bullet$$b_{2}$$\cdots$$\cdots$$\bullet$$b_{k}$$\bullet$$b_{n-2}$$\bullet$$b_{n-1}$$\bullet$$b_{n}$$\bullet$$a_{1}$$\bullet$$a_{2}$$\bullet$$a_{3}

Observe that the posets $X_{4,3}$ and $X_{4,2}$ are isomorphic to the posets $P^{443}_{4}$ and $P^{443}_{5}$ of [5, Fig.1].

We will prove that the spaces $X_{n,k}$ have the fixed point property for all $n,k$ as above. To this end we will compute $\mathscr{U}(X_{n,k})$. Let

[TABLE]

Observe that the connected components of $U_{a_{1}a_{2}}$ are $A$ and $F$, the connected components of $U_{a_{1}a_{3}}$ are $B$ and $E$, the connected components of $U_{a_{2}a_{3}}$ are $C$ and $D$ and that the connected components of $U_{a_{1}a_{2}a_{3}}$ are $D$, $E$ and $F$. Hence $\mathscr{U}(X_{n,k})$ is the finite space given by the following Hasse diagram

\bullet$$U_{a_{1}}$$\bullet$$U_{a_{2}}$$\bullet$$U_{a_{3}}$$\bullet$$A$$\bullet$$B$$\bullet$$C$$\bullet$$D$$\bullet$$E$$\bullet$$F

Suppose that $f\colon X_{n,k}\to X_{n,k}$ is a continuous map without fixed points. Since all the elements of $\mathscr{U}(X_{n,k})$ are contractible subspaces of $X_{n,k}$, then $\mathscr{U}(f)$ does not have fixed points by 4.1. Thus, by 5.1, $\mathscr{U}(f)(\{A,B,C\})=\{A,B,C\}$. Since $S=A\cup B\cup C$, we have that $f(S)=f(A\cup B\cup C)=f(A)\cup f(B)\cup f(C)\subseteq A\cup B\cup C=S$.

We will prove now that the restriction $f|\colon S\to S$ is not bijective. Note that

[TABLE]

where the first inequality holds since $n+k>5$. Thus, the sets $A$, $B$ and $C$ do not all have the same cardinality and since $\mathscr{U}(f)$ does not have fixed points and $\mathscr{U}(f)(\{A,B,C\})=\{A,B,C\}$ it follows that there exists $T\in\{A,B,C\}$ such that $\#\mathscr{U}(f)(T)<\#T$. Hence $\#f(T)<\#T$ and then $f$ is not bijective.

Finally, since $f|\colon S\to S$ is not bijective, it has a fixed point by 2.6 which entails a contradiction.

6. Beat points

Remark 6.1*.*

Note that the space $X$ of example 4.5 is not contractible but $\mathscr{U}(X)$ is. Thus $X$ and $\mathscr{U}(X)$ do not have the same homotopy type.

Proposition 6.2**.**

Let $X$ be a finite $T_{0}$–space and let $A$ be a bp-retract of $X$. Let $r\colon X\to A$ be the corresponding retraction.

1. (a)

If $C\in\mathscr{C}(X)$, then $r(C)=C\cap A$. 2. (b)

If $\textup{mxl}([)X]\subseteq A$ and $C\in\mathscr{U}(X)$, then $C\cap A\in\mathscr{U}(A)$. 3. (c)

If $\textup{mnl}([)X]\subseteq A$ and $C\in\mathscr{F}(X)$, then $C\cap A\in\mathscr{F}(A)$.

Proof.

($a$) Suppose that $A$ is a dbp-retract of $X$. Clearly, $C\cap A\subseteq r(C)$ and $r(C)\subseteq A$. It remains to prove that $r(C)\subseteq C$.

Suppose first that $C\in\mathscr{U}(X)$. Let $z\in C$. Since $r(z)\leq z$ and $C$ is an open subset of $X$ it follows that $r(z)\in C$.

Suppose now that $C\in\mathscr{F}(X)$. Let $z\in C$. Note that $z\geq y$ for all $y\in C^{\flat}$. Then $r(z)\geq r(y)=y$ for all $y\in C^{\flat}$, that is, $r(z)\in F_{C^{\flat}}$. Since $r(z)\leq z$ it follows that $C\cup\{r(z)\}$ is a connected subspace of $F_{C^{\flat}}$. Thus $C\cup\{r(z)\}\subseteq C$ and hence $r(z)\in C$.

