Draft lesson.

Serre quotients and local saturation

Written with GPT-6.1 Sol (OpenAI). Original text: CC0.

A quotient of an abelian category can erase a chosen class of objects while retaining exact sequences. Its morphisms allow both restriction of the source and removal of a negligible part of the target. Recovering an ambient representative requires a further operation: maps must extend uniquely across every arrow made invertible by the quotient.

We use the fraction and natural-transformation conventions of One-sided fractions and saturation, the subobject bounds of Generators and small quotient families, and the representability theorem of Solution sets and universal representations. All indexing sets, Hom sets and colimits called small belong to a fixed universe. A preliminary localization may be formed in a larger universe; we prove the needed small Hom bound before applying representability. Ordinary abelian-category kernels, images and exactness have the conventions retained in Trace coreflections and balanced nonabelian categories.

1. The exact quotient interface

Let $\mathcal A$ be abelian. A Serre subcategory $\mathcal T$ is a full subcategory, closed under isomorphisms, containing zero and closed under subobjects, quotients and extensions. In particular it is additive. Section 8 proves the exact quotient theorem, also treated in the Stacks Serre-quotient theorem: the arrows $S_{\mathcal T}=\{f:\ker f,\operatorname{coker}f\in\mathcal T\} \tag{1.1}$ admit both fraction calculi, and their localization $q:\mathcal A\longrightarrow\mathcal B=\mathcal A/\mathcal T \tag{1.2}$ is abelian and exact, with $qX=0$ exactly for $X\in\mathcal T$. Exact functors annihilating $\mathcal T$ factor through $q$. The complete construction, additive structure and exactness for these Serre denominators are proved in Section 8. Further reading for general two-sided denominator systems is the Stacks localization theorem and its additive lemma. The incoming and outgoing names refer to the diagrams, so do not require translating an author's left/right terminology.

These statements apply before any coproduct or generator hypothesis. A localization in a larger universe does not yet assert small Hom sets in the original universe.

We need the precise saturation consequence. Exactness gives $\begin{gathered} qf\text{ invertible}\\ \Longleftrightarrow q(\ker f)=q(\operatorname{coker}f)=0\\ \Longleftrightarrow f\in S_{\mathcal T}. \end{gathered} \tag{1.3}$ Here an arrow in an abelian category is invertible precisely when its kernel and cokernel vanish. Thus (1.1) is already the inverse image of all quotient isomorphisms. In particular, whenever any two of $f,g,gf$ belong to it, so does the third. It also satisfies the middle-arrow saturation condition: $gf,hg\in S_{\mathcal T}$ imply $g\in S_{\mathcal T}$, because $qg$ has both a left and a right inverse, which coincide.

Closure under kernels and cokernels of maps between objects of a subcategory is weaker than Serre closure under all ambient subobjects and quotients. The latter hypotheses are essential in (1.1); they must not be replaced by a different convention for “thick.”

2. A small formula for quotient morphisms

Assume now that $\mathcal A$ is locally small and that subobjects of each object form a small set of classes. For $X,Y$, use pairs $\begin{gathered} U\subseteq X,\quad X/U\in\mathcal T,\\ V\subseteq Y,\quad V\in\mathcal T. \end{gathered} \tag{2.1}$ A transition replaces $U$ by a smaller such subobject and $V$ by a larger one. Representatives can be chosen once for the subobject classes; the unique maps over $X$ or $Y$ make the resulting indexing poset unambiguous.

This poset is nonempty, using $(X,0)$, and directed. For two source choices use $U_1\cap U_2$: its quotient in $X$ embeds in $X/U_1\oplus X/U_2$, so belongs to $\mathcal T$. For two target choices use $V_1+V_2$, which is a quotient of $V_1\oplus V_2$ and belongs to $\mathcal T$.

Proposition 2.1. There is a natural bijection $\begin{gathered} \operatorname{Hom}_{\mathcal B}(qX,qY)\\ \simeq \underset{(U,V)\text{ in }(2.1)}{\operatorname{colim}} \operatorname{Hom}_{\mathcal A}(U,Y/V). \end{gathered} \tag{2.2}$ Consequently $\mathcal B$ has small Hom sets.

Proof. Write $i:U\hookrightarrow X$ for the inclusion and $p:Y\to Y/V$ for the quotient map. A representative $a:U\to Y/V$ determines $q(p)^{-1}q(a)q(i)^{-1}:qX\longrightarrow qY. \tag{2.3}$ Both inverted maps belong to (1.1). Restrictions and target quotients leave (2.3) unchanged.

For surjectivity, take an outgoing fraction $q(s)^{-1}q(f)$, where $f:X\to Z$ and $s:Y\to Z$ is a denominator. Put $V=\ker s$, and identify $Y/V$ with $\operatorname{im}s$. Let $U=f^{-1}(\operatorname{im}s)$. Then $X/U$ embeds into $\operatorname{coker}s$, so is negligible. The restricted map factors through $Y/V$; its image under (2.3) is the original fraction, since the ordinary square commutes.

For injectivity, suppose two representatives determine the same quotient arrow. Restrict both to $U=U_1\cap U_2$ and project their targets to $Y/(V_1+V_2)$. The difference $d$ of the resulting ordinary maps has $qd=0$, by cancelling the two denominator isomorphisms in (2.3). Exactness gives $q(\operatorname{im}d)=0$, hence $\operatorname{im}d\in\mathcal T$. Its inverse image $W$ in $Y$ is an extension of $\operatorname{im}d$ by $V_1+V_2$; therefore $W\in\mathcal T$. Projection to $Y/W$ makes the two maps equal. They thus agree at a common later index, exactly the equivalence relation in the directed colimit.

