Two cokernels, controlled rows and abelian completion

Written and self-checked with GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original text: CC0. The reconstructed version has no independent review.

An arrow of represented presheaves can have two different cokernels. A growing exact row can keep its kernel constant while its middle terms change. These two computational problems organize the lesson. The first leads to recognition and stagewise abelian structure; the second leads to controlled presentations, lifting covers and one-sided exact functors. Completion and extension closure then become reusable consequences with their own hypotheses.

Fix a Grothendieck universe $\mathcal U$, containing the natural numbers. Let $\mathcal C$ be an abelian category whose Hom sets are $\mathcal U$-small. Its objects need not form a $\mathcal U$-small set. Every formal index category, sum and colimit called small below is $\mathcal U$-small. A larger universe houses functor categories when necessary. Write $\iota X$ for the represented ind-object and

h_X(Z)=\operatorname{Hom}_{\mathcal C}(Z,X). \tag{0.1}

We use the recognition and Hom formulas from Ind-objects through their elements, finite diagram strictification and the full-subcategory criterion from Finite diagrams and comma objects, and the colimit and limit constructions from Limits and colimits of formal objects. In particular, a map between ind-objects has a common small filtered stage model, and represented objects are finitely presentable for small filtered colimits in the ind-category.

For universal objects in an arbitrary diagram category, retain the separate Pointwise abelian structure and natural splittings. That lesson is adapted from Tom Leinster and remains CC BY-NC-SA 4.0; following this link does not place its expression under the CC0 terms of this lesson. The additional additive-presheaf and ind-recognition arguments needed here are proved in Section 2.

1. Two cokernels for the same arrow

First problem. In $\mathsf{Ab}$, compare the cokernel of $h_{\mathbb Z}\xrightarrow{2}h_{\mathbb Z}$ taken in additive presheaves with the one taken in ind-objects. Is evaluation at a single free group enough to tell them apart?

In $\mathsf{Ab}$ the cokernel is $\mathbb Z/2$. By Theorem 2.2 the ind-cokernel is $h_{\mathbb Z/2}$. The pointwise presheaf cokernel $Q$ has

\begin{gathered} Q(Z)\\ {}=\operatorname{Hom}(Z,\mathbb Z)/ 2\operatorname{Hom}(Z,\mathbb Z). \end{gathered} \tag{1.1}

At $Z=\mathbb Z$, both $Q(Z)$ and $h_{\mathbb Z/2}(Z)$ are $\mathbb Z/2$; the induced comparison is the identity on the residue of a generator. At $Z=\mathbb Z/2$, one has $\operatorname{Hom}(Z,\mathbb Z)=0$, so $Q(Z)=0$, whereas $h_{\mathbb Z/2}(Z)\simeq\mathbb Z/2$. The two presheaves are therefore different. The last arrow of

0\longrightarrow h_{\mathbb Z}\xrightarrow{2}h_{\mathbb Z} \longrightarrow h_{\mathbb Z/2}\longrightarrow0 \tag{1.2}

is epic in the ind-category, but its value at $\mathbb Z/2$ is $0\to\mathbb Z/2$, which is not surjective. Kernels agree in the two categories; their cokernel universal properties test different collections of target presheaves. Formula (2.8) tests only the left exact ind-presheaves, so it gives precisely the internal cokernel claimed in (1.2).

This computation poses two proof obligations. Why are ind-presheaves precisely the allowed left exact targets, with a size qualification when the base is large? And why is the stagewise quotient their universal cokernel? The next section supplies both providers, rather than treating the pointwise quotient as an ind-object by default.

2. Recognition and universal stage computations

Let $\widehat{\mathcal C}_{\mathrm{add}}$ be the category of additive functors $\mathcal C^{\mathrm{op}}\to\mathsf{Ab}_{\mathcal U}$ and additive natural transformations. This category is abelian, with all kernels and cokernels computed at each object of $\mathcal C$. For example, the groups $\ker(P(Z)\to Q(Z))$ and $\operatorname{coker}(P(Z)\to Q(Z))$ carry induced maps for every arrow in $\mathcal C$. Their additivity follows by restricting, or descending, the additive maps of $P$ and $Q$. The pointwise universal properties give their natural universal properties. Finite biproducts and the coimage-to-image isomorphism are pointwise too. This argument takes place in the larger universe; it does not say that all Hom sets of this presheaf category are $\mathcal U$-small.