($b$) Note that $\textup{mxl}([)A]=\textup{mxl}([)X]$ and that $C\cap A$ is connected and non-empty by item ($a$). Since $C\in\mathscr{U}(X)$, $C$ is a connected component of $U^{X}_{C^{\sharp}}$. Then $C\cap A\subseteq U^{X}_{C^{\sharp}}\cap A=U^{A}_{C^{\sharp}}$.

Thus, there exists a connected component $C_{0}$ of $U^{A}_{C^{\sharp}}$ such that $C\cap A\subseteq C_{0}$. Then $C_{0}\subseteq U^{X}_{C^{\sharp}}$ and since $\varnothing\neq C\cap A\subseteq C\cap C_{0}$ and $C$ is a connected component of $U^{X}_{C^{\sharp}}$ we obtain that $C_{0}\subseteq C$. Hence, $C\cap A\subseteq C_{0}\subseteq C\cap A$. Therefore $C\cap A=C_{0}\in\mathscr{U}(A)$.

($c$) Analogous to that of item ($b$). ∎

Proposition 6.3**.**

Let $X$ be a finite $T_{0}$–space and let $A$ be a bp-retract of $X$. Let $r\colon X\to A$ be the corresponding retraction and let $i\colon A\to X$ be the inclusion map.

1. (a)

If $\textup{mxl}([)X]\subseteq A$ then $\mathscr{U}(r)(C)=r(C)$ for all $C\in\mathscr{U}(X)$ and $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(A)}$. 2. (b)

If $A$ is an ubp-retract of $X$ then $\mathscr{U}(r)$ is a homeomorphism. 3. (c)

If $\textup{mxl}([)X]\subseteq A$ and $A$ is a dbp-retract of $X$ then $\mathscr{U}(i)\mathscr{U}(r)\leq\textup{id}_{\mathscr{U}(X)}$.

Proof.

($a$) Let $C\in\mathscr{U}(X)$. Then $r(C)=C\cap A\in\mathscr{U}(A)$ by 6.2. It follows that $\mathscr{U}(r)(C)=r(C)$.

Let $D\in\mathscr{U}(A)$. Then

[TABLE]

by 3.18. Since $D$ is a connected component of $U^{A}_{D^{\sharp}}$ by 3.14 and $\mathscr{U}(r)\mathscr{U}(i)(D)$ is connected, it follows that $\mathscr{U}(r)\mathscr{U}(i)(D)=D$.

($b$) Since $A$ is an ubp-retract then $\textup{mxl}([)X]\subseteq A$ and hence $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(A)}$ by item ($a$). Let $C\in\mathscr{U}(X)$. Then

[TABLE]

Note that $ir(C)=C\cap A$ and since $C\in\mathscr{U}(X)$ it follows that $\mathscr{U}(ir)(C)\subseteq C$.

Let $z\in C$. Since $r(z)\geq z$, $r(z)\in ir(C)\subseteq\mathscr{U}(ir)(C)$ and $\mathscr{U}(ir)(C)$ is an open subset of $X$, we obtain that $z\in\mathscr{U}(ir)(C)$. Hence, $\mathscr{U}(ir)(C)=C$.

Therefore, $\mathscr{U}(i)\mathscr{U}(r)(C)=C$.

($c$) Let $C\in\mathscr{U}(X)$. Then

[TABLE]

by 3.23. ∎

Clearly, an analogous result holds for $\mathscr{F}$.

Proposition 6.4**.**

Let $X$ be a finite $T_{0}$–space and let $A$ be a bp-retract of $X$. Let $r\colon X\to A$ be the corresponding retraction and let $i\colon A\to X$ be the inclusion map.

1. (a)

If $\textup{mnl}([)X]\subseteq A$ then $\mathscr{F}(r)(C)=r(C)$ for all $C\in\mathscr{F}(X)$ and $\mathscr{F}(r)\mathscr{F}(i)=\textup{id}_{\mathscr{F}(A)}$. 2. (b)

If $A$ is a dbp-retract of $X$ then $\mathscr{F}(r)$ is a homeomorphism. 3. (c)

If $\textup{mnl}([)X]\subseteq A$ and $A$ is an ubp-retract of $X$ then $\mathscr{F}(i)\mathscr{F}(r)\geq\textup{id}_{\mathscr{F}(X)}$.