This proves the bijection. It is natural because (2.3) is defined by the localization functor and composition; equivalently, pulling back $U$ handles a map into $X$, and taking the image of $V$ handles a map out of $Y$. The index set and all its Hom sets are small, so their colimit is small. $\square$

The two alterations in (2.1) have different purposes. A map in the quotient can exist after restricting its source even when its target has no nonzero negligible subobject. This is why removing the largest negligible subobject alone cannot generally represent quotient maps.

3. The torsion part and preservation of colimits

From this point suppose that $\mathcal A$ is a Grothendieck category: it is locally small and abelian, has all small colimits, has exact filtered colimits, and has a generator $G$. Suppose also that $\mathcal T$ is Serre and closed under small coproducts. Closure under all small colimits follows, since a colimit is a cokernel of a map between coproducts.

In an abelian category the generator is both detecting and separating, and every epimorphism is strict. The retained subobject and kernel-pair bounds therefore make $\mathcal A$ wellpowered and co-wellpowered. Proposition 2.1 applies to its quotient. These are bounds in the same fixed universe.

For any $X$, sum its small family of subobjects belonging to $\mathcal T$: $tX=\sum_{\substack{V\subseteq X\\V\in\mathcal T}}V. \tag{3.1}$ The sum is the image of their coproduct, hence belongs to $\mathcal T$, and contains every such subobject. It is the largest torsion subobject. Images show that each $f:X\to Y$ restricts to $tf:tX\to tY$; these restrictions define an additive functor. Every map from a torsion object to $X$ factors uniquely through $tX$. Thus $t$ is right adjoint to the inclusion of $\mathcal T$.

Write $\overline X=X/tX$. This object is torsion-free, meaning it has no nonzero subobject in $\mathcal T$. Indeed a torsion subobject of $\overline X$ lifts to an extension of torsion objects in $X$, which must be contained in $tX$. Equivalently, $\operatorname{Hom}_{\mathcal A}(T,\overline X)=0 \qquad(T\in\mathcal T), \tag{3.2}$ because the image of such a map is torsion.

Proposition 3.1. The quotient has all small colimits, and $q$ preserves them.

Proof. We first prove preservation and existence of coproducts, without assuming a right adjoint. Small coproducts in $\mathcal A$ are exact: write them as filtered colimits of their finite subsums and use exact filtered colimits. In particular coproducts of monomorphisms are monic.

For a family $X_i$, any arrows $qX_i\to qY$ can, by (2.2), be represented by $a_i:U_i\longrightarrow Y/tY,\qquad X_i/U_i\in\mathcal T.$ Projecting the original target quotient further to $Y/tY$ permits this single common target. The map $\bigoplus U_i\to\bigoplus X_i$ is monic with torsion cokernel $\bigoplus(X_i/U_i)$. It is a denominator. The maps $a_i$ assemble into one map to $Y/tY$, so (2.3) gives the required map $q(\bigoplus X_i)\to qY$.

For uniqueness, subtract two possible maps. Represent their difference by $a:U\to Y/tY$, with $U\subseteq X=\bigoplus X_i$ and $X/U\in\mathcal T$. If its restrictions to every $qX_i$ vanish, set $U_i=X_i\cap U$. Then $X_i/U_i$ embeds into $X/U$, and is torsion. The ordinary map $a|_{U_i}$ has zero quotient image; its image is torsion and is a subobject of the torsion-free $Y/tY$. Thus $a|_{U_i}=0$.

The subobject $\bigoplus U_i\subseteq X$ lies in $U$. Moreover $U/\bigoplus U_i\ \subseteq\ X/\bigoplus U_i \simeq\bigoplus(X_i/U_i)$ is torsion. Hence $a$, which vanishes on $\bigoplus U_i$, factors through a torsion object and is zero by (3.2). The quotient difference is zero. The empty family is covered by $q0=0$. This proves the coproduct universal property.

Exactness of $q$ gives preservation of cokernels. In an additive category a coequalizer of $f,g$ is $\operatorname{coker}(f-g)$. Every small diagram has its colimit presented by the coproduct over arrows mapping to the coproduct over objects, with the usual difference of the two structure maps. The quotient now has these coproducts and cokernels, and $q$ preserves their presentation. Thus it has all small colimits, including diagrams whose arrows are already quotient arrows, and $q$ preserves all existing colimits in $\mathcal A$. $\square$

4. The section functor and its local objects

Theorem 4.1. The quotient functor has a fully faithful right adjoint $q\dashv s:\mathcal B\longrightarrow\mathcal A. \tag{4.1}$ Its counit $qs\to\operatorname{id}_{\mathcal B}$ is invertible. For its unit $\eta_X:X\to sqX$, $\ker\eta_X=tX,\qquad \operatorname{coker}\eta_X\in\mathcal T. \tag{4.2}$ Its values are exactly the objects $L$ with the following two properties:

$\operatorname{Hom}(T,L)=0$ for every $T\in\mathcal T$.
Every short exact sequence $0\to L\to E\to T\to0$, with $T\in\mathcal T$, splits.

We call these local objects.

Proof. For $B\in\mathcal B$, the small-set-valued functor $X\longmapsto\operatorname{Hom}_{\mathcal B}(qX,B) \tag{4.3}$ sends colimits to limits by Proposition 3.1. Theorem 3.1 of the representability lesson applies: $\mathcal A$ is cocomplete, co-wellpowered and has a separating generator. It represents (4.3). Yoneda defines the representatives' maps functorially in $B$, giving $s$ and both naturality variables of (4.1).

An object is local precisely when $\operatorname{Hom}(-,L)$ takes every denominator to a bijection. In one direction, a map $f:X\to L$ kills the torsion kernel of a denominator $d:X\to Y$, so descends to $I=\operatorname{im}d$. Push out $0\longrightarrow I\longrightarrow Y\longrightarrow \operatorname{coker}d\longrightarrow0$ along $I\to L$. Its new short exact sequence has torsion quotient and splits; a retraction to $L$ gives an extension $Y\to L$ of $f$. Two extensions differ by a map from $\operatorname{coker}d$ to $L$, hence coincide. Conversely, applying the bijection to $0\to T$ makes $\operatorname{Hom}(T,L)=0$. Applying it to $L\hookrightarrow E$ extends $\operatorname{id}_L$ to a retraction $E\to L$, so the stated extension splits.

