Types of Serre subcategories of Grothendieck categories

TL;DR

This paper classifies Serre subcategories of Grothendieck categories by assigning them unique types, explores their properties via recollements, and establishes the existence of subcategories with each type, revealing structural differences from triangulated categories.

Contribution

It introduces a classification of Serre subcategories in Grothendieck categories by types and demonstrates the existence of subcategories for each type, using recollements and torsion theories.

Findings

Each type corresponds to a unique Serre subcategory.

Recollements of abelian categories can be extended or split under certain conditions.

Not all properties hold in triangulated categories.

Abstract

Every Serre subcategory of an abelian category is assigned a unique type. The type of a Serre subcategory of a Grothendieck category is in the list: $$(0, 0), \ (0, -1), \ (1, -1), \ (0, -2), \ (1, -2), \ (2, -1), \ (+\infty, -\infty);$$ and for each $(m, -n)$ in this list, there exists a Serre subcategory such that its type is $(m, -n)$. This uses right (left) recollements of abelian categories, Tachikawa-Ohtake [TO] on strongly hereditary torsion pairs, and Geigle-Lenzing [GL] on localizing subcategories. If all the functors in a recollement of abelian categories are exact, then the recollement splits. Quite surprising, any left recollement of a Grothendieck category can be extended to a recollement; but this is not true for a right recollement. Thus, a colocalizing subcategory of a Grothendieck category is localizing; but the converse is not true. All these results do not hold in triangulated categories.

Equations

$$(0,0),(0,-1),(1,-1),(0,-2),(1,-2),(2,-1),(+\infty ,-\infty );$$

$$(F_m,\cdots ,F_1,F_0=i,F_{-1},\cdots ,F_{-n})\text{mbox}and(G_m,\cdots ,G_1,G_0=Q,G_{-1},\cdots ,G_{-n})$$

$$\displaystyle(F_{1},\$$

$$\displaystyle(G_{4},\ G_{3},\ G_{2},\ G_{1},\$$

$$(0,0),(0,-1),(1,-1),(0,-2),(1,-2),(2,-1),(+\infty ,-\infty );$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\to B_1⟶A⟶C⟶B_2\to 0$$

$$0\to B_1⟶C⟶A⟶B_2\to 0$$

$$i_{\ast }{i}^{!}{i}_{\ast }{B}^{\prime }\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\mathrm{Hom}(j^{\ast }X,j^{\ast }A)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$Q{j}_{\ast }F(QA)\cong Q{j}_{\ast }F(Q{j}_{\ast }{j}^{\ast }A)\cong Q{j}_{\ast }(j^{\ast }{j}_{\ast })j^{\ast }A\cong Q{j}_{\ast }{j}^{\ast }A\cong QA$$

$$0\to j^{\ast }{j}_{\ast }{C}_1\cong C_1\to j^{\ast }{j}_{\ast }C\cong C\to j^{\ast }{j}_{\ast }{C}_2\cong C_2\to j^{\ast }\mathrm{Coker}(j_{\ast }g)\to 0,$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\to i_{\ast }{i}^{!}{A}_1⟶i_{\ast }{i}^{!}A\stackrel{i_{\ast }{i}^{!}g}{⟶}{i}_{\ast }{i}^{!}{A}_2\to 0.$$

$$0\to i^{!}{A}_1⟶i^{!}A\stackrel{i^{!}g}{⟶}{i}^{!}{A}_2⟶\mathrm{Coker}(i^{!}g)\to 0,$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\mathcal{B}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$0\to T_1\stackrel{a}{⟶}G⟶C\stackrel{b}{⟶}{T}_2\to 0$$

$$0\to i_{\mathcal{G}}G⟶j_{!}{Q}_{\mathcal{G}}A⟶A⟶i_{\mathcal{G}}{i}^{\ast }A\to 0$$

$$i_1=-{\otimes }_{\Lambda }\Lambda /\Lambda e_2\Lambda =-{\otimes }_{\Lambda }\left(\begin{array}{c}R\\ 0\end{array}\right),$$

$$i_0={\mathrm{Hom}}_{\Lambda /\Lambda e_2\Lambda }(\Lambda /\Lambda e_2\Lambda ,-)={\mathrm{Hom}}_R(R,-)\cong -{\otimes }_{R}R,$$

$$i_{-1}={\mathrm{Hom}}_{\Lambda }(\Lambda /\Lambda e_2\Lambda ,-)={\mathrm{Hom}}_{\Lambda }(e_1\Lambda ,-)\cong -{\otimes }_{\Lambda }\Lambda e_1,$$

$$i_{-2}={\mathrm{Hom}}_{\Lambda /\Lambda e_2\Lambda }(\Lambda e_1,-)={\mathrm{Hom}}_R(\left(\begin{array}{c}R\\ M\end{array}\right),-),$$

$$j_1=-{\otimes }_{e_2\Lambda e_2}{e}_2\Lambda =-{\otimes }_S(M,S),$$

$$j_0={\mathrm{Hom}}_{\Lambda }(e_2\Lambda ,-)={\mathrm{Hom}}_{\Lambda }((M,S),-)\cong -{\otimes }_{\Lambda }\Lambda e_2,$$

$$j_{-1}={\mathrm{Hom}}_{e_2\Lambda e_2}(\Lambda e_2,-)={\mathrm{Hom}}_S(S,-)\cong -{\otimes }_{S}S,$$

$$j_{-2}={\mathrm{Hom}}_{\Lambda }(S,-),$$

$$(i_1,i_0,i_{-1},i_{-2}\cong j_1,j_0,j_{-1},j_{-2})$$

$$(3,0),(2,-1),(1,-2),(0,-3),(2,0),(1,-1),(0,-2),(1,0),(0,-1)(0,0).$$

$$(+\infty ,-\infty ),(2,-1),(1,-2),(1,-1),(0,-2),(0,-1),(0,0)$$

Full text

[TABLE]

and for each $(m,-n)$ in this list, there exists a Serre subcategory such that its type is $(m,-n)$. This uses right (left) recollements of abelian categories, Tachikawa-Ohtake [TO] on strongly hereditary torsion pairs, and Geigle-Lenzing [GL] on localizing subcategories. If all the functors in a recollement of abelian categories are exact, then the recollement splits. Quite surprising, any left recollement of a Grothendieck category can be extended to a recollement; but this is not true for a right recollement. Thus, a colocalizing subcategory of a Grothendieck category is localizing; but the converse is not true. All these results do not hold in triangulated categories. Key words and phrases.  the type of a Serre subcategory, right recollement, strongly hereditary torsion pair, quotient functor, localizing subcategory, Grothendieck category

2010 Mathematical Subject Classification.   18E40, 18E15, 16G10, 18E35.

∗The corresponding author.   fengjian008@sina.com $($J. Feng$)$,   pzhang@sjtu.edu.cn $($P. Zhang$)$.

Supported by the NSFC 11271251 and 11431010.

1. Introduction

Given a Serre subcategory $\mathcal{S}$ of an abelian category $\mathcal{A}$ with inclusion functor $i:\mathcal{S}\rightarrow\mathcal{A}$ and quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/\mathcal{S}$, it is fundamental to know when it is localizing (resp. colocalizing), i.e., $Q$ has a right (resp. left) adjoint ([S], [G]). By W. Geigle and H. Lenzing [GL], $\mathcal{S}$ is localizing if and only if there exists an exact sequence $0\rightarrow S_{1}\rightarrow A\rightarrow C\rightarrow S_{2}\rightarrow 0$ with $S_{1}\in\mathcal{S},\ S_{2}\in\mathcal{S},$ and $C\in\mathcal{S}^{\perp_{\leq 1}};$ and if and only if the restriction $Q:\mathcal{S}^{\perp_{\leq 1}}\rightarrow\mathcal{A}/\mathcal{S}$ is an equivalence of categories. In this case the right adjoint of $Q$ is fully faithful. There is a corresponding work for a thick triangulated subcategory of a triangulated category (A. Neeman [N, Chap. 9]).

It is then natural to describe Serre subcategories of a fixed abelian category via the length of two adjoint sequences where $i$ and $Q$ lie. A finite or an infinite sequence  $(\cdots,F_{1},F_{0},F_{-1},\cdots)$  of functors between additive categories is an adjoint sequence, if each pair $(F_{i},F_{i-1})$ is an adjoint pair. Each functor in an adjoint sequence is additive.

Let $\mathcal{S}$ be a Serre subcategory of an abelian category $\mathcal{A}$ with the inclusion functor $i:\mathcal{S}\rightarrow\mathcal{A}$ and the quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/\mathcal{S}$. The pair $(\mathcal{S},i)$ is of type $(m,-n)$, or in short, the Serre subcategory $\mathcal{S}$ is of type $(m,-n)$, where $m$ and $n$ are in the set $\mathbb{N}_{0}\cup\{+\infty\}$, and $\mathbb{N}_{0}$ is the set of non-negative integers, provided that there exist adjoint sequences

[TABLE]

such that $F_{m}$ and $G_{m}$ can not have left adjoints at the same time, and that $F_{-n}$ and $G_{-n}$ can not have right adjoints at the same time.

We stress that the type of $\mathcal{S}$ depends on the abelian category $\mathcal{A}$ in which $\mathcal{S}$ is a Serre subcategory. Since in an adjoint pair one functor uniquely determines the other, every Serre subcategory is of a unique type $(m,-n)$. We will see a Serre subcategory $\mathcal{S}$ of type $(1,-2)$, but with adjoint sequences

[TABLE]

See Remark 5.5.

A right recollement $(\mathcal{B},\ \mathcal{A},\ \mathcal{C},\ i_{*},\ i^{!},\ j^{*},\ j_{*})$ (see e.g. [P], [Kö]) of abelian categories is a diagram

\mathcal{B}$$\mathcal{A}$$\mathcal{C}(1.1)i_{\ast}$$i^{!}$$j^{\ast}$$j_{\ast}

of functors between abelian categories such that

${\rm(i)}$  $i_{*}$ and $j^{*}$ are exact functors;

${\rm(ii)}$  $i_{*}$ and $j_{*}$ are fully faithful;

${\rm(iii)}$   $(i_{*},i^{!})$ and $(j^{*},j_{*})$ are adjoint pairs; and

${\rm(iv)}$   ${\rm Im}i_{*}={\rm Ker}j^{*}$ (and thus $i^{!}j_{*}=0$).

In a right recollement $(1.1)$ the functor $i^{!}$ and $j_{*}$ are left exact but not exact, in general. A right recollement is also called a localization sequence e.g. in [S], [G], [IKM], and [Kr], and a step in [BGS].

A left recollement $(\mathcal{B},\ \mathcal{A},\ \mathcal{C},\ i^{*},\ i_{*},\ j_{!},\ j^{*})$ of abelian categories is a diagram

\mathcal{B}$$\mathcal{A}$$\mathcal{C}(1.2)i^{*}$$i_{*}$$j_{!}$$j^{*}

of functors between abelian categories such that

${\rm(i)}$    $i_{*}$ and $j^{*}$ are exact;

${\rm(ii)}$  $i_{*}$ and $j_{!}$ are fully faithful;

${\rm(iii)}$   $(i^{*},i_{*})$ and $(j_{!},j^{*})$ are adjoint pairs; and

${\rm(iv)}$   ${\rm Im}i_{*}={\rm Ker}j^{*}$ (and thus $i^{*}j_{!}=0$).