Proof.

Similar to the proof of 6.3. ∎

Remark 6.5*.*

Let $X$ be a finite $T_{0}$–space and let $A$ be a dbp-retract of $X$ such that $\textup{mxl}([)X]\subseteq A$. Let $r$ and $i$ be as in 6.3. Then $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(A)}$ and $\mathscr{U}(i)\mathscr{U}(r)\leq\textup{id}_{\mathscr{U}(X)}$. Thus, $\mathscr{U}(i)$ is an embedding, and applying 2.2 it follows that $\mathscr{U}(i)(\mathscr{U}(A))$ is a dbp-retract of $\mathscr{U}(X)$, that is, $\mathscr{U}(i)(\mathscr{U}(A))$ can be obtained from $\mathscr{U}(X)$ by removing down beat points. Observe that $\mathscr{U}(i)(\mathscr{U}(A))$ is homeomorphic to $\mathscr{U}(A)$.

The following examples show that several of the hypotheses of the previous propositions are necessary.

Example 6.6**.**

Let $X$ be the finite $T_{0}$–space given in figure 5(a). Note that 4 is a down beat point of $X$. The spaces $\mathscr{U}(X)$, $X-\{4\}$ and $\mathscr{U}(X-\{4\})$ are given in figures 5(b), 5(c) and 5(d), respectively.

Observe that $\{0,2,4\}\in\mathscr{U}(X)$ but $\{0,2\}\notin\mathscr{U}(X-\{4\})$. Thus, 6.2 might not hold if $\textup{mxl}([)X]\not\subseteq A$.

Let $i\colon X-\{4\}\to X$ be the inclusion map and let $r\colon X\to X-\{4\}$ be the retraction. Note that $\mathscr{U}(r)(\{0,2,4\})=\{0,1,2,3\}\neq r(\{0,2,4\})$ and that $\mathscr{U}(i)\mathscr{U}(r)(\{0,2,4\})=\{0,1,2,3\}\not\leq\{0,2,4\}$. Hence, items ($a$) and ($c$) of proposition 6.3 might not hold if $\textup{mxl}([)X]\not\subseteq A$.

Example 6.7**.**

Let $X$ be the finite $T_{0}$–space given in figure 6(a). Note that 1 is a down beat point of $X$. The spaces $\mathscr{U}(X)$, $X-\{1\}$ and $\mathscr{U}(X-\{1\})$ are given in figures 6(b), 6(c) and 6(d), respectively.

Note that $\mathscr{U}(X)$ and $\mathscr{U}(X-\{1\})$ are not homeomorphic. Hence, item ($b$) of 6.3 might not hold if $A$ is a dbp-retract of $X$.

Proposition 6.8**.**

Let $X$ be a finite $T_{0}$–space and let $a\in\textup{mxl}([)X]$ be a down beat point of $X$. Let $b=\max\widehat{U}_{a}$.

1. (a)

If $b\notin\textup{mxl}([)X-\{a\}]$ then $U_{a}$ is a down beat point of $\mathscr{U}(X)$ and $\mathscr{U}(X-\{a\})$ is a ubp-retract of $\mathscr{U}(X)-\{U_{a}\}$. 2. (b)

If $b\in\textup{mxl}([)X-\{a\}]$ then $\mathscr{U}(X)$ and $\mathscr{U}(X-\{a\})$ are homeomorphic.

Proof.

($a$) Since $b\notin\textup{mxl}([)X-\{a\}]$ then there exists $d\in\textup{mxl}([)X]$ such that $d>b$ and $d\neq a$. Hence $U_{b}=U_{ad}$ and thus $U_{b}\in\mathscr{U}(X)$. Clearly $U_{b}<U_{a}$. We will prove that $U_{b}=\max\widehat{U}^{\mathscr{U}(X)}_{U_{a}}$. Let $C\in\mathscr{U}(X)$ be such that $C<U_{a}$. Clearly $a\notin C$ (since otherwise $U_{a}\subseteq C$). Thus, $C\subseteq\widehat{U}_{a}=U_{b}$. Therefore, $U_{a}$ is a down beat point of $\mathscr{U}(X)$.