Adjunction makes every $sB$ local, since $qd$ is invertible. For a local $L$, formula (2.2) simplifies to $\operatorname{Hom}_{\mathcal A}(X,L) \simeq\operatorname{Hom}_{\mathcal B}(qX,qL). \tag{4.4}$ There is no nonzero torsion subobject $V$ of $L$, so its target choices are only zero. Restriction along every $U\hookrightarrow X$ in (2.1) is bijective by locality. The directed colimit therefore equals the ordinary Hom set, via the map induced by $q$.

For $L=sB$, combine (4.4) with adjunction. The induced bijection $\operatorname{Hom}_{\mathcal B}(qX,qsB) \longrightarrow\operatorname{Hom}_{\mathcal B}(qX,B)$ is postcomposition with the counit: on $qf$ it is the adjoint transpose $\epsilon_Bqf$. Every quotient object has the form $qX$. Yoneda makes $\epsilon_B$ invertible. The usual adjunction bijection, followed by this counit isomorphism, identifies $\operatorname{Hom}(sB,sB')$ with $\operatorname{Hom}(B,B')$, so $s$ is fully faithful.

The triangle identity $\epsilon_{qX}q\eta_X=\operatorname{id}_{qX}$ now makes $q\eta_X$ invertible. By (1.3) its kernel and cokernel are torsion. Every map from a torsion object to the local $sqX$ vanishes, so $tX\subseteq\ker\eta_X$; maximality gives equality. This proves (4.2).

Finally, for local $L$, (4.4) and adjunction identify $L$ and $sqL$ by their represented functors. The resulting map is $\eta_L$, so it is invertible. Thus all local objects occur as values of $s$, up to isomorphism. $\square$

The right adjoint is additive and left exact. For additivity, $q$ preserves biproducts; the adjunction and their Hom universal properties identify $s(B\oplus B')$ with $sB\oplus sB'$, with the prescribed projections and inclusions, and $s0=0$. The biproduct equations then give additivity on arrows. For any existing limit in $\mathcal B$, the adjunction takes its universal property to the limit universal property in $\mathcal A$. In particular $s$ preserves kernels and finite products, hence is left exact. Exercise 3 shows that right exactness can fail.

The object $sqX$ is the local saturation of $X$. Formula (4.2) factors its unit as $X\twoheadrightarrow X/tX\hookrightarrow sqX, \tag{4.5}$ where the second map has torsion cokernel. The correct representation is $\begin{gathered} \operatorname{Hom}_{\mathcal B}(qZ,qX) \\ \simeq\operatorname{Hom}_{\mathcal A}(Z,sqX). \end{gathered} \tag{4.6}$ Replacing $sqX$ by $X/tX$ is valid exactly when $X/tX$ is local.

For example let $\mathcal A=\operatorname{Ab}$ and let $\mathcal T$ consist of all torsion groups. It is closed under subgroups, quotients and coproducts. In an extension with torsion ends, a positive multiple of any element lands in the torsion kernel and a further positive multiple kills it; hence it is Serre. Rationalization is exact by Stacks, Proposition 10.9.12, and the retained module quotient equivalence gives $\operatorname{Ab}/\mathcal T\simeq\operatorname{Vect}_{\mathbf Q}.$ Here is the needed fullness interface. For any group $M$, the map $M\to\mathbf Q\otimes M$ has torsion kernel and cokernel, so is inverted by $q$. Between rational vector spaces, every additive map is rational linear, since multiplication by each positive integer is invertible. Such spaces are unchanged by rationalization. Thus any map between rationalizations is already represented by an ordinary map between these target representatives, and transporting along the inverted units proves fullness. Exactness identifies the killed objects as precisely torsion groups, so the induced functor is faithful by Stacks, Lemma 12.10.7. Every rational vector space rationalizes to itself, proving essential surjectivity.

The right adjoint is the underlying-group functor; its elementary adjunction sends $m/n$ to $f(m)/n$. Hence $sq\mathbf Z=\mathbf Q$, whereas $t\mathbf Z=0$. Moreover $\operatorname{Hom}(\mathbf Q,\mathbf Z)=0$: the integer $f(1)$ must be divisible by every positive integer and is zero, and then torsion-freeness gives $f(m/n)=0$. In the quotient, $\mathbf Z\hookrightarrow\mathbf Q$ is an invertible denominator with nonzero inverse, detected by rationalization. Thus ordinary maps into the torsion-free quotient $\mathbf Z$ cannot represent all quotient maps into $q\mathbf Z$.

5. The quotient is again Grothendieck

Theorem 5.1. Under the hypotheses of §3, $\mathcal B$ is Grothendieck.

Proof. It is locally small by (2.2), abelian by the canonical quotient theorem, and cocomplete by Proposition 3.1. The object $qG$ is a separating generator. Indeed two different quotient arrows become different under the faithful $s$; a map from $G$ distinguishes them. Adjunction turns that map into one from $qG$ distinguishing the original arrows. In an abelian category separation implies detection, as in the prerequisite generator lesson.