Note that in a left recollement $(1.2)$ the functor $i^{*}$ and $j_{!}$ are right exact but not exact, in general. Thus, given a right recollement $(\mathcal{B},\ \mathcal{A},\ \mathcal{C},\ i_{*},\ i^{!},\ j^{*},\ j_{*})$, the data $(\mathcal{C},\ \mathcal{A},\ \mathcal{B},\ j^{*},\ j_{*},\ \ i_{*},\ i^{!})$ is not a left recollement in general (similar remark for a left recollement).

A recollement is first introduced for triangulated categories by A. Beilinson, J. Bernstein, and P. Deligne [BBD]. A recollement of abelian categories appeared in [Ku] and [CPS]. A recollement $(\mathcal{B},\ \mathcal{A},\ \mathcal{C},\ i^{*},\ i_{*},\ i^{!},\ j_{!},\ j^{*},\ j_{*})$ of abelian categories is a diagram

\mathcal{B}$$\mathcal{A}$$\mathcal{C}$$(1.3)$$i_{\ast}$$j^{\ast}$$i^{!}$$j_{*}$$i^{*}$$j_{!}

of functors between abelian categories such that

${\rm(i)}$   $(i^{*},i_{*}),(i_{*},i^{!})$, $(j_{!},j^{*})$ and $(j^{*},j_{*})$ are adjoint pairs;

${\rm(ii)}$   $i_{*},j_{!}$ and $j_{*}$ are fully faithful; and

${\rm(iii)}$  ${\rm Im}i_{*}={\rm Ker}j^{*}$.

Thus in a recollement $(1.3)$ the functors $i_{*}$ and $j^{*}$ are exact. So $(1.3)$ is a recollement if and only if the upper two rows is a left recollement and the lower two rows is a right recollement. By V. Franjou and T. Pirashvili [FP], recollements of abelian categories have some different properties from recollement of triangulated categories. For example, ${\rm Ker}i^{*}\neq{\rm Im}j_{!}$ and ${\rm Ker}i^{!}\neq{\rm Im}j_{*}$ in general, and Parshall-Scott’s theorem on comparison between two recollements of triangulated categories ([PS, Thm. 2.5]) does not hold in general. See also [Ps, PV, GYZ].

In a recollement of abelian categories, if $i^{*}$ and $i^{!}$ are exact, then $j_{!}$ and $j_{*}$ are also exact (see Prop. 3.1 and 3.2). The following result describes recollements of abelian categories with exact functors.

Theorem 1.1**.**

Given a recollement $(1.3)$ of abelian categories. If $i^{*}$ and $i^{!}$ are exact, then $i^{*}\cong i^{!}$ and $j_{!}\cong j_{*},$ and $\mathcal{A}\cong\mathcal{B}\oplus\mathcal{C}$.

Quite surprising, we have

Theorem 1.2**.**

Assume that $\mathcal{A}$ is a Grothendieck category. Then any left recollement $(1.2)$ of abelian categories can be extended to a recollement of abelian categories.

As a consequence, a colocalizing subcategory of a Grothendieck category is localizing. We stress that a right recollement of abelian categories does not necessarily extend to a recollement, and that a localizing subcategory of a Grothendieck category is not necessarily colocalizing. See Subsection 5.2. On the other hand, W. Geigle and H. Lenzing [GL, Prop. 5.3] have proved that any Serre subcategory $\mathcal{S}$ of the finitely generated module category of an Artin algebra is always localizing and colocalizing.

Theorem 1.3**.**

The type of a Serre subcategory of a Grothendieck category $\mathcal{A}$ is in the list

[TABLE]

and for each $(m,-n)$ in this list, there exists a Serre subcategory such that its type is $(m,-n);$ and if a Serre subcategory $\mathcal{S}$ is of type $(+\infty,-\infty),$ then $\mathcal{A}\cong\mathcal{S}\oplus(\mathcal{A}/\mathcal{S})$ as categories.

The main tools for proving Theorem 1.3 are the work of strongly hereditary torsion pairs by H. Tachikawa and K. Ohtake [TO; O], the work of localizing subcategories by Geigle-Lenzing [GL], and the argument on right (left) recollements of abelian categories, especially Theorems 1.1 and 1.2. This result could also be reformulated in terms of the height of a ladder of a Grothendieck category (see [BGS], [AHKLY], [ZZZZ]).  Theorems 1.1, 1.2 and 1.3 do not hold in triangulated categories.

2. Preliminaries

Throughout $\mathcal{A}$ is an abelian category. A subcategory means a full subcategory closed under isomorphisms. We will use the properties of a Grothendieck category $\mathcal{A}$: it is well-powered ([M]) in the sense that for each object $A\in\mathcal{A}$, the class of the subobjects of $A$ forms a set; $\mathcal{A}$ has coproducts and products, enough injective objects; and the canonical morphism from a coproduct to the corresponding product is a monomorphism (see [F], [Mit]).

2.1. Serre subcategories

For Serre subcategories and quotient categories we refer to [G], [Pop], and [GL]. A subcategory $\mathcal{S}$ of $\mathcal{A}$ is a Serre subcategory if $\mathcal{S}$ is closed under subobjects, quotient objects, and extensions. If $\mathcal{S}$ is a Serre subcategory of $\mathcal{A}$ with the inclusion functor $i:\mathcal{S}\rightarrow\mathcal{A}$, then we have the quotient category $\mathcal{A}/\mathcal{S}$ which is abelian, and the quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/\mathcal{S}$ is exact with $Qi=0$, and $Q$ has the universal property in the sense that if $F:\mathcal{A}\rightarrow\mathcal{C}$ is an exact functor between abelian categories with $Fi=0$, then there exists a unique exact functor $G:\mathcal{A}/\mathcal{S}\rightarrow\mathcal{C}$ such that $F=GQ$.

A Serre subcategory $\mathcal{S}$ is localizing, if the quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/\mathcal{S}$ has a right adjoint $j_{*}$. In this case, $j_{*}$ is fully faithful ([GL, Prop. 2.2]). Dually, a Serre subcategory $\mathcal{S}$ is colocalizing, if $Q$ has a left adjoint $j_{!}$. In this case, $j_{!}$ is fully faithful (the dual of [GL, Prop. 2.2]).

2.2. Exact sequences of abelian categories

A sequence $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\longrightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\longrightarrow}}\mathcal{C}\rightarrow 0$ of exact functors between abelian categories is an exact sequence, provided that there exists a Serre subcategory $\mathcal{S}$ of an abelian category $\mathcal{A}^{\prime}$ such that there is a commutative diagram

[TABLE]

A sequence $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\longrightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\longrightarrow}}\mathcal{C}\rightarrow 0$ of exact functors between abelian categories is an exact sequence if and only if $i_{*}$ is fully faithful, $i_{*}\mathcal{B}$ is a Serre subcategory of $\mathcal{A}$, $j^{*}i_{*}=0$, and $j^{*}$ has also the universal property. In this case, we have ${\rm Im}i_{*}={\rm Ker}j^{*}$.

2.3. Torsion pairs

For torsion pairs in an abelian category we refer to [D], [J], and [TO]. A pair $(\mathcal{T},\mathcal{F})$ of subcategories of $\mathcal{A}$ is a torsion pair ([D]), if ${\rm Hom}(T,F)=0$ for $T\in\mathcal{T}$ and $F\in\mathcal{F}$, and for each object $A\in\mathcal{A}$, there is an exact sequence $0\rightarrow T\rightarrow A\rightarrow F\rightarrow 0$ with $T\in\mathcal{T}$ and $F\in\mathcal{F}$. In this case, the exact sequence is called the $t$-decomposition of $A$ with respect to $(\mathcal{T},\mathcal{F})$. A subcategory $\mathcal{T}$ (resp. $\mathcal{F}$) is a torsion class (resp. a torsionfree class) if there exists a subcategory $\mathcal{F}$ (resp. $\mathcal{T}$) such that $(\mathcal{T},\mathcal{F})$ is a torsion pair. If $(\mathcal{T},\mathcal{F})$ is a torsion pair, then $\mathcal{F}=\mathcal{T}^{\perp_{0}}$ and $\mathcal{T}=\ ^{\perp_{0}}\mathcal{F},$ where $\mathcal{T}^{\perp_{0}}:=\{A\in\mathcal{A}\ |\ {\rm Hom}(T,A)=0,\ \forall\ T\in\mathcal{T}\}$, and ${}^{\perp_{0}}\mathcal{F}$ is dually defined. By S. E. Dickson [D, Thm.2.3], if $\mathcal{A}$ is a well-powered abelian category with coproducts and products, then a subcategory $\mathcal{T}$ (resp. $\mathcal{F}$) is a torsion class (resp. a torsionfree class) if and only if $\mathcal{T}$ (resp. $\mathcal{F}$) is closed under quotient objects, extensions, and coproducts (resp. under subobjects, extensions, and products).

A subcategory $\mathcal{B}$ is weakly localizing, provided that for each object $A$ of $\mathcal{A}$, there exists an exact sequence

[TABLE]

with $B_{1}\in\mathcal{B},\ B_{2}\in\mathcal{B},$ and $C\in\mathcal{B}^{\perp_{\leq 1}}:=\{A\in\mathcal{A}\ |\ {\rm Hom}(B,A)=0={\rm Ext}^{1}(B,A),\ \forall\ B\in\mathcal{B}\}.$ By W. Geigle and H. Lenzing [GL, Prop. 2.2], a Serre subcategory $\mathcal{S}$ is localizing if and only if it is weakly localizing. Dually, a subcategory $\mathcal{B}$ is weakly colocalizing, provided that for each object $A$ of $\mathcal{A}$, there exists an exact sequence

[TABLE]

with $B_{1}\in\mathcal{B},\ B_{2}\in\mathcal{B},$ and $C\in\ ^{\perp_{\leq 1}}\mathcal{B}:=\{A\in\mathcal{A}\ |\ {\rm Hom}(A,B)=0={\rm Ext}^{1}(A,B),\ \forall\ B\in\mathcal{B}\}.$ By the dual of [GL, Prop. 2.2], a Serre subcategory $\mathcal{S}$ is colocalizing if and only if it is weakly colocalizing.

Following H. Tachikawa and K. Ohtake [TO], a torsion pair $(\mathcal{T},\mathcal{F})$ is hereditary (resp. cohereditary), if $\mathcal{T}$ (resp. $\mathcal{F}$) is closed under subobjects (resp. quotient objects); and it is strongly hereditary (resp. strongly cohereditary), if $\mathcal{T}$ (resp. $\mathcal{F}$) is weakly localizing (resp. weakly colocalizing). Every strongly hereditary (resp. strongly cohereditary) torsion pair is hereditary (resp. cohereditary) (see [TO, Prop. 1.7∗, 1.7]). K. Ohtake [O, Thm. 2.6, 1.6] has proved that if $\mathcal{A}$ has enough injective objects (resp. enough projective objects), then every hereditary (resp. cohereditary) torsion pair is also strongly hereditary (resp. strongly cohereditary) (see also [TO, Thm. 1.8∗, 1.8]).

