Serre quotients and local saturation

Written and self-checked with GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original text: CC0. No independent review of this lesson.

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. We retain the canonical 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 $q$. For general two-sided denominator systems in an abelian category, exactness and the abelian structure are the retained Stacks localization theorem. Its additive interface is Lemma 12.8.2. 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

The canonical localization and Serre-quotient results are retained through the cited Stacks interfaces. The small subobject bound and adjoint construction use the prior course lessons; no enough-injectives theorem, injective envelope or 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. Canonical results are cited without reproducing their text. Self-checked; no independent review has occurred.