It remains to prove exactness of filtered colimits, rather than infer it from existence of an adjoint. For a small filtered diagram of short exact sequences $0\to B'_i\to B_i\to B''_i\to0$, apply the left exact $s$, and put $E_i=\operatorname{coker}(sB_i\to sB''_i)$. These cokernels form a diagram and give exact sequences $\begin{gathered} 0\longrightarrow sB'_i\longrightarrow sB_i \\ \longrightarrow sB''_i\longrightarrow E_i\longrightarrow0. \end{gathered} \tag{5.1}$ The counit and exactness of $q$ show $qE_i=0$, so $E_i\in\mathcal T$. Exact filtered colimits in $\mathcal A$ preserve (5.1): factor it into two short exact sequences using its functorial images. Applying the exact, colimit-preserving $q$, and using the natural counit $qs\simeq\operatorname{id}$, gives the desired sequence of colimits in $\mathcal B$. Its final term is zero because $\operatorname{colim}E_i\in\mathcal T$. This proves exact filtered colimits. $\square$

6. Graded exercises with complete solutions

Exercise 1 (warm-up: denominators at one prime). Fix a prime $p$. Let $\mathcal T_p$ be the groups whose individual elements are killed by a power of $p$. Determine the torsion part and saturation of $\mathbf Z\oplus\mathbf Z/p^r$, for $r\geq1$. Characterize all local groups. Is every torsion-free group local?

Solution. Subgroups and quotients retain the elementwise property. In an extension, successive powers killing an element's image and then its multiple in the kernel give one power killing that element. Coproducts retain it because each element has finite support. Thus §3 applies. Exact localization $M\mapsto M[1/p]$ kills precisely $\mathcal T_p$. Its unit has kernel the elements killed by a power of $p$, and its cokernel has the same property: the class of $m/p^n$ is killed by $p^n$.

A local group must invert multiplication $p:\mathbf Z\to\mathbf Z$, a denominator, on Hom sets into it. This says multiplication by $p$ on the group is bijective. Conversely such a group has a unique $\mathbf Z[1/p]$-module structure; maps into it extend uniquely from $M$ to $M[1/p]$ by $m/p^n\mapsto p^{-n}f(m)$. Exact localization takes every denominator to an isomorphism, so this Hom adjunction proves locality. As for rationalization in §4, transporting along the units proves that the quotient is the category of $\mathbf Z[1/p]$-modules and its section functor forgets that structure.

The stated group's torsion part is $0\oplus\mathbf Z/p^r$, and its saturation is $\mathbf Z[1/p]$. The torsion-free quotient is $\mathbf Z$. It is not local, since multiplication by $p$ is not surjective. Thus even absence of all ordinary torsion does not ensure locality.

Exercise 2 (moderate: a split quotient). Let $R=R_1\times R_2$, with unital rings that need not be commutative. For left $R$-modules, let $\mathcal T$ consist of those killed by the central idempotent $e_1=(1,0)$. Compute the quotient, its section functor, and the torsion and saturation of an arbitrary module. Prove that its torsion-free quotient is already local.

Solution. Put $e_2=(0,1)$. Every module has a natural decomposition $M=e_1M\oplus e_2M$, since the idempotents sum to one and multiply to zero. Maps preserve it. Conversely a pair of left $R_i$-modules gives such an $R$-module, so the ambient category is the product of the two module categories. Exactness and colimits are componentwise. Therefore $\mathcal T$, the objects with zero first component, is Serre and coproduct-closed.

The projection $M\mapsto e_1M$ is exact and has fully faithful right adjoint $N\mapsto(N,0)$. The map $M\to(e_1M,0)$ is a denominator, and maps between these latter objects are exactly $R_1$-linear maps. Transporting along those maps proves fullness and essential surjectivity of the induced quotient functor; its killed objects are exactly $\mathcal T$, so the canonical faithful-quotient criterion gives faithfulness. The quotient is thus $R_1$-modules and the indicated embedding is its section functor.

Here $tM=e_2M$, since every torsion subobject lies in that component. Both $M/tM$ and $sqM$ are $(e_1M,0)$. Directly, a map from $(0,T)$ to this object is zero, and every extension of $(0,T)$ by $(N,0)$ splits componentwise. These are precisely the locality conditions. The section functor is exact in this example.

Exercise 3 (hard: a section functor that loses epimorphisms). Over any field $k$, put $A=k[x,y]$, $\mathfrak a=(x,y)$. Let $\mathcal T$ consist of modules whose individual elements are killed by some power of $\mathfrak a$. Show that $\mathcal T$ is Serre and coproduct-closed. Prove that $A_x,A_y,A_{xy}$ are local, and that $q(A_x\oplus A_y)\xrightarrow{\,q(u-v)\,}qA_{xy}$ is epic although its image under the section functor is not epic.

Solution. Submodules and quotients preserve the condition, as do coproducts by finite support. For an extension with torsion ends, take $m$ in its middle term. Some $\mathfrak a^n m$ lies in its torsion kernel. The ideal $\mathfrak a^n$ is finitely generated by its $n+1$ monomials; choose a common exponent $r$ killing their multiples of $m$. Then $\mathfrak a^{r+n}m=0$. This proves extension closure, including the needed finite-generation step.

Every $T\in\mathcal T$ has $T_x=T_y=0$, since $\mathfrak a^n t=0$ implies $x^nt=y^nt=0$. Exact module localization therefore takes every denominator in (1.1) to an isomorphism after inverting $x$, or after inverting $y$. For any $A_x$-module $L$, ordinary maps $M\to L$ extend uniquely to $M_x\to L$, by $m/x^n\mapsto x^{-n}f(m)$. This adjunction makes $\operatorname{Hom}(-,L)$ invert all denominators, hence makes $L$ local. Apply it to $A_x$ and $A_{xy}$; the argument with $y$ gives $A_y$. Their finite direct sum is also local, since both Hom conditions are componentwise.

The cokernel of $d(u,v)=u-v$ is $H=A_{xy}/(A_x+A_y).$ A class represented by $f/(x^ry^s)$, with $r,s\geq1$, is killed by $\mathfrak a^{r+s-1}$. Each generating monomial of that degree has $x$-degree at least $r$ or $y$-degree at least $s$; multiplying cancels one whole denominator and puts the result in $A_y$ or $A_x$. Representatives with $r=0$ or $s=0$ already vanish. Thus $H\in\mathcal T$, and exactness makes $qd$ epic.