Let $i\colon X-\{a\}\to X$ be the inclusion map and let $r\colon X\to X-\{a\}$ be the retraction which corresponds to the removal of the beat point $a$. Note that $\mathscr{U}(X-\{a\})\subseteq\mathscr{U}(X)-\{U_{a}\}$, and hence $\mathscr{U}(i)(D)=D$ for all $D\in\mathscr{U}(X-\{a\})$. Then $\mathscr{U}(i)\colon\mathscr{U}(X-\{a\})\to\mathscr{U}(X)$ is the inclusion map.

Let $D\in\mathscr{U}(X-\{a\})$. Note that $\mathscr{U}(r)\mathscr{U}(i)(D)=\mathscr{U}(r)(D)=D$ since $a\notin D$. Hence $\mathscr{U}(r)\mathscr{U}(i)=\textup{id}_{\mathscr{U}(X-\{a\})}$.

Let $C\in\mathscr{U}(X)-\{U_{a}\}$. Applying 6.2 we obtain that $C=C-\{a\}=r(C)\subseteq\mathscr{U}(r)(C)=\mathscr{U}(i)\mathscr{U}(r)(C)$. From 2.2 it follows that $\mathscr{U}(X-\{a\})$ is a ubp-retract of $\mathscr{U}(X)-\{U_{a}\}$.

($b$) Since $b\in\textup{mxl}([)X-\{a\}]$, $b$ is an up beat point of $X$. Hence $\mathscr{U}(X)$ is homeomorphic to $\mathscr{U}(X-\{b\})$ by 6.3, which is homeomorphic to $\mathscr{U}(X-\{a\})$ since $X-\{b\}$ and $X-\{a\}$ are homeomorphic. ∎

Proposition 6.9**.**

Let $X$ be a finite $T_{0}$–space. Then $\mathscr{U}(X)$ does not have up beat points and $\mathscr{F}(X)$ does not have down beat points.

Proof.

Suppose that $C$ is an up beat point of $\mathscr{U}(X)$. Let $D=\min\widehat{F}^{\mathscr{U}(X)}_{C}$. Then

[TABLE]

Thus, by 3.8,

[TABLE]

Then, by 3.14, $C$ and $D$ are connected components of $U_{C^{\sharp}}$ and since $C\subseteq D$, it follows that $C=D$, which entails a contradiction.

The second part of the proposition can be proved in a similar way. ∎

7. Properties of $\mathscr{U}(X)$

Proposition 7.1**.**

Let $X$ be a finite $T_{0}$–space such that every $C\in\mathscr{U}(X)$ has a maximum element. Let $X^{\prime}=\{\max C\mid C\in\mathscr{U}(X)\}$. Then $X^{\prime}$ is an ubp-retract of $X$.

Proof.

Let $i\colon X^{\prime}\to X$ be the inclusion map. For each $x\in X$ let $C(x)=\min\{C\in\mathscr{U}(X)\mid x\in C\}$. Let $r\colon X\to X^{\prime}$ be defined by $r(x)=\max C(x)$.

We claim that $r$ is continuous. Indeed, let $x_{1},x_{2}\in X$ such that $x_{1}\leq x_{2}$. Since $C(x_{2})$ is an open subset of $X$ we obtain that $x_{1}\in C(x_{2})$ and thus $C(x_{1})\subseteq C(x_{2})$. Hence, $r(x_{1})\leq r(x_{2})$.

Clearly, $ir\geq\textup{id}_{X}$. We will prove now that $ri=\textup{id}_{X^{\prime}}$. Let $z\in X^{\prime}$. Then there exists $C\in\mathscr{U}(X)$ such that $z=\max C$. Since $C(z)$ is an open subset of $X$ we obtain that $U_{z}\subseteq C(z)\subseteq C\subseteq U_{z}$. It follows that $ri(z)=z$.

Therefore, $X^{\prime}$ is an ubp-retract of $X$ by 2.2. ∎

Remark 7.2*.*

Observe that, under the hypotheses of the previous proposition, from 6.3 and 7.1 we obtain that $\mathscr{U}(X^{\prime})\cong\mathscr{U}(X)$.