2.4.

Let $F:\mathcal{C}\rightarrow\mathcal{A}$ be a fully faithful functor between abelian categories. We say that $F$ is Giraud if $F$ has a left adjoint which is an exact functor. Dually, $F$ is coGiraud if $F$ has a right adjoint which is exact. By [TO, Coroll. 3.8, 2.8], $F$ is Giraud (resp. coGiraud) if and only if $F\mathcal{C}$ is a Giraud subcategory (a coGiraud subcategory) of $\mathcal{A}$ in the sense of [TO].

Lemma 2.1**.**

Let $\mathcal{A}$ be an abelian category.

${\rm(1)}$*  Given a Giraud functor $j_{*}:\mathcal{C}\rightarrow\mathcal{A}$ with an exact left adjoint $j^{*}:\mathcal{A}\rightarrow\mathcal{C}$, there exists a functor $i^{!}:\mathcal{A}\rightarrow{\rm Ker}j^{*}$, such that $({\rm Ker}j^{*},\ \mathcal{A},\ \mathcal{C},\ i,\ i^{!},\ j^{*},\ j_{*})$ is a right recollement, where $i:{\rm Ker}j^{*}\rightarrow\mathcal{A}$ is the inclusion functor.*

${\rm(1^{\prime})}$*  Given a coGiraud functor $j_{!}:\mathcal{C}\rightarrow\mathcal{A}$ with exact right adjoint $j^{*}:\mathcal{A}\rightarrow\mathcal{C}$, there exists a functor $i^{*}:\mathcal{A}\rightarrow{\rm Ker}j^{*}$, such that $({\rm Ker}j^{*},\ \mathcal{A},\ \mathcal{C},\ i^{*},\ i,\ j_{!},\ j^{*})$ is a left recollement, where $i:{\rm Ker}j^{*}\rightarrow\mathcal{A}$ is the inclusion functor.*

$(2)$*  Let $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\longrightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\longrightarrow}}\mathcal{C}\rightarrow 0$ be an exact sequence of abelian categories. If $j^{*}$ has a right adjoint $j_{*}$, then $j_{*}$ is fully faithful, and there exists a functor $i^{!}:\mathcal{A}\rightarrow{\rm Ker}j^{*}$ such that $({\rm Ker}j^{*},\ \mathcal{A},\ \mathcal{C},\ i,\ i^{!},\ j^{*},\ j_{*})$ is a right recollement, where $i:{\rm Ker}j^{*}\rightarrow\mathcal{A}$ is the inclusion functor.*

${\rm(2^{\prime})}$*  Let $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\longrightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\longrightarrow}}\mathcal{C}\rightarrow 0$ be an exact sequence of abelian categories. If $j^{*}$ has a left adjoint $j_{!}$, then $j_{!}$ is fully faithful, and there exists a functor $i^{*}:\mathcal{A}\rightarrow{\rm Ker}j^{*}$ such that $({\rm Ker}j^{*},\ \mathcal{A},\ \mathcal{C},\ i^{*},\ i,\ j_{!},\ j^{*})$ is a left recollement, where $i:{\rm Ker}j^{*}\rightarrow\mathcal{A}$ is the inclusion functor.*

${\rm(3)}$*  If $(\mathcal{B},\mathcal{C})$ is a strongly hereditary torsion pair in $\mathcal{A}$. Then $\mathcal{B}$ is a Serre subcategory of $\mathcal{A}$ and there is a right recollement of abelian category*

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i$$i^{!}$$Q$$j_{\ast}

with ${\rm Im}j_{*}=B^{\perp_{\leq 1}}$, where $i$ is the inclusion functor and $Q$ is the quotient functor.

${\rm(3^{\prime})}$*  If $(\mathcal{B},\mathcal{C})$ is a strongly cohereditary torsion pair in $\mathcal{A}$. Then $\mathcal{C}$ is a Serre subcategory of $\mathcal{A}$ and there is a left recollement of abelian category*

\mathcal{C}$$\mathcal{A}$$\mathcal{A}/\mathcal{C}$$i^{*}$$i$$j_{!}$$Q

with ${\rm Im}j_{!}\cong\ ^{\perp_{\leq 1}}C$ as categories, where $i$ is the inclusion functor and $Q$ is the quotient functor.

Proof. We only prove $(1)$, (2) and $(3)$. The assertions $(1^{\prime})$, (2’) and $(3^{\prime})$ can be dually proved.

$(1)$  By assumption $j^{*}$ is exact, thus ${\rm Ker}j^{*}$ is an abelian category and the inclusion functor $i:{\rm Ker}j^{*}\rightarrow\mathcal{A}$ is exact. We claim that $i$ admits a right adjoint $i^{!}:\mathcal{A}\rightarrow{\rm Ker}j^{*}$. In fact, for any $A\in\mathcal{A}$, there is an exact sequence $0\rightarrow{\rm Ker}\zeta_{A}\rightarrow A\stackrel{{\scriptstyle\zeta_{A}}}{{\rightarrow}}j_{*}j^{*}A$, where $\zeta:{\rm Id}_{\mathcal{A}}\rightarrow j_{*}j^{*}$ is the unit of the adjoint pair $(j^{*},j_{*})$. Put $i^{!}A:={\rm Ker}\zeta_{A}$. Since $j^{*}$ is exact and $j^{*}\zeta_{A}$ is an isomorphism, ${\rm Ker}\zeta_{A}\in{\rm Ker}j^{*}$. Thus, $i^{!}:\mathcal{A}\rightarrow{\rm Ker}j^{*}$ defines a functor. For $B\in{\rm Ker}j^{*}$ and $A\in\mathcal{A}$, since ${\rm Hom}_{\mathcal{A}}(B,j_{*}j^{*}A)\cong{\rm Hom}_{\mathcal{C}}(j^{*}B,j^{*}A)=0$, by applying the left exact functor ${\rm Hom}_{\mathcal{A}}(B,-)$ to the exact sequence we get an isomorphism ${\rm Hom}(B,{\rm Ker}\zeta_{A})\cong{\rm Hom}(iB,A)$. This proves the claim, and hence $({\rm Ker}j^{*},\ \mathcal{A},\ \mathcal{C},\ i,\ i^{!},\ j^{*},\ j_{*})$ a right recollement.

$(2)$  Since $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\longrightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\longrightarrow}}\mathcal{C}\rightarrow 0$ is an exact sequence of abelian categories, without loss of generality, one may regard $j^{*}$ just as the quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/i_{*}\mathcal{B}$. By [GL, Prop. 2.2], $j_{*}$ is fully faithful. Thus $j_{*}:\mathcal{C}\rightarrow\mathcal{A}$ is a Giraud functor with the exact left adjoint $j^{*}:\mathcal{A}\rightarrow\mathcal{C}$, and hence the assertion follows from ${\rm(1)}$.

$(3)$  By [TO, Prop. 1.7∗], $(\mathcal{B},\mathcal{C})$ is a hereditary torsion pair. Thus $\mathcal{B}$ is a Serre subcategory. Since by assumption $\mathcal{B}$ is weakly localizing, by [GL, Prop. 2.2], the quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/\mathcal{B}$ admits a right adjoint $j_{*}:\mathcal{A}/\mathcal{B}\rightarrow\mathcal{A}$ which is fully faithful, and ${\rm Im}j_{*}=B^{\perp_{\leq 1}}$. So $j_{*}$ is a Giraud functor with an exact left adjoint $Q$. By $(1)$, there exists a functor $i^{!}:\mathcal{A}\rightarrow{\rm Ker}Q=\mathcal{B}$, such that $(\mathcal{B},\ \mathcal{A},\ \mathcal{A}/\mathcal{B},\ i,\ i^{!},\ Q,\ j_{*})$ is a right recollement. $\hfill\blacksquare$

3. **Recollements of

abelian categories with exact functors**

3.1.

The following proposition gives the properties of a right recollement of abelian categories we need. Some of them are well-known for recollements of abelian categories (see [FP], [Ps], [PV]).

Proposition 3.1**.**

Let $(1.1)$ be a right recollement of abelian categories. Then

$(1)$*  ${\rm Im}i_{*}$ is a weakly localizing subcategory. Explicitly, for each object $A\in\mathcal{A}$, there is an exact sequence $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A\longrightarrow i_{*}B\rightarrow 0$ for some object $B\in\mathcal{B}$, with $j_{*}j^{*}A\in({\rm Im}i_{*})^{\perp_{\leq 1}}$, where $\omega$ the counit and $\zeta$ is the unit.*

$(2)$*  $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\rightarrow}}\mathcal{C}\rightarrow 0$ is an exact sequence of abelian categories.*

$(3)$*  ${\rm Ker}i^{!}=({\rm Im}i_{*})^{\perp_{0}};$ $({\rm Im}i_{*},{\rm Ker}i^{!})$ is a strongly hereditary torsion pair in $\mathcal{A},$ and $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\rightarrow{\rm Coker}\omega_{A}\rightarrow 0$ is the $t$-decomposition of $A$.*

${\rm(4)}$*  The following are equivalent$:$*

${\rm(i)}$*   $i^{!}$ is exact;*

${\rm(ii)}$*   $i^{!}$ and $j_{*}$ are exact;*

${\rm(iii)}$*   $0\rightarrow\mathcal{C}\stackrel{{\scriptstyle j_{*}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle i^{!}}}{{\rightarrow}}\mathcal{B}\rightarrow 0$ is an exact sequence of abelian categories;*

${\rm(iv)}$*   the sequence $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A\rightarrow 0$ is exact for each object $A\in\mathcal{A};$*

${\rm(v)}$*   ${\rm Im}j_{*}={\rm Ker}i^{!}$, $({\rm Im}i_{*},{\rm Im}j_{*})$ is a cohereditary torsion pair in $\mathcal{A},$ and $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A\rightarrow 0$ is the $t$-decomposition of $A;$*

${\rm(vi)}$*   $({\rm Im}i_{*},{\rm Im}j_{*})$ is a hereditary and cohereditary torsion pair in $\mathcal{A};$*

${\rm(vii)}$*  $({\rm Im}i_{*},{\rm Im}j_{*})$ is a strongly hereditary and strongly cohereditary torsion pair in $\mathcal{A};$*

${\rm(viii)}$*  $({\rm Im}i_{*},{\rm Im}j_{*})$ is a strongly cohereditary torsion pair in $\mathcal{A}$.*

Proof. $(1)$   Applying the exact functor $j^{*}$ to the exact sequence $0\rightarrow{\rm Ker}\zeta_{A}\rightarrow A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A\rightarrow{\rm Coker}\zeta_{A}\rightarrow 0$, we get an exact sequence $0\rightarrow j^{*}{\rm Ker}\zeta_{A}\rightarrow j^{*}A\stackrel{{\scriptstyle j^{*}\zeta_{A}}}{{\longrightarrow}}j^{*}j_{*}j^{*}A\rightarrow j^{*}{\rm Coker}\zeta_{A}\rightarrow 0.$ Since $j_{*}$ fully faithful, $j^{*}\zeta_{A}$ is an isomorphism. So $j^{*}{\rm Ker}\zeta_{A}=0=j^{*}{\rm Coker}\zeta_{A}$, and then ${\rm Ker}\zeta_{A}\cong i_{*}B^{\prime}$ and ${\rm Coker}\zeta_{A}\cong i_{*}B$ for some $B^{\prime}\in\mathcal{B}$ and $B\in\mathcal{B}$. Applying the left exact functor $i_{*}i^{!}$ to the exact sequence $0\rightarrow i_{*}B^{\prime}\rightarrow A\stackrel{{\scriptstyle\zeta_{A}}}{{\rightarrow}}j_{*}j^{*}A$, we get a commutative diagram

[TABLE]

Since $i_{*}$ is fully faithful, $\omega_{i_{*}B^{\prime}}$ is an isomorphism, and hence ${\rm Ker}\zeta_{A}\cong i_{*}B^{\prime}\cong i_{*}i^{!}A.$ So we get the desired exact sequence.