But $H\neq0$. In the Laurent polynomial basis of $k[x^{\pm1},y^{\pm1}]$, every monomial in $A_x+A_y$ has at least one nonnegative exponent. The monomial $x^{-1}y^{-1}$ has neither, so its class is nonzero. Hence $d$ is not epic in $A$-modules. Since both its source and target are local, the unit identifies $s(qd)$ with $d$. This proves that the right adjoint need not preserve epimorphisms or be right exact, over every field.

Exercise 4 (expert: two successive localizations). Let $\mathcal T\subseteq\mathcal U$ be coproduct-closed Serre subcategories of a Grothendieck category $\mathcal A$. Show that the full subcategory $\overline{\mathcal U}$ of $\mathcal A/\mathcal T$ represented by objects of $\mathcal U$ is again coproduct-closed Serre. Prove the equivalence $(\mathcal A/\mathcal T)/\overline{\mathcal U} \simeq\mathcal A/\mathcal U,$ and identify the section functor for the composite quotient.

Solution. Write $q_T,q_U$ for the two quotient functors. Exact universal descent factors $q_U=Fq_T$, with $F$ exact. An object $q_TX$ is killed by $F$ precisely when $q_UX=0$, that is $X\in\mathcal U$. Thus $\overline{\mathcal U}=\ker F$. Kernels of exact functors are Serre: apply exactness to subobjects and quotients of a killed object and to an extension with killed ends.

To check arbitrary coproducts, first observe that $s_Tq_TX\in\mathcal U$ whenever $X\in\mathcal U$. By (4.2), it is an extension of $X/t_TX\in\mathcal U$ by an object of $\mathcal T\subseteq\mathcal U$. For any small family of objects in $\overline{\mathcal U}$, their section values therefore belong to $\mathcal U$. Their coproduct in $\mathcal A$ belongs to $\mathcal U$, and its image under $q_T$ is their coproduct in the quotient, using Proposition 3.1 and the counit. This proves coproduct closure. Theorems 4.1 and 5.1 apply to this second quotient; denote its quotient and section by $r$ and $j$.

Both $rq_T$ and $q_U$ are exact functors killing exactly the objects of $\mathcal U$. More strongly, they have the same universal descent property. Any exact functor annihilating $\mathcal U$ first descends through $q_T$, and the descended functor annihilates $\overline{\mathcal U}$, so descends through $r$. Conversely every functor factoring through $rq_T$ annihilates $\mathcal U$. Apply these properties to the two quotient functors themselves. They produce comparison functors both ways. Their composites restrict to the respective quotient functors on $\mathcal A$; full faithfulness of restriction on natural transformations for the two localizations lifts the identity comparisons to natural isomorphisms. Thus the comparison functors are inverse equivalences.

For $X\in\mathcal A$ and $B$ in the second quotient, the two adjunctions give $\begin{gathered} \operatorname{Hom}(rq_TX,B) \simeq\operatorname{Hom}(q_TX,jB)\\ \simeq\operatorname{Hom}(X,s_TjB). \end{gathered}$ These identifications are natural in both variables. The composite section is $s_Tj$; under the proved equivalence it is naturally isomorphic to $s_U$, by uniqueness of a right adjoint with its adjunction. Its saturation is therefore $s_Tjrq_TX$, with both local extension conditions retained. Replacing either section by removal of a largest torsion subobject would discard the condition proved in §4.

7. References and scope

Section 8 proves the ordinary Serre quotient theorem. The small subobject bound and the first adjoint construction use the prior course lessons; that argument does not require injective envelopes. Section 9 gives an alternative construction from an envelope. No derived construction is needed. Grothendieck categories are also treated in Pierre Schapira, An Introduction to Categories and Homological Algebra, lecture notes, version of 1 March 2026, Section 5.4. A torsion-free quotient formula requires the local saturation proved in §4.

Original exposition and solutions are dedicated to CC0 1.0. The cited references provide further reading.

8. Constructing the exact quotient

This section proves the quotient assertions in §1 for an arbitrary abelian category and Serre subcategory. The coproduct and generator assumptions of §3 are not used. The universe convention in the introduction permits forming the initial morphism classes; the bound in §2 then gives small Hom sets when its hypotheses hold.

Lemma 8.1 (denominators). The class in (1.1) contains identities and isomorphisms, is closed under composition, satisfies two out of three, and is preserved by pullbacks and pushouts. It has the following two cancellation properties: an equality after postcomposition by a denominator can be made an ordinary equality by a denominator into the source; an equality after precomposition by a denominator can be made an ordinary equality by a denominator out of the target.

Proof. Put $S=S_{\mathcal T}$. For composable $f:X\to Y$ and $g:Y\to Z$, restriction of $f$ and passage to quotient objects give exact sequences $\begin{gathered} 0\longrightarrow\ker f\longrightarrow\ker(gf) \longrightarrow\ker g,\\ \operatorname{coker}f\longrightarrow\operatorname{coker}(gf) \longrightarrow\operatorname{coker}g\longrightarrow0. \end{gathered} \tag{8.1}$ There is no surjectivity assertion at the end of the first row and no injectivity assertion at the beginning of the second. The image in the first row is $\ker g\cap\operatorname{im}f$. In the second, the kernel of the last arrow is the image of the map induced by $g$. These descriptions follow by factoring through the coimage and using the kernel or cokernel universal property. Serre closure under subobjects, quotients and extensions proves that $f,g\in S$ implies $gf\in S$.