Proposition 7.3**.**

Let $X$ be a finite $T_{0}$–space such that every $C\in\mathscr{U}(X)$ has a maximum element. Let $X^{\prime}=\{\max C\mid C\in\mathscr{U}(X)\}$. Then there exists mutually inverse isomorphisms $\varphi\colon\mathscr{U}(X)\to X^{\prime}$ and $\psi\colon X^{\prime}\to\mathscr{U}(X)$ given by $\varphi(C)=\max C$ and $\psi(z)=U^{X}_{z}$.

Proof.

Observe that if $C\in\mathscr{U}(X)$ and $\max C=z$ then, as in the proof of 7.1, we obtain that $C=U^{X}_{z}$. The result follows. ∎

Lemma 7.4**.**

Let $X$ be a finite $T_{0}$–space. Let $B\subseteq\textup{mxl}([)\mathscr{U}(X)]$ be a non-empty subset. Let $A=\{a\in\textup{mxl}([)X]\mid U^{X}_{a}\in B\}$. Then

1. (a)

$\textup{mxl}([)U^{\mathscr{U}(X)}_{B}]=\mathcal{C}(U^{X}_{A})$, and 2. (b)

$\mathcal{C}(U^{\mathscr{U}(X)}_{B})=\{U_{M}\mid M\in\mathcal{C}(U^{X}_{A})\}$.

Proof.

($a$) First note that $B=\{U^{X}_{a}\mid a\in A\}$ by 3.8.

Let $D\in\textup{mxl}([)U^{\mathscr{U}(X)}_{B}]$. Then $D\subseteq U^{X}_{A}$. Since $D$ is connected, there exists a connected component $D_{0}\in\mathcal{C}(U^{X}_{A})$ such that $D\subseteq D_{0}$. It follows that $D_{0}\in U^{\mathscr{U}(X)}_{B}$. From the maximality of $D$ we obtain that $D=D_{0}$ and hence $D\in\mathcal{C}(U^{X}_{A})$.

Now, let $D\in\mathcal{C}(U^{X}_{A})$. Then, $D\in U^{\mathscr{U}(X)}_{B}$. Let $C\in U^{\mathscr{U}(X)}_{B}$ such that $D\subseteq C$. Then $C\subseteq U^{X}_{A}$, and since $C$ is connected it follows that $C\subseteq D$. Thus, $C=D$. Therefore, $D\in\textup{mxl}([)U^{\mathscr{U}(X)}_{B}]$.

($b$) From ($a$) we obtain that

[TABLE]

Now, observe that if $M_{1},M_{2}\in\mathcal{C}(U^{X}_{A})$ are such that $M_{1}\neq M_{2}$ then $U^{\mathscr{U}(X)}_{M_{1}}\cap U^{\mathscr{U}(X)}_{M_{2}}=\varnothing$ since $M_{1}\cap M_{2}=\varnothing$. The result follows. ∎

Proposition 7.5**.**

Let $X$ be a finite $T_{0}$–space. Then there exists mutually inverse isomorphisms $\varphi\colon\mathscr{U}(\mathscr{U}(X))\to\mathscr{U}(X)$ and $\psi\colon\mathscr{U}(X)\to\mathscr{U}(\mathscr{U}(X))$ given by $\varphi(D)=\max D$ and $\psi(C)=U^{\mathscr{U}(X)}_{C}$.

Proof.

First observe that every $C\in\mathscr{U}(\mathscr{U}(X))$ has a maximum element by item ($b$) of 7.4. Now, if $C\in\mathscr{U}(X)$ then $C$ is a connected component of $U^{X}_{C^{\sharp}}$ and hence $U_{C}$ is a connected component of $U^{\mathscr{U}(X)}_{\{U_{a}\mid a\in C^{\sharp}\}}$ by item ($b$) of 7.4. Therefore, $\{\max D\mid D\in\mathscr{U}(\mathscr{U}(X))\}=\mathscr{U}(X)$. The result follows from 7.3. ∎

Corollary 7.6**.**

Let $X$ be a finite $T_{0}$–space. Then there exists a finite $T_{0}$–space $Y$ such that $X\cong\mathscr{U}(Y)$ if and only if $X\cong\mathscr{U}(X)$.