To see ${\rm Im}i_{*}$ is weakly localizing, by the exact sequence just established, it suffices to prove $j_{*}j^{*}A\in({\rm Im}i_{*})^{\perp_{\leq 1}}$ for each object $A\in\mathcal{A}$. We only need to show ${\rm Ext}^{1}(i_{*}B,j_{*}j^{*}A)=0$ for $B\in\mathcal{B}$. Let $0\rightarrow j_{*}j^{*}A\stackrel{{\scriptstyle a}}{{\longrightarrow}}X\rightarrow i_{*}B\rightarrow 0$ be an exact sequence. Applying the exact functor $j^{*}$ we get an isomorphism $j^{*}a:j^{*}j_{*}j^{*}A\cong j^{*}X$. By the commutative diagram

[TABLE]

we see that $\operatorname{Hom}(a,j_{*}j^{*}A):\operatorname{Hom}(X,j_{*}j^{*}A)\rightarrow\operatorname{Hom}(j_{*}j^{*}A,j_{*}j^{*}A)$ is an isomorphism, which implies that the exact sequence $0\rightarrow j_{*}j^{*}A\stackrel{{\scriptstyle a}}{{\longrightarrow}}X\rightarrow i_{*}B\rightarrow 0$ splits.

$(2)$   Since ${\rm Im}i_{*}={\rm Ker}j^{*}$ and $j^{*}$ is exact, $i_{*}\mathcal{B}$ is a Serre subcategory of $\mathcal{A}$. Since $j^{*}i_{*}=0$, by the universal property of the quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/i_{*}\mathcal{B}$ we get a unique exact functor $F:\mathcal{A}/i_{*}\mathcal{B}\rightarrow\mathcal{C}$ such that $FQ\cong j^{*}$. We claim that $Qj_{*}$ is a quasi-inverse of $F$. In fact, $FQj_{*}\cong j^{*}j_{*}\cong{\rm Id}_{\mathcal{C}}$; on the other hand, for each object $A\in\mathcal{A}$, by $(1)$ there is a functorial isomorphism $QA\cong Qj_{*}j^{*}A$, and hence for each object $QA\in\mathcal{A}/i_{*}\mathcal{B}$ there are functorial isomorphisms

[TABLE]

in $\mathcal{A}/\mathcal{B}$, i.e., $Qj_{*}F\cong{\rm Id}_{\mathcal{A}/i_{*}\mathcal{B}}$.

(3)  For $A\in({\rm Im}i_{*})^{\perp_{0}}$, we have $\operatorname{Hom}_{\mathcal{B}}(i^{!}A,i^{!}A)\cong\operatorname{Hom}_{\mathcal{B}}(i_{*}i^{!}A,A)=0$, thus $i^{!}A=0$. So $({\rm Im}i_{*})^{\perp_{0}}\subseteq{\rm Ker}i^{!}.$ Conversely, if $A\in{\rm Ker}i^{!}$, then for each $B\in\mathcal{B}$ we have $\operatorname{Hom}_{\mathcal{A}}(i_{*}B,A)\cong\operatorname{Hom}_{\mathcal{B}}(B,i^{!}A)=0.$ So ${\rm Ker}i^{!}\subseteq({\rm Im}i_{*})^{\perp_{0}}.$ This shows ${\rm Ker}i^{!}=({\rm Im}i_{*})^{\perp_{0}}$.

For $A\in\mathcal{A},$ considering the exact sequence $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A$ given in (1), we get an exact sequence $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\rightarrow{\rm Coker}\omega_{A}\rightarrow 0$. To see $({\rm Im}i_{*},{\rm Ker}i^{!})$ is a torsion pair in $\mathcal{A},$ it suffices to show ${\rm Coker}\omega_{A}\in{\rm Ker}i^{!}.$ We see this by applying the left exact functor $i^{!}$ to the exact sequence $0\rightarrow{\rm Coker}\omega_{A}\rightarrow j_{*}j^{*}A$, and using $i^{!}j_{*}=0$. By (1), ${\rm Im}i_{*}$ is a weakly localizing subcategory. Thus $({\rm Im}i_{*},{\rm Ker}i^{!})$ is a strongly hereditary torsion pair.

(4)  ${\rm(i)}\Rightarrow{\rm(ii)}:$ Applying the left exact functor $j_{*}$ to a given exact sequence $0\rightarrow C_{1}\stackrel{{\scriptstyle f}}{{\longrightarrow}}C\stackrel{{\scriptstyle g}}{{\longrightarrow}}C_{2}\rightarrow 0$ in $\mathcal{C}$, we get an exact sequence $0\rightarrow j_{*}C_{1}\stackrel{{\scriptstyle j_{*}f}}{{\longrightarrow}}j_{*}C\stackrel{{\scriptstyle j_{*}g}}{{\longrightarrow}}j_{*}C_{2}\rightarrow{\rm Coker}(j_{*}g)\rightarrow 0$; then by applying the exact functor $i^{!}$ we see $i^{!}{\rm Coker}(j_{*}g)=0$ (since $i^{!}j_{*}C_{2}=0$). Applying the exact functor $j^{*}$ we get an exact sequence

[TABLE]

and thus $j^{*}{\rm Coker}(j_{*}g)=0$. So ${\rm Coker}(j_{*}g)=i_{*}B$ for some $B\in\mathcal{B}$, and hence $0=i^{!}{\rm Coker}(j_{*}g)=i^{!}i_{*}B\cong B$. Thus ${\rm Coker}(j_{*}g)=i_{*}B=0$, which proves the exactness of $j_{*}$.

${\rm(ii)}\Rightarrow{\rm(iii)}:$ We first claim ${\rm Im}j_{*}={\rm Ker}i^{!}$. It is clear that ${\rm Im}j_{*}\subseteq{\rm Ker}i^{!}$. For each object $A\in{\rm Ker}i^{!}$, by $(1)$ we have an exact sequence $0\rightarrow A\rightarrow j_{*}j^{*}A\rightarrow i_{*}B\rightarrow 0;$ applying the exact functor $i^{!}$ we see that $i^{!}i_{*}B=0,$ and hence $B\cong i^{!}i_{*}B=0$. So $A\cong j_{*}j^{*}A\in{\rm Im}j_{*}$. This proves the claim. Thus ${\rm Im}j_{*}={\rm Ker}i^{!}$ is a Serre subcategory. It remains to prove that $i^{!}$ has the universal property. For this, assume that $F:\mathcal{A}\rightarrow\mathcal{B}^{\prime}$ is an exact functor such that $Fj_{*}=0$. Applying $F$ to the exact sequence in $(1)$ we get a functorial isomorphism $Fi_{*}i^{!}(A)\cong F(A)$ for each object $A\in\mathcal{A}$, i.e., $(Fi_{*})i^{!}\cong F$. If $G:\mathcal{B}\rightarrow\mathcal{B}^{\prime}$ is an exact functor such that $Gi^{!}\cong F$, then $Gi^{!}\cong(Fi_{*})i^{!}$ and hence $G\cong Fi_{*}$ since $i^{!}$ is dense.

${\rm(iii)}\Rightarrow{\rm(iv)}:$  For each object $A\in\mathcal{A}$, applying the exact functor $i^{!}$ to the exact sequence in $(1)$, we get an exact sequence $0\rightarrow i^{!}i_{*}i^{!}A\rightarrow i^{!}A\rightarrow i^{!}j_{*}j^{*}A\rightarrow i^{!}i_{*}B\rightarrow 0.$ Since $i^{!}j_{*}j^{*}A=0$, $i^{!}i_{*}B=0$. Thus $B\cong i^{!}i_{*}B=0$ and hence we get the desired exact sequence.

${\rm(iv)}\Rightarrow{\rm(v)}:$ From the given exact sequence one easily see ${\rm Im}j_{*}={\rm Ker}i^{!}$, and hence $({\rm Im}i_{*},{\rm Im}j_{*})$ is a torsion pair by (3). It remains to prove that ${\rm Im}j_{*}={\rm Ker}i^{!}$ is closed under quotient objects. For this, let $0\rightarrow A_{1}\rightarrow A\longrightarrow A_{2}\rightarrow 0$ be an exact sequence with $A\in{\rm Ker}i^{!}$. Then we get a commutative diagram with exact rows and columns:

[TABLE]

Applying the Snake Lemma to the two columns on the right, we get an exact sequence $0\rightarrow i_{*}i^{!}A_{1}\longrightarrow i_{*}i^{!}A\longrightarrow i_{*}i^{!}A_{2}\rightarrow 0,$ and hence $i_{*}i^{!}A_{2}=0.$ Since $i_{*}$ is fully faithful, $i^{!}A_{2}=0$.

${\rm(v)}\Rightarrow{\rm(vi)}:$ Since ${\rm Im}i_{*}={\rm Ker}j^{*}$ and $j^{*}$ is exact, ${\rm Im}i_{*}$ is closed under subobjects. So $({\rm Im}i_{*},{\rm Im}j_{*})$ is a hereditary and cohereditary torsion pair.

${\rm(vi)}\Rightarrow{\rm(vii)}$ follows from [TO, Thm. 4.1] (we stress that this step does not need that $\mathcal{A}$ has enough injective objects).

${\rm(vii)}\Rightarrow{\rm(viii)}$ is clear.

${\rm(viii)}\Rightarrow{\rm(i)}:$ For each object $A\in\mathcal{A}$, by $(1)$ we have an exact sequence $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A\rightarrow i_{*}B\rightarrow 0$ for some $B\in\mathcal{B}$. By assumption $({\rm Im}i_{*},{\rm Im}j_{*})$ is a strongly cohereditary torsion pair, it follows from [TO, Prop. 1.7] that $({\rm Im}i_{*},{\rm Im}j_{*})$ is a cohereditary torsion pair. Thus ${\rm Im}j_{*}$ is closed under quotient objects. So $i_{*}B\in{\rm Im}i_{*}\cap{\rm Im}j_{*}=\{0\},$ and hence we get the exact sequence $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A\rightarrow 0$.