If $f,gf\in S$, the intersection $\ker g\cap\operatorname{im}f$ is a quotient of $\ker(gf)$, whereas its quotient in $\ker g$ embeds into $Y/\operatorname{im}f$. Thus $\ker g\in\mathcal T$; also $\operatorname{coker}g$ is a quotient of $\operatorname{coker}(gf)$. If instead $g,gf\in S$, then $\ker f\subseteq\ker(gf)$. The kernel of the induced map $Y/\operatorname{im}f\to Z/\operatorname{im}(gf)$ is $(\ker g+\operatorname{im}f)/\operatorname{im}f$, a quotient of $\ker g$; its image is a subobject of $\operatorname{coker}(gf)$. This kernel description follows by taking the inverse image of $g(\operatorname{im}f)$ under the epimorphism $Y\to\operatorname{im}g$. It proves $\operatorname{coker}f\in\mathcal T$. These are the two remaining cases of two out of three. Identities and isomorphisms have zero kernel and cokernel.

For completeness, pullbacks of epimorphisms in an abelian category are epimorphisms. If $e:D\twoheadrightarrow C$ and $a:B\to C$, their pullback is the kernel of $[e,-a]:D\oplus B\to C$. This last map is epic. Any $h:B\to H$ vanishing on the pullback projection makes $[0,h]$ vanish on that kernel, hence factors as $v[e,-a]$. Its first component gives $ve=0$, so $v=0$ and $h=0$. The cokernel of the projection is therefore zero, proving the claim.

Now pull back $s:D\to C$ in $S$ along $a:B\to C$, and denote the new projection by $p:P\to B$. Its kernel is $\ker s$. Its image is $a^{-1}(\operatorname{im}s)$: after restricting to this inverse image, the remaining projection is a pullback of the epimorphism $D\to\operatorname{im}s$. Consequently $\operatorname{coker}p$ is the image of $B\to\operatorname{coker}s$. Both defects belong to $\mathcal T$, so $p\in S$. Applying this proof to the opposite abelian category proves the pushout statement: the new kernel is a quotient of $\ker s$, and the new cokernel is $\operatorname{coker}s$.

Finally, if $su=sv$ with $s\in S$, then $\operatorname{im}(u-v)\subseteq\ker s$ is torsion. The inclusion $\ker(u-v)\hookrightarrow\operatorname{dom}u$ is a denominator and equalizes $u,v$. If $us=vs$, the difference factors through $\operatorname{coker}s$; its image is torsion. Quotienting the common target by that image is a denominator equalizing $u,v$. This proves both cancellation assertions. $\square$

Theorem 8.2 (roofs and the exact universal property). There is an abelian category with the objects of $\mathcal A$ and the localization functor (1.2). It is exact, kills exactly $\mathcal T$, and has the exact universal property stated in §1.

Proof: the category of roofs. An arrow from $X$ to $Y$ is represented by $X\xleftarrow{s}W\xrightarrow{f}Y, \qquad s\in S. \tag{8.2}$ Two roofs $(s,f)$ and $(s',f')$ are identified if there are $a:R\to W$ and $b:R\to W'$, both in $S$, with $sa=s'b$ and $fa=f'b$. Reflexivity and symmetry are immediate. To check transitivity, pull back the two refinement arrows into the middle roof. Both pullback projections are denominators by Lemma 8.1; composing the corresponding arrows gives a common refinement of the first and third roofs with both equalities. Thus this is an equivalence relation. One may equivalently allow refinement arrows for which just the common composite denominator belongs to $S$: two out of three then makes each refinement arrow a denominator.

Compose (8.2) with $Y\xleftarrow{t}V\xrightarrow{g}Z$ by forming $P=W\times_YV$, with projections $p:P\to W$ and $r:P\to V$. Since $p$ is a pullback of $t$, it is a denominator. Define the composite to be $X\xleftarrow{sp}P\xrightarrow{gr}Z. \tag{8.3}$ Replacing the first roof by a refinement induces a map from its new pullback to $P$ which is the pullback of that refinement along $p$; it is a denominator. Replacing the second induces the pullback of its refinement along $r$. The resulting numerator and denominator squares commute, so each replacement refines (8.3). Common refinements therefore prove well-definedness. The two associations of three roofs have apex canonically isomorphic to the iterated fibre product $W\times_YV\times_ZD$; the canonical isomorphism respects both outer maps and is a denominator. This proves associativity. The roof $(\operatorname{id}_X,\operatorname{id}_X)$ is the identity. Ordinary arrows define a functor $qf=(\operatorname{id}_X,f)$.

For a denominator $s:W\to X$, the roof $(s,\operatorname{id}_W)$ is inverse to $qs$. One composite is $(s,s)$, which refines the identity roof. The other uses $W\times_XW$, whose projections $p,r$ are denominators and satisfy $sp=sr$. Incoming cancellation gives a denominator $e$ with $pe=re$. Refining by $e$ makes the two legs identical, again the identity. Every roof is thus $qf(qs)^{-1}$. In particular, directly from the refinement relation, $qf=qg\quad\Longleftrightarrow\quad fe=ge\text{ for some }e\in S \tag{8.4}$ for ordinary parallel arrows.

If a functor $H:\mathcal A\to\mathcal C$ inverts $S$, its extension must send a roof to $Hf(Hs)^{-1}$. Refinement leaves this value unchanged. For composition, the pullback square gives $tr=fp$; applying $H$ and inverting $Ht,Hp$ proves that this value respects (8.3), without requiring $H$ to preserve pullbacks. Identities are respected, so this is the unique extension. A natural transformation between two such functors is natural on ordinary arrows and their inverses, hence on every roof; it too extends uniquely.

The dual construction, using the pushout and outgoing cancellation already proved, has the same universal property. The two unique extension functors are inverse and identity on objects. Consequently every arrow also has an outgoing expression $(qs)^{-1}qf$, where $f:X\to W$ and $s:Y\to W$ is a denominator. Concretely, push out the two legs of an incoming roof; the new arrow out of its target is a denominator and the square supplies this outgoing expression. The dual of (8.4) holds as well.