Proof.

Follows easily from 3.20 and 7.5. ∎

8. Other properties of the $\mathscr{C}$–construction

In this section we will develop more results about the $\mathscr{C}$–construction. Not all of them are related to the fixed point property but are interesting on their own and permit a better understanding of this construction. In addition, these results show that the $\mathscr{C}$–construction is an interesting object of study.

In the following proposition we will see that if $X$ is a finite $T_{0}$–space, then $X$ is homotopy equivalent to the Grothendieck construction on a functor from $\mathscr{U}(X)$ to the category of finite posets, or equivalently to the the non-Hausdorff homotopy colimit of a $\mathscr{U}(X)$–diagram of finite posets [4].

Recall that if $A$ is a finite poset, $\mathcal{P}$ is the category of finite posets and $F\colon A\to\mathcal{P}$ a functor, then the Grothendieck construction for $F$ is the poset $\int F$ with underlying set $\bigcup\limits_{a\in A}(\{a\}\times F(a))$ and with partial order defined as follows: if $a,b\in A$, $x\in F(a)$, $y\in F(b)$ then $(a,x)\leq(b,y)$ if and only if $a\leq b$ and $F(a\leq b)(x)\leq y$.

Definition 8.1**.**

Let $X$ be a finite $T_{0}$–space. We define the functor $\mathcal{I}_{X}\colon\mathscr{U}(X)\to\mathcal{P}$ by $\mathcal{I}_{X}(C)=C$ and $\mathcal{I}_{X}(C\subseteq C^{\prime})$ as the inclusion map $C\hookrightarrow C^{\prime}$.

We also define maps $\rho_{X}\colon\int\mathcal{I}_{X}\to X$ and $\iota_{X}\colon X\to\int\mathcal{I}_{X}$ by $\rho_{X}(C,x)=x$ for all $(C,x)\in\int\mathcal{I}_{X}$ and $\iota(x)=({\bf C}_{\mathscr{U}}(x),x)$ for all $x\in X$.

Proposition 8.2**.**

Let $X$ be a finite $T_{0}$–space. Then, the maps $\rho_{X}\colon\int\mathcal{I}_{X}\to X$ and $\iota_{X}\colon X\to\int I_{X}$ are continuous, $\rho_{X}\iota_{X}=\textup{id}_{X}$ and $\iota_{X}\rho_{X}\leq\textup{id}_{\int\mathcal{I}_{X}}$.

In particular, $\iota_{X}(X)$ is a dbp–retract of $\int\mathcal{I}_{X}$ and $X$ is homotopy equivalent to $\int\mathcal{I}_{X}$.

Proof.

Let $(C_{1},x_{1}),(C_{2},x_{2})\in\int\mathcal{I}_{X}$ such that $(C_{1},x_{1})\leq(C_{2},x_{2})$. Then $C_{1}\subseteq C_{2}$ and $x_{1}=\textup{inc}(x_{1})\leq x_{2}$ in $C_{2}$. Hence $\rho_{X}(C_{1},x_{1})=x_{1}\leq x_{2}=\rho_{X}(C_{2},x_{2})$ in $X$. Thus, $\rho_{X}$ is continuous.

Now, let $x_{1},x_{2}\in X$ such that $x_{1}\leq x_{2}$. Since $x_{2}\in{\bf C}_{\mathscr{U}}(x_{2})$ and ${\bf C}_{\mathscr{U}}(x_{2})$ is an open subset of $X$, then $x_{1}\in{\bf C}_{\mathscr{U}}(x_{2})$. Hence, ${\bf C}_{\mathscr{U}}(x_{1})\leq{\bf C}_{\mathscr{U}}(x_{2})$ and thus $\iota_{X}(x_{1})=({\bf C}_{\mathscr{U}}(x_{1}),x_{1})\leq({\bf C}_{\mathscr{U}}(x_{2}),x_{2})=\iota(x_{2})$. Therefore, $\iota_{X}$ is continuous.

Now, for all $x\in X$, $\rho_{X}\iota_{X}(x)=\rho_{X}({\bf C}_{\mathscr{U}}(x),x)=x$. Hence $\rho_{X}\iota_{X}=\textup{id}_{X}$.