Let $0\rightarrow A_{1}\rightarrow A\stackrel{{\scriptstyle g}}{{\longrightarrow}}A_{2}\rightarrow 0$ be an exact sequence in $\mathcal{A}$. Then we get a commutative diagram (3.1) with exact rows and columns. Applying the Snake Lemma to the two columns on the right, we get an exact sequence

[TABLE]

Applying the left exact functor $i^{!}$ to $0\rightarrow A_{1}\rightarrow A\stackrel{{\scriptstyle g}}{{\longrightarrow}}A_{2}\rightarrow 0$ we get the exact sequence

[TABLE]

and hence we have the exact sequence $0\rightarrow i_{*}i^{!}A_{1}\longrightarrow i_{*}i^{!}A\stackrel{{\scriptstyle i_{*}i^{!}g}}{{\longrightarrow}}i_{*}i^{!}A_{2}\longrightarrow i_{*}{\rm Coker}(i^{!}g)\rightarrow 0.$ Thus $i_{*}{\rm Coker}(i^{!}g)=0$. Since $i_{*}$ is fully faithful, ${\rm Coker}(i^{!}g)=0$. This proves the exactness of $i^{!}$. $\hfill\blacksquare$

3.2.

We state the dual result on left recollements without proofs.

Proposition 3.2**.**

Let $(1.2)$ be a left recollement of abelian categories. Then

$(1)$*  ${\rm Im}i_{*}$ is a weakly colocalizing subcategory. Explicitly, for each object $A\in\mathcal{A}$, there is an exact sequence $0\rightarrow i_{*}B\longrightarrow j_{!}j^{*}A\stackrel{{\scriptstyle\epsilon_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\eta_{A}}}{{\longrightarrow}}i_{*}i^{*}A\rightarrow 0$ for some $B\in\mathcal{B}$, with $j_{!}j^{*}A\in\ ^{\perp_{\leq 1}}({\rm Im}i_{*})$, where $\epsilon$ the counit and $\eta$ the unit.*

$(2)$*  $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\rightarrow}}\mathcal{C}\rightarrow 0$ is an exact sequence of abelian categories.*

${\rm(3)}$*   ${\rm Ker}i^{*}=\ ^{\perp_{0}}({\rm Im}i_{*});$ and $({\rm Ker}i^{*},{\rm Im}i_{*})$ is a strongly cohereditary torsion pair, and $0\rightarrow{\rm Ker}\eta_{A}\longrightarrow A\stackrel{{\scriptstyle\eta_{A}}}{{\longrightarrow}}i_{*}i^{*}A\rightarrow 0$ is the $t$-decomposition of $A$.*

${\rm(4)}$*  The following are equivalent$:$*

${\rm(i)}$*   $i^{*}$ is exact;*

${\rm(ii)}$*   $i^{*}$ and $j_{!}$ are exact;*

${\rm(iii)}$*   $0\rightarrow\mathcal{C}\stackrel{{\scriptstyle j_{!}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle i^{*}}}{{\rightarrow}}\mathcal{B}\rightarrow 0$ is an exact sequence of abelian categories;*

${\rm(iv)}$*   the sequence $0\rightarrow j_{!}j^{*}A\stackrel{{\scriptstyle\epsilon_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\eta_{A}}}{{\longrightarrow}}i_{*}i^{*}A\rightarrow 0$ is exact for each object $A\in\mathcal{A};$*

${\rm(v)}$*    ${\rm Im}j_{!}={\rm Ker}i^{*};$ $({\rm Im}j_{!},{\rm Im}i_{*})$ is a hereditary torsion pair of $\mathcal{A},$ and $0\rightarrow j_{!}j^{*}A\stackrel{{\scriptstyle\epsilon_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\eta_{A}}}{{\longrightarrow}}i_{*}i^{*}A\rightarrow 0$ is the $t$-decomposition of $A;$*

${\rm(vi)}$*   $({\rm Im}j_{!},{\rm Im}i_{*})$ is a hereditary and cohereditary torsion pair of $\mathcal{A};$*

${\rm(vii)}$*  $({\rm Im}j_{!},{\rm Im}i_{*})$ is a strongly hereditary and strongly cohereditary torsion pair of $\mathcal{A};$*

${\rm(viii)}$*  $({\rm Im}j_{!},{\rm Im}i_{*})$ is a strongly hereditary torsion pair of $\mathcal{A}$.*

Remark 3.3**.**

By Lemma 2.1$(1)$ $($resp. Lemma 2.1(1^{\prime})$$), there is a bijective correspondence between right $($resp. left$)$ recollements and Giraud functors $($resp. coGiraud functors$)$.

By Lemma 2.1$(3)$ and Proposition 3.1$(3)$ $($resp. Lemma 2.1$(3^{\prime})$ and Proposition 3.2(3)$$), there is a bijective correspondence between right $($resp. left$)$ recollements and strongly hereditary $($resp. strongly cohereditary$)$ torsion pairs.

3.3.

To prove Theorem 1.1 we use the following fact, in which the first assertion is just [TO, Lemmas 4.2, 4.2∗]. For the use of the second assertion, we include a proof.

Lemma 3.4**.**

Let $(\mathcal{U},\mathcal{V})$ and $(\mathcal{V},\mathcal{W})$ be torsion pairs in $\mathcal{A}$. Assume that $\mathcal{U}$ is closed under subobjects and $\mathcal{W}$ is closed under quotient objects. Then $\mathcal{U}=\mathcal{W}$.

For each object $A\in\mathcal{A}$, let $0\rightarrow A_{\mathcal{U}}\rightarrow A\stackrel{{\scriptstyle g}}{{\rightarrow}}A^{\mathcal{V}}\rightarrow 0$ be the $t$-decomposition of $A$ with respect to $(\mathcal{U},\mathcal{V})$, and $0\rightarrow A_{\mathcal{V}}\rightarrow A\stackrel{{\scriptstyle h}}{{\rightarrow}}A^{\mathcal{W}}\rightarrow 0$ the $t$-decomposition of $A$ with respect to $(\mathcal{V},\mathcal{W})$. Then $A_{\mathcal{V}}\cong A^{\mathcal{V}}$ and $A^{\mathcal{W}}\cong A_{\mathcal{U}}$,  $A\cong A_{\mathcal{U}}\oplus A^{\mathcal{V}}$, and $A\mapsto(A_{\mathcal{U}},A^{\mathcal{V}})$ gives an equivalence $\mathcal{A}\cong\mathcal{U}\oplus\mathcal{V}$ of categories.

Proof. Consider the push-out of $g$ and $h$, we get a commutative diagram with exact rows and columns

[TABLE]

Since $\mathcal{V}$ and $\mathcal{W}$ are closed under quotient objects, $B\in\mathcal{V}\cap\mathcal{W}=\{0\}$. Since $\mathcal{U}$ and $\mathcal{V}$ are closed under subobjects, $E\in\mathcal{U}\cap\mathcal{V}=\{0\}$.

Thus $A_{\mathcal{V}}\cong C\cong A^{\mathcal{V}}$, and hence $g$ is a splitting epimorphism $A\cong A_{\mathcal{U}}\oplus A^{\mathcal{V}}$. Also, $A_{\mathcal{U}}\cong D\cong A^{\mathcal{W}}$, and hence $h$ is a splitting epimorphism. Taking $A\in\mathcal{U}$ and $A\in\mathcal{W}$, respectively, we see that $\mathcal{U}=\mathcal{W}$. It is routine to verify that $\mathcal{A}\rightarrow\mathcal{U}\oplus\mathcal{V}$ given by $A\mapsto(A_{\mathcal{U}},A^{\mathcal{V}})$ is an equivalence of categories. $\hfill\blacksquare$

Proof of Theorem 1.1. By Proposition 3.2$(4){\rm(v)}$, $({\rm Im}j_{!},{\rm Im}i_{*})$ is a torsion pair with $t$-decomposition $0\rightarrow j_{!}j^{*}A\stackrel{{\scriptstyle\epsilon_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\eta_{A}}}{{\longrightarrow}}i_{*}i^{*}A\rightarrow 0$ and ${\rm Im}j_{!}={\rm Ker}i^{*}$ is closed under subobjects. By Proposition 3.1$(4){\rm(v)}$, $({\rm Im}i_{*},{\rm Im}j_{*})$ is a torsion pair with $t$-decomposition $0\rightarrow i_{*}i^{!}A\stackrel{{\scriptstyle\omega_{A}}}{{\longrightarrow}}A\stackrel{{\scriptstyle\zeta_{A}}}{{\longrightarrow}}j_{*}j^{*}A\rightarrow 0$ and ${\rm Im}j_{*}={\rm Ker}i^{!}$ is closed under quotient objects. It follows from Lemma 3.4 that ${\rm Im}j_{!}={\rm Im}j_{*}$ and $\mathcal{A}\cong{\rm Im}i_{*}\oplus{\rm Im}j_{!}\cong\mathcal{B}\oplus\mathcal{C}$. For $A\in\mathcal{A}$, by Lemma 3.4, $i_{*}i^{*}A\cong i_{*}i^{!}A$ and $j_{!}j^{*}A\cong j_{*}j^{*}A$. Since $i_{*}$ is fully faithful, $i^{*}A\cong i^{!}A$ and hence $i^{*}\cong i^{!}$. Since $j^{*}$ is dense, it follows that $j_{*}\cong j_{!}$. $\hfill\blacksquare$

4. Proof of Theorem 1.2

Lemma 4.1**.**

Let $\mathcal{A}$ be a Grothendieck category, and $(1.2)$ a left recollement of abelian categories. Then $(i_{*}\mathcal{B},(i_{*}\mathcal{B})^{\perp_{0}})$ is a torsion pair.

Proof. By Proposition 3.2(3), $({\rm Ker}i^{*},i_{*}\mathcal{B})$ is a torsion pair. So the torsionfree class $i_{*}\mathcal{B}$ is closed under subobjects and products ([D, Thm. 2.3]). Since $\mathcal{A}$ is a Grothendieck category, any coproduct is a subobject of the corresponding product, $i_{*}\mathcal{B}$ is also closed under coproducts. On the other hand, since $i_{*}\mathcal{B}={\rm Ker}j^{*}$ and $j^{*}$ is exact, $i_{*}\mathcal{B}$ is closed under quotient objects and extensions. Thus by [D, Thm. 2.3] $i_{*}\mathcal{B}$ is a torsion class, and hence $(i_{*}\mathcal{B},(i_{*}\mathcal{B})^{\perp_{0}})$ is a torsion pair. $\hfill\blacksquare$

Proof of Theorem 1.2. Given a left recollement $(1.2)$, by Lemma 4.1, $(i_{*}\mathcal{B},(i_{*}\mathcal{B})^{\perp_{0}})$ is a torsion pair. Since $i_{*}\mathcal{B}={\rm Ker}j^{*}$ and $j^{*}$ is exact, $i_{*}\mathcal{B}$ is closed under subobjects, i.e., $(i_{*}\mathcal{B},(i_{*}\mathcal{B})^{\perp_{0}})$ is a hereditary torsion pair. Since $\mathcal{A}$ is a Grothendieck category, $\mathcal{A}$ has enough injective objects. By [TO, Thm. 1.8∗], $(i_{*}\mathcal{B},(i_{*}\mathcal{B})^{\perp_{0}})$ is a strongly hereditary torsion pair. Applying Lemma 2.1(3) we get a right recollement

i_{*}\mathcal{B}$$\mathcal{A}$$\mathcal{A}/i_{*}\mathcal{B}$${i}$$\widetilde{i^{!}}$$Q$$j

where $i$ is the inclusion functor and $Q$ is the quotient functor. By Proposition 3.2(2), $0\rightarrow\mathcal{B}\stackrel{{\scriptstyle i_{*}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle j^{*}}}{{\rightarrow}}\mathcal{C}\rightarrow 0$ is an exact sequence of abelian categories. By the universal property of the functors $j^{*}$ and $Q$, we get a commutative diagram

[TABLE]

and hence we get a recollement

\mathcal{B}$$\mathcal{A}$$\mathcal{C}$$i_{\ast}$$j^{\ast}$$i^{!}$$j_{*}$$i^{*}$$j_{!}

with $i^{!}=F^{-1}\widetilde{i^{!}}$   and $j_{*}=jG$. $\hfill\blacksquare$

Corollary 4.2**.**

A colocalizing subcategory of a Grothendieck category is localizing.