Proof: addition and zero. The image of the ordinary zero object is a zero object. For a roof out of zero, $s:W\to0$ is inverted and its inverse equals $q(0\to W)$, since their composite is identity on zero. Its numerator therefore gives the ordinary zero arrow. For a roof into zero the numerator is $0:W\to0$; composing $q(0:X\to0)$ with $qs$ gives precisely that numerator, proving uniqueness. In particular $q$ preserves ordinary zero arrows.

The object $q(A\oplus B)$, with the ordinary projections, is a product. To pair two quotient arrows out of $X$, pull back their incoming denominators into $X$ to obtain a common denominator, then pair their ordinary numerators. For uniqueness, represent the difference-free comparison of two candidate arrows over a common incoming denominator. Equality after each projection supplies, by (8.4), a further denominator making those projected numerators equal. Pull back these two further denominators to get one denominator on which both projections agree; the ordinary product property makes the numerators equal there. The roofs coincide. The dual argument, with a common pushout of outgoing denominators and the dual of (8.4), proves the coproduct property. The ordinary inclusion/projection equations are preserved, including zero, so the canonical coproduct-to-product comparison is identity. Thus this is a biproduct.

These biproducts give addition by pairing two arrows, using the diagonal of the source and the codiagonal of the target, as proved in Recovering addition from products and coproducts. To recall why composition is bilinear, the product property gives $\Delta_Xa=(a\oplus a)\Delta_D$, and the coproduct property gives $b\nabla_Y=\nabla_E(b\oplus b)$; substitution in the addition formula proves both distributive laws. The threefold biproduct gives associativity, the interchange isomorphism gives commutativity, and pairing with a zero arrow gives the identity law. Since $q$ preserves the defining diagrams, it preserves ordinary addition. Hence $q(-\operatorname{id}_Y)\alpha$ is an additive inverse to every roof $\alpha=qf(qs)^{-1}$, by distributivity and $q(-f)+qf=0$. The Hom sets are abelian groups and $q$ is additive.

Proof: kernels, cokernels and exactness. Before using exactness we establish the useful criterion $qh=0\quad\Longleftrightarrow\quad \operatorname{im}h\in\mathcal T \quad(h\text{ ordinary}). \tag{8.5}$ If $qh=0$, (8.4) gives $he=0$ for a denominator $e$. The image of $h$ is then a quotient of $\operatorname{coker}e$, hence torsion. Conversely, if its image is torsion, quotienting the target by that image is a denominator which kills $h$. Its image under $q$ is invertible, so $qh=0$.

If $i$ is an ordinary monomorphism and $qi\alpha=0$, write $\alpha=qf(qs)^{-1}$. Then $\operatorname{im}(if)\in\mathcal T$ by (8.5), and $\operatorname{im}f\simeq\operatorname{im}(if)$. Thus $qf=0$ and $\alpha=0$. Applying this to differences proves $qi$ monic. Dually, if $e$ is an ordinary epimorphism and $\beta qe=0$, use an outgoing expression $\beta=(qs)^{-1}qg$. Here $\operatorname{im}(ge)=\operatorname{im}g$; (8.5) makes $qg=0$, so $qe$ is epic.

Let $a:X\to Y$, and write $k:K\to X$ for its ordinary kernel. If $\alpha=qf(qs)^{-1}:qD\to qX$ satisfies $qa\alpha=0$, then $\operatorname{im}(af)\in\mathcal T$. The inclusion $j:\ker(af)\hookrightarrow\operatorname{dom}f$ is a denominator. The map $fj$ factors uniquely as $kb$. Therefore $\alpha=qk\,qb\,(q(sj))^{-1}. \tag{8.6}$ The factor is unique because $qk$ is monic. This proves that $qk$ is the kernel of $qa$.

For the ordinary cokernel $c:Y\to C$, suppose $\beta=(qs)^{-1}qg:qY\to qD$ kills $qa$, where $g:Y\to W$, $s:D\to W$. Then $\operatorname{im}(ga)$ is torsion. The projection $\pi:W\to W/\operatorname{im}(ga)$ is a denominator, and $\pi g$ kills $a$, so it factors as $hc$. We obtain $\beta=(q(\pi s))^{-1}qh\,qc. \tag{8.7}$ This factor is unique because $qc$ is epic. Thus $q$ preserves kernels and cokernels of ordinary arrows. Every quotient arrow is an ordinary image composed with a source isomorphism, so transporting these constructions gives kernels and cokernels for every arrow. The comparison of coimage with image transports in the same way; for an ordinary image it is the image under $q$ of the ordinary abelian comparison isomorphism. It is therefore invertible. Together with the additive structure and finite biproducts, this proves that the quotient is abelian and $q$ exact.

An object $qX$ is zero exactly when its identity is zero, which by (8.5) is equivalent to $X\in\mathcal T$. An exact functor $H$ annihilating $\mathcal T$ takes every denominator to an arrow with zero kernel and cokernel, hence to an isomorphism. Its unique roof extension preserves biproducts and addition. The kernel and cokernel formulas above show that it preserves kernels and cokernels of ordinary images, because $H$ does; transport across source isomorphisms proves this for all quotient arrows. It is exact. This completes the universal property and the theorem. Formula (1.3) now follows from proved exactness, with no use of it in the construction. $\square$

Example 8.3 (finite groups and rational endomorphisms). In abelian groups, take $\mathcal T$ to be the finite groups. Subgroups and quotients of a finite group are finite, and an extension of two finite groups has order the product of their orders. Thus this is a Serre subcategory. It is not closed under infinite coproducts: a countable direct sum of copies of $\mathbb Z/2$ is infinite. Nevertheless Theorem 8.2 applies.