Passing to underlying set-valued presheaves is fully faithful. To see the part that needs checking, let $v$ be a natural transformation of the underlying set functors of two additive presheaves $P,Q$. An additive functor carries a zero object to a zero group, so naturality for $Z\to0$ gives $v_Z(0)=0$. The two inclusions $Z\to Z\oplus Z$ show that, under the biproduct identifications, $v_{Z\oplus Z}$ acts on a pair componentwise. Naturality for the diagonal $Z\to Z\oplus Z$ then gives

v_Z(a+b)=v_Z(a)+v_Z(b). \tag{2.1}

Thus every component is a group homomorphism.

Filtered colimits of represented presheaves are additive. They identify $\operatorname{Ind}_{\mathcal U}(\mathcal C)$ with a full subcategory of $\widehat{\mathcal C}_{\mathrm{add}}$. Its Hom groups are $\mathcal U$-small, even when those of the ambient presheaf category need not be: for two small filtered presentations the formula is

\begin{gathered} A=\mathop{\mathrm{colim}}_i^{\rm form}\iota X_i,\\ B=\mathop{\mathrm{colim}}_j^{\rm form}\iota Y_j,\\ \operatorname{Hom}(A,B)\\ {}\simeq\lim_i\mathop{\mathrm{colim}}_j \operatorname{Hom}_{\mathcal C}(X_i,Y_j). \end{gathered} \tag{2.2}

This is a small limit of small filtered colimits of small groups. Composition is bilinear, as it is on representatives at a common stage.

Proposition 2.1. An additive presheaf $P$ is an ind-object precisely when it is left exact and its category of elements is cofinally small. If $\mathcal C$ is small, the second condition is automatic, and

\operatorname{Ind}(\mathcal C) \simeq\operatorname{Lex}(\mathcal C^{\mathrm{op}},\mathsf{Ab}_{\mathcal U}). \tag{2.3}

Here the category of elements is formed from the underlying set presheaf. Cofinal smallness means that it has a small cofinal category; it is not an extra assertion that the whole element category is small.

Proof. Proposition 1.1 of Limits and colimits of formal objects says that, for a category with finite colimits, a set-valued presheaf is an ind-object exactly when it carries finite colimits to finite limits and has a cofinally small element category. Additive functors preserve finite biproducts. In an additive category a coequalizer of $f,g$ is a cokernel of $f-g$, so the finite-colimit condition is exactly left exactness on $\mathcal C^{\mathrm{op}}$. Equivalently, every right exact sequence in $\mathcal C$ gives

\begin{gathered} X\to Y\to Z\to0\\ \Downarrow\\ 0\to P(Z)\to P(Y)\to P(X). \end{gathered} \tag{2.4}

exact. The underlying-set fullness proved above identifies the resulting morphisms with the additive ones. If $\mathcal C$ is small, its element category is small, since each value of $P$ is small. This proves (2.3). $\square$

The additive Yoneda embedding is fully faithful and left exact: $\operatorname{Hom}(Z,-)$ preserves finite limits. It generally fails to be right exact as a functor to additive presheaves. Section 1 gives an explicit failure. Its embedding into the ind-category will nevertheless be exact.

For a small family $(X_s)_{s\in S}$ of objects of $\mathcal C$, define the formal sum using the directed poset of finite subsets of $S$:

\mathop{\bigoplus}_{s\in S}^{\rm form}\iota X_s =\mathop{\mathrm{colim}}_{F\subset S,\ F\text{ finite}}^{\rm form} \iota\bigl(\bigoplus_{s\in F}X_s\bigr). \tag{2.5}

Evaluation at $Z$ gives $\bigoplus_{s\in S}\operatorname{Hom}(Z,X_s)$: every element has finite support. This is a coproduct in the ind-category. Indeed, for an ind-presheaf $L$, the Hom formula identifies maps from (2.5) to $L$ with compatible choices of an element of $L(\bigoplus_{s\in F}X_s)$ for each finite $F$. Additivity identifies this group with the finite product of the $L(X_s)$, and compatibility is exactly an arbitrary family of maps $\iota X_s\to L$. The finite version proves that finite coproducts are biproducts; the zero object is $\iota0$.