On the other hand, if $(C,x)\in\int\mathcal{I}_{X}$, then $x\in C$ and $\iota_{X}\rho_{X}(C,x)=\iota_{X}(x)=({\bf C}_{\mathscr{U}}(x),x)\leq(C,x)$. Hence $\iota_{X}\rho_{X}\leq\textup{id}_{\int\mathcal{I}_{X}}$. ∎

Remark 8.3*.*

Note that if $X$ is a finite $T_{0}$–space then $\textup{mxl}([)\int\mathcal{I}_{X}]=\{(U_{a},a)\mid a\in\textup{mxl}([)X]\}$.

Lemma 8.4**.**

Let $X$ be a finite $T_{0}$–space. For each $D\in\mathscr{U}(X)$ let $\widetilde{D}=\{(C,x)\in\int\mathcal{I}_{X}\mid C\subseteq D\}$.

1. (a)

Let $B\subseteq\textup{mxl}([)\int\mathcal{I}_{X}]$. Let $A=\{a\in\textup{mxl}([)X]\mid(U_{a},a)\in B\}$. Then

[TABLE] 2. (b)

Let $C\in\mathscr{U}(X)$. Then $\mathscr{U}(\iota_{X})(C)=\widetilde{C}$.

Proof.

($a$) Note that $B=\{(U_{a},a)\mid a\in A\}$ by the previous remark.

We will prove first that $U_{B}^{\int\mathcal{I}_{X}}=\bigcup\limits_{D\in\mathcal{C}(U_{A}^{X})}\widetilde{D}$.

Let $(C,x)\in U_{B}^{\int\mathcal{I}_{X}}$. Then $C\subseteq U_{a}$ for all $a\in A$. Hence, $C\subseteq U^{X}_{A}$, and since $C$ is connected, there exists $D\in\mathcal{C}(U_{A}^{X})$ such that $C\subseteq D$. Thus, $(C,x)\in\widetilde{D}$.

Now, let $D\in\mathcal{C}(U_{A}^{X})$ and let $(C,x)\in\widetilde{D}$. Then $D\subseteq U_{a}$ for all $a\in A$. Hence $(C,x)\leq(U_{a},a)$ for all $a\in A$. Thus, $(C,x)\in U_{B}^{\int\mathcal{I}_{X}}$.

We claim that if $D_{1},D_{2}$ are distinct elements of $\mathcal{C}(U_{A}^{X})$ then $\widetilde{D}_{1}\cap\widetilde{D}_{2}=\varnothing$. Indeed, if there exists $(C,x)\in\widetilde{D}_{1}\cap\widetilde{D}_{2}$, then $C\subseteq D_{1}\cap D_{2}$ which entails a contradiction since $D_{1}$ and $D_{2}$ are different connected components of $U^{X}_{A}$.

We will prove now that $\widetilde{D}$ is connected for all $D\in\mathcal{C}(U_{A}^{X})$. Let $D\in\mathcal{C}(U_{A}^{X})$. Note that

[TABLE]

Now, for each $x\in D$, $U^{\int\mathcal{I}_{X}}_{(D,x)}\cup\{D\}\times D$ is a connected subspace of $\int\mathcal{I}_{X}$ since $(D,x)\in U^{\int\mathcal{I}_{X}}_{(D,x)}\cap\{D\}\times D$ and $U^{\int\mathcal{I}_{X}}_{(D,x)}$ and $\{D\}\times D$ are connected. Therefore, $\widetilde{D}$ is also connected.

Finally, note that $\widetilde{D}$ is an open subset of $\int\mathcal{I}_{X}$ for all $D\in\mathcal{C}(U_{A}^{X})$. Since $\int\mathcal{I}_{X}$ is a finite space, the result follows.

($b$) Let $C\in\mathscr{U}(X)$. If $x\in C$ then $\iota_{X}(x)=({\bf C}_{\mathscr{U}}(x),x)\in\widetilde{C}$ since ${\bf C}_{\mathscr{U}}(x)\subseteq C$. Thus, $\iota_{X}(C)\subseteq\widetilde{C}$.