For $X=Y=\mathbb Z$, the source choices in (2.1) are $n\mathbb Z$, $n\geq1$, and the target choice is only zero. A map $n\mathbb Z\to\mathbb Z$ is specified by the integer $a$ which is its value at $n$. Restriction to $\ell\mathbb Z$, for $n\mid\ell$, sends $a$ to $(\ell/n)a$. Associate the rational number $a/n$. Every rational occurs, and two representatives are equal exactly when restriction to a common multiple makes their values equal. Therefore $\operatorname{End}_{\mathcal A/\mathcal T}(q\mathbb Z) \simeq\mathbb Q. \tag{8.8}$ The roof is multiplication by $a$ followed by the inverse of multiplication by $n$. These integer maps commute; their inverses commute too. Composition is rational multiplication, while addition is rational addition by the common-multiple calculation. This is an isomorphism of rings. It uses only the arbitrary Serre quotient; the coproduct hypothesis needed for §4 is deliberately absent.

9. Saturation constructed from an injective envelope

Return to the Grothendieck and coproduct-closed torsion hypotheses of §3. The essential-extension and injective-envelope theorem is proved in Testing injectivity on generators and constructing enough injectives. We use its categorical existence proof, rather than just the module case.

Theorem 9.1. For $M\in\mathcal A$, put $\overline M=M/tM$ and choose an injective envelope $\overline M\hookrightarrow E$. If $C=E/\overline M$ and $P$ is the inverse image of $tC$ in $E$, the composite $\eta_M:M\to\overline M\hookrightarrow P$ is a universal arrow from $M$ to the local objects of §4. In particular $P\simeq sqM$, naturally with its unit. The local object with this unit is unique up to a unique isomorphism, although the injective-envelope comparison need not be unique.

Proof. The quotient $\overline M$ is torsion-free by (3.2). A torsion subobject of $E$ intersects $\overline M$ in a torsion subobject and hence in zero. Essentiality of the envelope forces that subobject to be zero. Thus $tE=0$. Forming the inverse image gives $\begin{gathered} \overline M\subseteq P\subseteq E,\\ P/\overline M\simeq tC\in\mathcal T, \qquad E/P\simeq C/tC. \end{gathered} \tag{9.1}$ The first quotient identification uses the pullback of the epimorphism $E\to C$; the second follows by taking its further quotient by $tC$. Hence $P$ and $E/P$ are torsion-free. In particular every map from a torsion object to either is zero, since its image would be a torsion subobject.

To prove locality, take any short exact sequence $0\to P\xrightarrow{i}D\to T\to0$ with $T\in\mathcal T$. Injectivity of $E$ extends $P\hookrightarrow E$ along $i$ to $u:D\to E$. The composite $D\to E/P$ kills $P$, so factors through $T$, and is zero by torsion-freeness. Thus $u$ factors through $P$; its factor $r:D\to P$ satisfies $ri=\operatorname{id}_P$. The sequence splits. Together with $\operatorname{Hom}(T,P)=0$, these are exactly the local-object conditions of §4, whose equivalence to extension across denominators was proved there without assuming a section functor.

The map $\eta_M$ has kernel $tM$ and cokernel $tC$ by (9.1), so it is a denominator. For every local $L$, the denominator criterion of §4 now gives a bijection $\operatorname{Hom}_{\mathcal A}(P,L) \xrightarrow{\ \circ\eta_M\ } \operatorname{Hom}_{\mathcal A}(M,L). \tag{9.2}$ This is the asserted universal arrow. Choose such a $P_M$ for each object. For $a:M\to N$, (9.2) gives a unique arrow $S(a):P_M\to P_N$ satisfying $S(a)\eta_M=\eta_Na$. Uniqueness proves the identity and composition laws and naturality of $\eta$. The bijection in (9.2) is a homomorphism of abelian groups; applying it to a sum proves $S(a+b)=S(a)+S(b)$.

For two choices of envelope the corresponding local universal arrows give unique maps in both directions respecting $\eta_M$. Their composites respect that same unit, so uniqueness makes them identities. The comparison between the local objects is unique and satisfies the compatibility law for three choices. This does not assert uniqueness of the envelopes themselves. Finally, $sqM$ is local by §4, and adjunction identifies maps from it to any local $L$ with maps $M\to L$, using (4.4). Its unit has the same universal property as (9.2). The unique unit-respecting comparison gives $P_M\simeq sqM$; uniqueness also makes this comparison natural in $M$. The construction provides an alternative existence argument for local representatives, without the representability step used earlier. $\square$

Example 9.2 (localizing integers at one prime). Let $p$ be prime and let $\mathcal T_p$ consist of groups in which every element is killed by a power of $p$. Exercise 1 verifies the Serre and coproduct conditions. For $M=\mathbb Z$, the torsion part is zero. The inclusion $\mathbb Z\subseteq\mathbb Q$ is essential: a nonzero subgroup contains a nonzero rational $a/b$ and hence its nonzero integer multiple $a$. The group $\mathbb Q$ is divisible; the group case of the injectivity test in the linked lesson proves it injective. Thus it is an injective envelope of $\mathbb Z$.

An element of $\mathbb Q/\mathbb Z$, written as a reduced rational fraction modulo integers, is killed by a power of $p$ exactly when its denominator divides a power of $p$. Consequently $\begin{gathered} t_p(\mathbb Q/\mathbb Z)=\mathbb Z[1/p]/\mathbb Z,\\ P_{\mathbb Z}=\mathbb Z[1/p],\qquad \eta_{\mathbb Z}:\mathbb Z\hookrightarrow\mathbb Z[1/p]. \end{gathered} \tag{9.3}$ This calculates the saturated local representative directly. It need not be injective as an ordinary abelian group. For a different prime $\ell$, the map $\ell\mathbb Z\to\mathbb Z[1/p]$ sending $\ell$ to $1$ cannot extend across $\ell\mathbb Z\subseteq\mathbb Z$: such an extension would supply an element whose $\ell$-multiple is $1$, which no fraction with a power of $p$ in the denominator can do. The ambient envelope is injective; the constructed local representative can fail this stronger property.