Proof. Let $\mathcal{S}$ be a colocalizing subcategory of a Grothendieck category $\mathcal{A}$. That is, the quotient functor $Q:\mathcal{A}\rightarrow\mathcal{A}/\mathcal{S}$ has a left adjoint, denoted by $j_{!}:\mathcal{A}/\mathcal{S}\rightarrow\mathcal{A}.$ By the dual of [GL, Prop. 2.2], $j_{!}$ is fully faithful. So $j_{!}$ is a coGiraud functor with exact right adjoint $Q$. By Lemma 2.1(1’) there exists a functor $i^{*}:\mathcal{A}\rightarrow{\rm Ker}Q=\mathcal{S}$, such that $(\mathcal{S},\ \mathcal{A},\ \mathcal{A}/\mathcal{S},\ i^{*},\ i,\ j_{!},\ Q)$ is a left recollement, where $i:\mathcal{S}\rightarrow\mathcal{A}$ is the inclusion functor. Then by Theorem 1.2 this left recollement can be extended to be a recollement, so $Q$ has a right adjoint, i.e., $\mathcal{S}$ is localizing. $\hfill\blacksquare$

5. Proof of Theorem 1.3

5.1. **Serre subcategories of

type $(0,0)$**

For a ring $R$, let ${\rm Mod}R$ be the category of right $R$-modules. If $R$ is a right noetherian, let ${\rm mod}R$ be the category of finitely generated right $R$-modules.

Lemma 5.1**.**

Let R be a right noetherian ring. Then ${\rm mod}R$ is a Serre subcategory of type $(0,0)$.

Proof. It is clear that ${\rm mod}R$ is a Serre subcategory of ${\rm Mod}R$. Let $i:{\rm mod}R\rightarrow{\rm Mod}R$ and $Q:{\rm Mod}R\rightarrow{\rm Mod}R/{\rm mod}R$ be the inclusion functor and the quotient functor, respectively. Assume that the type of ${\rm mod}R$ is not $(0,0)$. Then there exist either adjoint pairs $(i_{1},i)$ and $(j_{1},Q)$, or adjoint pairs $(i,i_{-1})$ and $(Q,j_{-1})$.

In the first case, by Lemma 2.1$(2^{\prime})$ we get a left recollement

{\rm mod}R$${\rm Mod}R$${\rm Mod}R/{\rm mod}R.$$i$$Q$$j_{1}$$i_{1}

By Proposition 3.2$(3)$ we have a torsion pair $({\rm Ker}i_{1},\ {\rm mod}R)$ in ${\rm Mod}R$. Thus the torsionfree class ${\rm mod}R$ is closed under products, which is absurd.

The dual argument shows that the second case is also impossible. We give a direct proof. For each $X\in{\rm Mod}R/{\rm mod}R$ we have ${\rm Hom}_{R}(M,j_{-1}X)\cong{\rm Hom}_{{\rm Mod}R/{\rm mod}R}(QM,X)=0$ for all $M\in{\rm mod}R$. So $j_{-1}X$ has no non-zero finitely generated submodule. Thus $j_{-1}X=0$. Since $j_{-1}$ is fully faithful, ${\rm Mod}R/{\rm mod}R=0$, i.e., ${\rm mod}R={\rm Mod}R$, which is absurd. $\hfill\blacksquare$

5.2. Serre subcategories of type $(0,-1)$

Lemma 5.2**.**

Let ${\rm Ab}_{t}$ be the category of the torsion abelian groups. Then ${\rm Ab}_{t}$ is a Serre subcategory of type $(0,-1)$.

Proof.  Let ${\rm Ab}_{f}$ be the category of the abelian groups in which every non-zero element is of infinite order. Then $({\rm Ab}_{t},{\rm Ab}_{f})$ is a torsion pair in ModZ. Let $i:{\rm Ab}_{t}\rightarrow{\rm Mod}Z$ and $Q:{\rm Mod}Z\rightarrow{\rm Mod}Z/{\rm Ab}_{t}$ be the inclusion functor and the quotient functor, respectively. Since ${\rm Ab}_{t}$ is closed under submodules, it follows from [TO, Thm. 1.8*] (also [O, Thm. 2.6]) that $({\rm Ab}_{t},{\rm Ab}_{f})$ is a strongly hereditary torsion pair. By Lemma 2.1$(3)$ we get a right recollement

{\rm Ab}_{t}$${\rm Mod}Z$${\rm Mod}Z/{\rm Ab}_{t}$$i$$i_{-1}$$Q$$j_{-1}

Thus $({\rm Ab}_{t},{\rm Ker}i_{-1})$ is a torsion pair, by Proposition 3.1$(3)$. Comparing with the torsion pair $({\rm Ab}_{t},{\rm Ab}_{f})$ we get ${\rm Ab}_{f}={\rm Ker}i_{-1}$.

Assume that the type of ${\rm Ab}_{t}$ is not $(0,-1)$. Then there exist either adjoint pairs $(i_{1},i)$ and $(j_{1},Q)$, or adjoint pairs $(i_{-1},i_{-2})$ and $(j_{-1},j_{-2})$.

In the first case, by Lemma 2.1$(2^{\prime})$ we get a left recollement $({\rm Ab}_{t},{\rm Mod}Z,{\rm Mod}Z/{\rm Ab}_{t},\ i_{1},\ i,\ j_{1},\ Q)$, and hence $({\rm Ker}i_{1},{\rm Ab}_{t})$ is a torsion pair, by Proposition 3.2$(3)$. So the torsionfree class ${\rm Ab}_{t}$ is closed under products, which is absurd.

In the second case, the functor $i_{-1}$ is exact, and hence ${\rm Ab}_{f}={\rm Ker}i_{-1}$ is closed under quotient groups, which is absurd. $\hfill\blacksquare$

Remark.  The above argument also shows that there is a right recollement of abelian categories which can not be extended to a recollement (cf. Theorem 1.2), and that a localizing subcategory is not necessarily colocalizing (cf. Corollary 4.2).

5.3. **Serre subcategories

of type $(0,-2)$ and $(1,-1)$**

Lemma 5.3**.**

Let $\mathcal{A}$ be a Grothendieck category. Assume that both $(\mathcal{T},\mathcal{G})$ and $(\mathcal{G},\mathcal{F})$ are hereditary torsion pairs in $\mathcal{A}$, such that $\mathcal{T}$ is not a torsionfree class. Then $\mathcal{T}$ is a Serre subcategory of type $(0,-2)$, and $\mathcal{G}$ is a Serre subcategory of type $(1,-1)$.

Proof. It is clear that $\mathcal{T}$ and $\mathcal{G}$ are Serre subcategories. Since $\mathcal{A}$ is a Grothendieck category, $\mathcal{A}$ has enough injective objects. Since $(\mathcal{T},\mathcal{G})$ is a hereditary torsion pair, it follows from [TO, Thm. 1.8∗] that $(\mathcal{T},\mathcal{G})$ is strongly hereditary. By Lemma 2.1(3) there is a right recollement

\mathcal{T}$$\mathcal{A}$$\mathcal{A}/\mathcal{T}$$i_{\mathcal{T}}$$i_{-1}$$Q_{\mathcal{T}}$$j_{-1}

with ${\rm Im}j_{-1}=\mathcal{T}^{\perp_{\leq 1}}$, where $i_{\mathcal{T}}$ and $Q_{\mathcal{T}}$ are respectively the inclusion functor and the quotient functor. We claim $\mathcal{G}=\mathcal{T}^{\perp_{\leq 1}}.$ In fact, $T^{\perp_{\leq 1}}\subseteq T^{\perp_{0}}=\mathcal{G}$. For each object $G\in\mathcal{G}$, since $\mathcal{T}$ is a weakly localizing subcategory, by definition there exists an exact sequence

[TABLE]

such that $T_{1}\in\mathcal{T},\ T_{2}\in\mathcal{T},$ and $C\in\mathcal{T}^{\perp_{\leq 1}}.$ But $(\mathcal{T},\mathcal{G})$ is a torsion pair, $a=0$ and $T_{1}=0$. Since $\mathcal{G}$ is closed under quotient objects, ${\rm Im}b\in\mathcal{G}$. Since by assumption $\mathcal{T}$ is closed under subobjects, ${\rm Im}b\in\mathcal{T}\cap\mathcal{G}=\{0\}$. So $G\cong C\in\mathcal{T}^{\perp_{\leq 1}}.$ This proves the claim.

Since $(\mathcal{T},\mathcal{G})$ is a hereditary and cohereditary torsion pair, it follows from [TO, Thm. 4.1] that $(\mathcal{T},\mathcal{G})$ is a strongly cohereditary torsion pair. By Lemma 2.1(3’) there is a left recollement

\mathcal{G}$$\mathcal{A}$$\mathcal{A}/\mathcal{G}$$i^{*}$$i_{\mathcal{G}}$$j_{!}$$Q_{\mathcal{G}}

with ${\rm Im}j_{!}=\ ^{\perp_{\leq 1}}\mathcal{G}$, where $i_{\mathcal{G}}$ and $Q_{\mathcal{G}}$ are respectively the inclusion functor and the quotient functor. Since we have shown $\mathcal{G}=\mathcal{T}^{\perp_{\leq 1}},$ it follows that $\mathcal{T}\subseteq\ ^{\perp_{\leq 1}}\mathcal{G}\subseteq\ ^{\perp_{0}}\mathcal{G}=\mathcal{T}.$ Thus $\mathcal{T}=\ ^{\perp_{\leq 1}}\mathcal{G}$.