We will prove now that $\iota_{X}(C)^{\sharp}=\widetilde{C}^{\sharp}$. Clearly, $\iota_{X}(C)^{\sharp}\supseteq\widetilde{C}^{\sharp}$ since $\iota_{X}(C)\subseteq\widetilde{C}$. Let $M\in\iota_{X}(C)^{\sharp}$. Then there exists $a\in\textup{mxl}([)X]$ such that $M=(U_{a},a)$. Let $x\in C$. Then $\iota_{X}(x)=({\bf C}_{\mathscr{U}}(x),x)\leq(U_{a},a)$ and hence $x\in{\bf C}_{\mathscr{U}}(x)\subseteq U_{a}$. Therefore, $C\subseteq U_{a}$. Now, let $(C_{1},x_{1})\in\widetilde{C}$. Then $x_{1}\in C_{1}\subseteq C\subseteq U_{a}$. Hence, $(C_{1},x_{1})\leq(U_{a},a)=M$. Thus, $M\in\widetilde{C}^{\sharp}$.

Now, by 3.14, $\widetilde{C}$ is a connected component of $U^{\int\mathcal{I}_{X}}_{\widetilde{C}^{\sharp}}$. Since $\iota_{X}(C)^{\sharp}=\widetilde{C}^{\sharp}$ and $\iota_{X}(C)\subseteq\widetilde{C}$ we obtain that $\mathscr{U}(\iota_{X})(C)=\widetilde{C}$ by 3.13. ∎

Proposition 8.5**.**

Let $X$ be a finite $T_{0}$–space. Then $\mathscr{U}(\int\mathcal{I}_{X})\cong\mathscr{U}(X)$. Moreover, $\mathscr{U}(\iota_{X})\colon\mathscr{U}(X)\to\mathscr{U}(\int\mathcal{I}_{X})$ is a homeomorphism with inverse $\mathscr{U}(\rho_{X})$.

Proof.

Let $C\in\mathscr{U}(X)$. Then $\mathscr{U}(\iota_{X})(C)=\{(D,x)\in\int\mathcal{I}_{X}\mid D\subseteq C\}$ by the previous lemma. Thus,

[TABLE]

Hence, $\mathscr{U}(\rho_{X})(\mathscr{U}(\iota_{X})(C))=C$. Therefore $\mathscr{U}(\rho_{X})\mathscr{U}(\iota_{X})=\textup{id}_{\mathscr{U}(X)}$.

Observe that $\mathscr{U}(\iota_{X})$ is surjective by the previous lemma. The result follows. ∎

Definition 8.6**.**

Let $X$ be a finite $T_{0}$–space. We define $q_{X}\colon\int\mathcal{I}_{X}\to\mathscr{U}(X)$ by $q_{X}(C,x)=C$.

Proposition 8.7**.**

Let $X$ be a finite $T_{0}$–space.

1. (1)

$q_{X}$* is a quotient map.* 2. (2)

Let $\varphi$ and $\psi$ be the isomorphisms of proposition 7.5. Then the following diagram commutes

[TABLE]

From 7.1 and 7.3 we obtain that $\mathscr{U}(X)$ and $X$ are homotopy equivalent if every $C\in\mathscr{U}(X)$ has a maximum element. The following proposition provides a version of this result for weak equivalences.

Proposition 8.8**.**

Let $X$ be a finite $T_{0}$–space. Suppose that every $C\in\mathscr{U}(X)$ is weakly contractible. Then there exists a weak homotopy equivalence $X\to\mathscr{U}(X)$.

Proof.

Let $\mathcal{J}\colon\mathscr{U}(X)\to\mathcal{P}$ be the constant functor with value $\ast$, the one-element poset. Let $\alpha\colon\mathcal{I}_{X}\Rightarrow\mathcal{J}$ be the unique natural transformation. It follows that $\alpha_{C}$ is a weak homotopy equivalence for every $C\in\mathscr{U}(X)$ and hence, from [4, Corollary 2.6], we obtain that there exists a weak homotopy equivalence $\eta\colon\int\mathcal{I}_{X}\to\int\mathcal{J}$. Clearly $\int\mathcal{J}\cong\mathscr{U}(X)$. Besides, by 8.2, $X$ is homotopy equivalent to $\int\mathcal{I}_{X}$. The result follows. ∎