Theorem 2.2. Suppose a morphism $f: A\to B$ has a common small filtered presentation $(f_i: X_i\to Y_i)_{i\in I}$. Then

\begin{gathered} \ker f\simeq\mathop{\mathrm{colim}}_i^{\rm form}\iota(\ker f_i),\\ \operatorname{coker}f\simeq \mathop{\mathrm{colim}}_i^{\rm form}\iota(\operatorname{coker}f_i). \end{gathered} \tag{2.6}

The first kernel agrees with the kernel in additive presheaves. The second cokernel is the cokernel in the ind-category, and need not be the pointwise presheaf cokernel. These constructions are natural in commutative squares of stage maps.

Proof. Kernel and cokernel universal properties in $\mathcal C$ supply the transition arrows, with composition and identities forced by uniqueness. On evaluating the first formal object at $Z$, we obtain

\begin{gathered} \mathop{\mathrm{colim}}_i\operatorname{Hom}(Z,\ker f_i)\\ {}\simeq\mathop{\mathrm{colim}}_i \ker\bigl(h_{X_i}(Z)\to h_{Y_i}(Z)\bigr)\\ {}\simeq\ker\bigl(A(Z)\to B(Z)\bigr). \end{gathered} \tag{2.7}

Filtered colimits commute with finite limits of groups; the needed underlying-set result is Stacks, Tag 04AX. The resulting identification is natural in $Z$, so it proves the ambient and internal kernel claims.

Put $Q_i=\operatorname{coker}f_i$ and $Q=\mathop{\mathrm{colim}}_i^{\rm form}\iota Q_i$. For any ind-presheaf $L$, write $f^*$ for precomposition by $f$. Proposition 2.1 and the Hom formula give

\begin{gathered} \operatorname{Hom}(Q,L)\simeq\lim_i L(Q_i)\\ {}\simeq\lim_i\ker\bigl(L(Y_i)\to L(X_i)\bigr)\\ {}\simeq\ker f^*. \end{gathered} \tag{2.8}

Limits commute with limits, so the last equality holds for this small diagram without an exactness assumption on $\lim$. These are natural bijections, with the prescribed map $B\to Q$; they prove the cokernel universal property. Every construction respects a square by its universal property, which proves naturality. Independence of a stage model follows from uniqueness of kernels and cokernels, rather than a choice of a preferred model. $\square$

We have now shown that the ind-category is additive with all kernels and cokernels. The Yoneda embedding into it preserves both, by applying (2.6) to a constant arrow. The inclusion into additive presheaves preserves kernels and finite biproducts, hence is left exact.

3. Abelianity from a small model of each arrow

The kernel and cokernel computations also determine the canonical comparison from coimage to image. First consider a small abelian category $\mathcal D$. Present any ind-morphism by a small filtered family $f_i:X_i\to Y_i$. Write $k_i:\ker f_i\to X_i$ and $q_i:Y_i\to\operatorname{coker}f_i$. Their natural transition maps give stage models for the kernel inclusion and cokernel projection, by Theorem 2.2. Applying that theorem to these two models gives

\begin{aligned} \operatorname{coim}f &\simeq\mathop{\mathrm{colim}}_i^{\rm form} \iota\bigl(\operatorname{coker}k_i\bigr),\\ \operatorname{im}f &\simeq\mathop{\mathrm{colim}}_i^{\rm form} \iota\bigl(\ker q_i\bigr). \end{aligned}

For each stage, the factorization of $f_i$ through these objects is the coimage-to-image comparison in $\mathcal D$, hence an isomorphism. Its naturality follows from the kernel and cokernel universal properties: a commuting square of the $f_i$ induces both factorizations, and uniqueness makes the resulting square commute. Taking their formal colimit therefore gives an isomorphism. It is the canonical comparison for $f$, since its composite with the coimage projection and image inclusion is $f$; the same universal properties characterize that comparison uniquely. Thus $\operatorname{Ind}(\mathcal D)$ is abelian. This argument uses the proved stage formulas, without a sheafification or representability theorem from outside the course.

The following reduction supplies the size generality that this application needs.

Lemma 3.1. Every small family of objects of $\mathcal C$ lies in a small full abelian subcategory $\mathcal D$ whose inclusion into $\mathcal C$ is exact.

Proof. Begin with the given objects and a zero object. At step $n+1$, add chosen finite biproducts of the objects at step $n$, and chosen kernels and cokernels of every morphism between objects at that step. Each new family is small: there are a small family of finite tuples and, by local smallness, a small family of those morphisms. The union over $n\in\mathbb N$ remains small. Take the full subcategory on this union. Any finite tuple or morphism appears at some step, and the requested construction appears at the next step. Thus $\mathcal D$ is additive and has the kernels and cokernels computed in $\mathcal C$. Its coimage-to-image maps are isomorphisms, since they are those in $\mathcal C$, and fullness includes their inverses. It is therefore abelian, and its inclusion preserves short exact sequences. Its objects and arrows both form small sets. $\square$

Theorem 3.2 (abelian structure). For every locally small abelian $\mathcal C$ with the universe convention above, $\operatorname{Ind}_{\mathcal U}(\mathcal C)$ is abelian and locally small. The constant embedding is exact and fully faithful. Essential smallness of $\mathcal C$ is not required.