Put $\widetilde{j_{!}}$ to be the equivalence $\mathcal{A}/\mathcal{G}\rightarrow j_{!}(\mathcal{A}/\mathcal{G})=\mathcal{T}$, and $\widetilde{j_{-1}}$ to be the equivalence $\mathcal{A}/\mathcal{T}\rightarrow j_{-1}(\mathcal{A}/\mathcal{T})=\mathcal{G}.$ We claim the diagram of functors

\mathcal{T}$$\mathcal{A}$$\mathcal{A}/\mathcal{T}$$(5.1)$$i_{\mathcal{T}}$$\widetilde{j_{!}}Q_{\mathcal{G}}$$Q_{\mathcal{T}}$$i_{\mathcal{G}}\widetilde{j_{-1}}

is a right recollement. In fact, since $j_{-1}=i_{\mathcal{G}}\widetilde{j_{-1}}$, $(Q_{\mathcal{T}},i_{\mathcal{G}}\widetilde{j_{-1}})$ is an adjoint pair. By Proposition 3.2(1), for each object $A\in\mathcal{A}$ there is an exact sequence

[TABLE]

for some $G\in\mathcal{G}$. Since $j_{!}Q_{\mathcal{G}}A\in\mathcal{T}$ and $\mathcal{T}$ is closed under subobjects, $i_{\mathcal{G}}G\in\mathcal{T}\cap\mathcal{G}=\{0\}$. Thus for $T\in\mathcal{T}$ we have ${\rm Hom}(T,j_{!}Q_{\mathcal{G}}A)\cong{\rm Hom}(T,A).$ This shows that $(i_{\mathcal{T}},\widetilde{j_{!}}Q_{\mathcal{G}})$ is an adjoint pair. This justifies the claim.

Again by [TO, Thm. 1.8∗], $(\mathcal{G},\mathcal{F})$ is a strongly hereditary torsion pair. By Lemma 2.1(3) there is a right recollement

\mathcal{G}$$\mathcal{A}$$\mathcal{A}/\mathcal{G}.$$i_{\mathcal{G}}$$i_{-2}$$Q_{\mathcal{G}}$$j_{-2}

Rewrite this we get a diagram of functors

\mathcal{A}/\mathcal{G}$$\mathcal{A}$$\mathcal{G}.$$Q_{\mathcal{G}}$$j_{-2}$$i_{\mathcal{G}}$$i_{-2}

(note that this is not a left recollement, since $i_{-2}$ and $j_{-2}$ are not exact). Hence we have a diagram of functors

\mathcal{T}$$\mathcal{A}$$\mathcal{A}/\mathcal{T}.$$(5.2)$$\widetilde{j_{!}}Q_{\mathcal{G}}$$j_{-2}\widetilde{j_{!}}^{-1}$$\widetilde{j_{-1}}^{-1}i_{-2}$$i_{\mathcal{G}}\widetilde{j_{-1}}$$\widetilde{j_{-1}}^{-1}i_{-2}

Putting $(5.1)$ and $(5.2)$ together we get a diagram of functors

\mathcal{T}$$\mathcal{A}$$(5.3)$$\mathcal{A}/\mathcal{T}$$i_{\mathcal{T}}$$\widetilde{j_{!}}Q_{\mathcal{G}}$$Q_{\mathcal{T}}$$i_{\mathcal{G}}\widetilde{j_{-1}}$$j_{-2}\widetilde{j_{!}}^{-1}$$\widetilde{j_{-1}}^{-1}i_{-2}

such that $(i_{\mathcal{T}},\ \widetilde{j_{!}}Q_{\mathcal{G}},\ j_{-2}\widetilde{j_{!}}^{-1})$ and $(Q_{\mathcal{T}},\ i_{\mathcal{G}}\widetilde{j_{-1}},\ \widetilde{j_{-1}}^{-1}i_{-2})$ are adjoint sequences.

Assume that the type of $\mathcal{T}$ is not $(0,-2)$. Then there exist either adjoint pairs $(i_{1},i_{\mathcal{T}})$ and $(j_{1},Q_{\mathcal{T}})$, or adjoint pairs $(j_{-2}\widetilde{j_{!}}^{-1},i_{-3})$ and $(\widetilde{j_{-1}}^{-1}i_{-2},j_{-3})$.

In the first case, by Lemma 2.1$(2^{\prime})$ we get a left recollement $(\mathcal{T},\mathcal{A},\mathcal{A}/\mathcal{T},i_{1},i_{\mathcal{T}},j_{1},Q_{\mathcal{T}})$, and hence $({\rm Ker}i_{1},\mathcal{T})$ is a torsion pair, by Proposition 3.2$(3)$. This contradicts the assumption that $\mathcal{T}$ is not a torsionfree class.

In the second case, all the functors in (5.3) are exact, and hence (5.3) is a recollement $(\mathcal{A}/\mathcal{T},\ \mathcal{A},\ \mathcal{T})$. By Theorem 1.1 we have $i_{\mathcal{T}}\cong j_{-2}\widetilde{j_{!}}^{-1}$ and $Q_{\mathcal{T}}\cong\widetilde{j_{-1}}^{-1}i_{-2}$, and hence both $i_{\mathcal{T}}$ and $Q_{\mathcal{T}}$ have left adjoints. This goes to the first case.

Thus the type of $\mathcal{T}$ is $(0,2)$. This also proves the type of $\mathcal{G}$ is $(1,-1)$. $\hfill\blacksquare$

Example 5.4**.**

Let $K$ be a field, $R:=\prod\limits_{i=1}^{\infty}K_{i}$ and $I:=\bigoplus\limits_{i=1}^{\infty}K_{i}$ with each $K_{i}=K$. Then $R$ is a commutative ring and $I$ is an idempotent ideal of $R$. Put $\mathcal{G}:=\{M\in{\rm Mod}R\ |\ MI=0\}$. Then $\mathcal{G}$ is a TTF-class in ${\rm Mod}R$, i.e., $\mathcal{G}$ is a torsion and torsion-free class, since $\mathcal{G}$ is subobjects, quotient objects, extensions, coproducts and products. So we have a TTF-triple $(\mathcal{T},\mathcal{G},\mathcal{F})$.

It is clear that $\mathcal{T}=\{M\in{\rm Mod}R\ |\ MI=M\}$. In fact, for any $R$-module $M_{1}$ with $M_{1}I=M_{1}$ and $M_{2}\in\mathcal{G}$, we have ${\rm Hom}(M_{1},M_{2})=0;$ and for any $M\in{\rm Mod}R$, we have an exact sequence $0\rightarrow MI\rightarrow M\rightarrow M/MI\rightarrow 0$ with $(MI)I=MI$ and $M/MI\in\mathcal{G}$.

We claim that $\mathcal{T}$ is closed under subobjects. By [D, Thm. 2.9] this is equivalent to say that $\mathcal{G}$ is closed under taking injective envelopes. Thus, it suffices to prove that for any $M\in\mathcal{G}$, the injective envelope $E(M)$ of $M$ satisfies $E(M)I=0$. Otherwise, there is an $m\in E(M)$ with $mI\neq 0$. Set $L:=\{b\in I\ |\ mb\neq 0\}.$ Then $L\neq\emptyset$. Choosing $b\in L$ such that the number of nonzero components is smallest. We may assume that each nonzero component of $b$ is $1_{k}$, the identity of $K$. In fact, if the nonzero components of $b$ are exactly $b_{i_{1}},\cdots,b_{i_{t}}$, where $b_{i}$ is the $i$-th component of $b$, then we use $bb^{\prime}$ to replace $b$, where the nonzero components of $b^{\prime}$ are exactly $b_{i_{1}}^{-1},\cdots,b_{i_{t}}^{-1}$ $($note that $mbb^{\prime}\neq 0:$ otherwise mb=mbb^{\prime}b=0$$). By the choice of $b$ we know that $mbI$ is a nonzero submodule of $E(M)$. Since $E(M)$ is an essential extension of $M$, $mbI\cap M\neq 0.$ So there is a nonzero element $r\in I$ with $mbr\neq 0$ and $mbr\in M$. Note that the support of $br\in I$ is contained in the support of $b$ $($by definition the support of $b$ is the set of $i$, such that the $i$-th component of $b$ is not zero$)$. By the choice of $b$, the support of $br\in I$ is just the support of $b$. Let $b^{\prime}_{i_{1}},\cdots,b^{\prime}_{i_{t}}$ be the nonzero components of $br$, and $d\in I$ the element with the nonzero components exactly ${b^{\prime}_{i_{1}}}^{-1},\cdots,{b^{\prime}_{i_{t}}}^{-1}$. Then we get the desired contradiction $0\neq mb=mbrd\in MI=0$. This proves the claim.

Since $K_{i}\in\mathcal{T}$ but $R\notin\mathcal{T}$, $\mathcal{T}=\{M\in{\rm Mod}R\ |\ MI=M\}$ is not closed under products. Thus $\mathcal{T}$ is not a torsion-free class. By Lemma 5.3, the type of $\mathcal{T}$ is $(0,-2)$ and the type of $\mathcal{G}$ is $(1,-1)$.

5.4. Serre subcategories of type $(1,-2)$ and $(2,-1)$

Let $R$ and $S$ be rings, ${}_{S}M_{R}$ a non-zero $S$-$R$-bimodule, and $\Lambda=\left(\begin{smallmatrix}R&0\\ M&S\end{smallmatrix}\right)$ the triangular matrix ring. A right $\Lambda$-module is identified with a triple $(X,Y,f)$, where $X$ is a right $R$-module, $Y$ a right $S$-module, and $f:Y\otimes_{S}M\rightarrow X$ a right $R$-map; and a left $\Lambda$-module is identified with a triple $\left(\begin{smallmatrix}U\\ V\end{smallmatrix}\right)_{g}$, where $U$ is a left $R$-module, $V$ a left $S$-module, and $g:M\otimes_{R}U\rightarrow V$ a left $S$-map ([ARS, p.71]). Put $e_{1}=\left(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix}\right)$ and $e_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right)$. It is well-known that there is a ladder of abelian categories (see [CPS, Sect. 2], [PV, 2.10]; also [H, 2.1], [AHKLY, Exam. 3.4])

{\rm Mod}R$${\rm Mod}\Lambda$${\rm Mod}S$$i_{0}$$j_{0}$$j_{-1}$$j_{-2}$$j_{1}$$i_{1}$$i_{-1}$$i_{-2}

i.e., the upper three rows form a recollement of abelian categories, and $(i_{-1},i_{-2})$ and $(j_{-1},j_{-2})$ are adjoint pairs, where

[TABLE]

where the right $\Lambda$-module $R$ is given by $r\left(\begin{smallmatrix}r^{\prime}&0\\ m&s\end{smallmatrix}\right):=rr^{\prime}$ and the right $\Lambda$-module $S$ is given by $s\left(\begin{smallmatrix}r&0\\ m&s^{\prime}\end{smallmatrix}\right):=ss^{\prime}$. Note that ${\rm Mod}R$ is a Serre subcategory of ${\rm Mod}\Lambda$ and $0\rightarrow{\rm Mod}R\stackrel{{\scriptstyle i_{0}}}{{\rightarrow}}{\rm Mod}\Lambda\stackrel{{\scriptstyle j_{0}}}{{\rightarrow}}{\rm Mod}S\rightarrow 0$ is an exact sequence of abelian categories.

We claim that the type of ${\rm Mod}R$ is $(1,-2)$.

In fact, since $M\neq 0$, $\left(\begin{smallmatrix}R\\ 0\end{smallmatrix}\right)$ is not flat as a left $\Lambda$-module, $i_{1}=-\otimes_{\Lambda}\left(\begin{smallmatrix}R\\ 0\end{smallmatrix}\right)$ is not exact, and hence $i_{1}$ has no left adjoint. Also, since $M\neq 0$, $S$ is not projective as a right $\Lambda$-module. So $j_{-2}={\rm Hom}_{\Lambda}(S,-)$ is not exact, and hence $j_{-2}$ has no right adjoint. This proves the claim. (We include another proof. If both $i_{1}$ and $j_{1}$ have left adjoints, then $i_{1}$ and $j_{1}$ are exact, and hence $i_{1}\cong i_{-1}$ by Theorem 1.1, i.e., $-\otimes_{\Lambda}\left(\begin{smallmatrix}R\\ 0\end{smallmatrix}\right)\cong-\otimes_{\Lambda}\left(\begin{smallmatrix}R\\ M\end{smallmatrix}\right)$. But this is not true, since $M\neq 0$. Similarly, if both $i_{-2}$ and $j_{-2}$ have right adjoints, then $i_{-2}$ and $j_{-2}$ are exact, and hence $i_{0}\cong i_{-2}$ by Theorem 1.1, i.e., ${\rm Hom}_{R}(R,-)\cong{\rm Hom}_{R}(\left(\begin{smallmatrix}R\\ M\end{smallmatrix}\right),-)$, which is absurd.)

The argument above also shows that $0\rightarrow{\rm Mod}S\stackrel{{\scriptstyle j_{-1}}}{{\rightarrow}}{\rm Mod}\Lambda\stackrel{{\scriptstyle i_{-1}}}{{\rightarrow}}{\rm Mod}R\rightarrow 0$ is an exact sequence of abelian categories, and that the type of ${\rm Mod}S$ is $(2,-1)$.

Remark 5.5**.**

Consider $\Lambda=T_{2}(R):=\left(\begin{smallmatrix}R&0\\ R&R\end{smallmatrix}\right)$, then $i_{-2}\cong j_{1}$ and hence we have adjoint sequences

[TABLE]

such that $i_{1}-\otimes_{\Lambda}\left(\begin{smallmatrix}R\\ 0\end{smallmatrix}\right)$ has no left adjoint, and $j_{-2}={\rm Hom}_{\Lambda}(R,-)$ has no right adjoint. Graphically we have

{\rm Mod}R$${\rm Mod}\Lambda$${\rm Mod}R$$i_{0}$$j_{0}$$j_{-1}$$j_{-2}$$j_{1}$$i_{1}$$i_{-1}$$i_{-2}\cong j_{1}$$j_{0}$$j_{-1}$$j_{-2}$$i_{-1}$$i_{0}$$i_{1}

5.5. Serre subcategories of type $(+\infty,-\infty)$

Let $\mathcal{S}$ and $\mathcal{T}$ be abelian categories. Then as a subcategory of $\mathcal{S}\oplus\mathcal{T}$, $\mathcal{S}$ is a Serre subcategory of type $(+\infty,-\infty)$. In fact, it is clear that $(p_{1},i_{1},p_{1})$ and $(i_{2},p_{2},i_{2})$ are adjoint sequences, where $i_{1}$ and $i_{2}$ are embeddings, and $p_{1}$ and $p_{2}$ are projections. On the other hand, if $\mathcal{S}$ is a Serre subcategory of $\mathcal{A}$ and the type of $\mathcal{S}$ is $(+\infty,-\infty)$, then by Theorem 1.1 it is easy to see $\mathcal{A}\cong\mathcal{S}\oplus\mathcal{(}\mathcal{A}/\mathcal{S})$.

5.6. Proof of Theorem 1.3.

Let $\mathcal{B}$ be a Serre subcategory of type $(m,-n)$, where $m$ and $n$ are in the set $\{+\infty,0,1,2,\cdots\}$. Denote by $i:\mathcal{B}\rightarrow\mathcal{A}$ the inclusion functor and $Q:\mathcal{A}\rightarrow\mathcal{A}/\mathcal{B}$ the quotient functor. Put $h:=m+n+1$.

Claim 1. If $h\geq 5$, then $(m,-n)=(+\infty,-\infty)$.

Assume that there is a diagram of functors

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i_{2}$$j_{2}$$i_{1}$$j_{1}$$i$$Q$$i_{3}$$j_{3}$$i_{4}$$j_{4}

such that $(i_{4},i_{3},i_{2},i_{1},i)$ and $(j_{4},j_{3},j_{2},j_{1},Q)$ are adjoint sequences. Then $i_{1},i_{2},i_{3},j_{1},j_{2},j_{3}$ are exact. Since a left adjoint of $Q$ is fully faithful (the dual of [GL, Prop. 2.2]), $j_{1}$ is fully faithful. Thus the two rows at the bottom form a left recollement. By Proposition 3.2$(4){\rm(iii)}$, $0\rightarrow\mathcal{A}/\mathcal{B}\stackrel{{\scriptstyle j_{1}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle i_{1}}}{{\rightarrow}}\mathcal{B}\rightarrow 0$ is an exact sequence of abelian categories and ${\rm Ker}i_{1}={\rm Im}j_{1}$. It follows from Lemma 2.1$(2^{\prime})$ that $i_{2}$ is fully faithful. Thus

\mathcal{A}/\mathcal{B}$$\mathcal{A}$$\mathcal{B}$$j_{1}$$i_{1}$$Q$$i$$j_{2}$$i_{2}

is recollement such that $j_{2}$ and $Q$ are exact. It follows from Theorem 1.1 that $j_{2}\cong Q$ and $i_{2}\cong i$, and hence the type is $(+\infty,-\infty)$.

Assume that there is a diagram of functors

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i_{1}$$j_{1}$$i$$Q$$i_{-1}$$j_{-1}$$i_{2}$$j_{2}$$i_{3}$$j_{3}

such that $(i_{3},i_{2},i_{1},i,i_{-1})$ and $(j_{3},j_{2},j_{1},Q,j_{-1})$ are adjoint sequences. Then the four functors $i_{1},i_{2},j_{1},j_{2}$ are exact. By the dual of [GL, Prop. 2.2], $j_{1}$ is fully faithful. By the same argument as above we know that the type is $(+\infty,-\infty)$.

Assume that there is a diagram of functors

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i$$Q$$i_{-1}$$j_{-1}$$i_{-2}$$j_{-2}$$i_{1}$$j_{1}$$i_{2}$$j_{2}

such that $(i_{2},i_{1},i,i_{-1},i_{-2})$ and $(j_{2},j_{1},Q,j_{-1},j_{-2})$ are adjoint sequences. Then $i_{1},i_{-1},j_{1},j_{-1}$ are exact. By [GL, Prop. 2.2] and its dual, $j_{-1}$ and $j_{1}$ are fully faithful. Thus the three rows in the middle form a recollement such that $i_{1}$ and $i_{-1}$ are exact. It follows from Theorem 1.1 that $i_{1}\cong i_{-1}$ and $j_{1}\cong j_{-1}$, and hence the type is $(+\infty,-\infty)$.

Assume that there is a diagram of functors

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i_{-1}$$j_{-1}$$i_{-2}$$j_{-2}$$i_{-3}$$j_{-3}$$i$$Q$$i_{1}$$j_{1}

such that $(i_{1},i,i_{-1},i_{-2},i_{-3})$ and $(j_{1},Q,j_{-1},j_{-2},j_{-3})$ are adjoint sequences. Then $i_{-1},i_{-2},j_{-1},j_{-2}$ are exact, and $j_{-1}$ is fully faithful. So

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i$$i_{-1}$$Q$$j_{-1}

is a right recollement. By Proposition 3.1$(4){\rm(iii)}$, $0\rightarrow\mathcal{A}/\mathcal{B}\stackrel{{\scriptstyle j_{-1}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle i_{-1}}}{{\rightarrow}}\mathcal{B}\rightarrow 0$ is an exact sequence of abelian categories and ${\rm Ker}i_{-1}={\rm Im}j_{-1}$. It follows from Lemma 2.1$(2)$ that $i_{-2}$ is fully faithful. Thus

\mathcal{A}/\mathcal{B}$$\mathcal{A}$$\mathcal{B}$$j_{-1}$$i_{-1}$$j_{-2}$$i_{-2}$$Q$$i

is recollement such that $Q$ and $j_{-2}$ are exact. It follows from Theorem 1.1 that $Q\cong j_{-2}$ and $i\cong i_{-2}$, and hence the type is $(+\infty,-\infty)$.

Assume that there is a diagram of functors

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i_{-2}$$j_{-2}$$i_{-3}$$j_{-3}$$i_{-4}$$j_{-4}$$i_{-1}$$j_{-1}$$i$$Q

such that $(i,i_{-1},i_{-2},i_{-3},i_{-4})$ and $(Q,j_{-1},j_{-2},j_{-3},j_{-4})$ are adjoint sequences. Then the six functors $i_{-1},i_{-2},i_{-3},j_{-1},j_{-2},j_{-3}$ are exact. By the same argument as above we know that the type is $(+\infty,-\infty)$.

Up to now we have proved Claim 1. So, from now on we assume that $h\leq 4$, i.e., $m+n\leq 3$. Then the type $(m,-n)$ of $\mathcal{B}$ is in the list

[TABLE]

Assume that there is a diagram of functors

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i_{1}$$i$$j_{1}$$Q

such that $(i_{1},i)$ and $(j_{1},Q)$ are adjoint pairs. Then $j_{1}$ is fully faithful, by the dual of [GL, Prop. 2.2]. So it is a left recollement, and hence by Theorem 1.2 it can be extended to be recollement. This shows that the type of $\mathcal{B}$ is not in the set $\{(3,0),\ (2,0),\ (1,0)\}.$

Claim 2. The type of $\mathcal{B}$ can not be $(0,-3)$.

Otherwise, there is a diagram of functors

\mathcal{B}$$\mathcal{A}$$\mathcal{A}/\mathcal{B}$$i_{-2}$$j_{-2}$$i_{-3}$$j_{-3}$$i_{-1}$$j_{-1}$$i$$Q

such that $(i,i_{-1},i_{-2},i_{-3})$ and $(Q,j_{-1},j_{-2},j_{-3})$ are adjoint sequences. Then $i_{-1},i_{-2},j_{-1},j_{-2}$ are exact, and $j_{-1}$ is fully faithful. So, the upper two rows form a right recollement. Thus, by Proposition 3.1(4)(iii), $0\rightarrow\mathcal{A}/\mathcal{B}\stackrel{{\scriptstyle j_{-1}}}{{\rightarrow}}\mathcal{A}\stackrel{{\scriptstyle i_{-1}}}{{\rightarrow}}\mathcal{B}\rightarrow 0$ is an exact sequence of abelian categories, and hence $i_{-2}$ is fully faithful, by Lemma 2.1(2). So the upper three rows form a recollement $(\mathcal{A}/\mathcal{B},\mathcal{A},\mathcal{B},Q,j_{-1},j_{-2},i,i_{-1},i_{-2})$, and then $i\cong i_{-2}$ and $Q\cong j_{-2}$ by Theorem 1.1. Thus the type of $\mathcal{B}$ is $(+\infty,-\infty)$, which is absurd.

It remains to prove that for each $(m,-n)$ in the list

[TABLE]

there exists a Serre subcategory of $\mathcal{A}$ such that its type is $(m,-n)$. This is true by Subsections 5.1-5.5. This completes the proof of Theorem 1.3.