Proof. Given $f$, choose its small common stage model. Lemma 3.1 places all its stage objects in a small full abelian $\mathcal D$. The functor

\operatorname{Ind}(\mathcal D)\longrightarrow\operatorname{Ind}(\mathcal C) \tag{3.1}

is fully faithful, by (2.2), and preserves kernels and cokernels, by (2.6) and the exact inclusion of $\mathcal D$. The canonical coimage-to-image map for $f$ is therefore the image under (3.1) of that map in $\operatorname{Ind}(\mathcal D)$. The stage comparison just proved makes it an isomorphism. This proves abelianity for every $f$, including when $\mathcal C$ is not essentially small. The exact Yoneda claim follows from Section 2 and full faithfulness. $\square$

Smallness was imposed on the chosen family presenting an arrow, not on the collection of all objects of $\mathcal C$. The separate completion theorem in Section 6 proves the colimit, AB5, limit and generator conclusions.

4. A constant kernel inside a growing exact row

Second problem. Let $p$ be prime and, for this example, let $\mathcal C$ be the category of finitely generated abelian groups. Can the row $0\to\mathbb Z\to\mathbb Z[1/p]\to P_p\to0$ be built while retaining exactly the constant kernel? Can a map from $\mathbb Z/p$ to $P_p$ lift to the middle group? Use the stage rows $\begin{gathered} 0\longrightarrow\mathbb Z\xrightarrow{p^n}\mathbb Z \longrightarrow\mathbb Z/p^n\longrightarrow0,\\ n\geq1, \end{gathered} \tag{4.1}$ with identity transitions on kernels and multiplication by $p$ on middle and quotient terms.

The first square commutes because $p\cdot p^n=p^{n+1}$. The quotient transition sends the class of $z$ modulo $p^n$ to the class of $pz$ modulo $p^{n+1}$; this is well defined, and commutes with the middle transition. Thus these are actual short exact stage rows with the prescribed constant kernel.

The category of finitely generated abelian groups is the category of finitely presented $\mathbb Z$-modules. Here is the needed finiteness fact: a subgroup of $\mathbb Z^m$ is finitely generated, by induction on $m$. Project to the last coordinate; its image is $d\mathbb Z$, with one chosen lift if $d\ne0$. The subgroup in the first $m-1$ coordinates is finitely generated by induction, and those generators together with the lift generate the original subgroup. Taking a finite free surjection onto a finitely generated group now gives a finite presentation. Kernels and cokernels stay finitely generated by the same fact, so this $\mathcal C$ is abelian. Formal colimits and compact presentations, Theorem 4.1 identifies $\operatorname{Ind}(\mathcal C)$ with $\mathsf{Ab}$, naturally by ordinary filtered colimit.

Map the $n$-th middle term to $\mathbb Z[1/p]$ by $z\mapsto z/p^n$. These maps respect the multiplication-by-$p$ transitions, are injective, and their images have union $\mathbb Z[1/p]$. Map the quotient stage by

\begin{gathered} \mathbb Z/p^n\longrightarrow\mathbb Z[1/p]/\mathbb Z,\\ \overline z\longmapsto z/p^n+\mathbb Z. \end{gathered} \tag{4.2}

It is well defined and injective, respects the transitions, and its images exhaust the quotient. The kernel map in the colimit sends $z$ to $p^nz/p^n=z$. Writing $P_p=\mathbb Z[1/p]/\mathbb Z$, the sequence is

0\to\mathbb Z\to\mathbb Z[1/p]\to P_p\to0. \tag{4.3}

The last group is the Prüfer $p$-group. Define $y: \mathbb Z/p\to\mathbb Z[1/p]/\mathbb Z$ by $\overline1\mapsto1/p+\mathbb Z$. Every map $\mathbb Z/p\to\mathbb Z[1/p]$ is zero, since the latter is torsion free; therefore $y$ has no lift with the identity cover. The quotient epimorphism $e: \mathbb Z\to\mathbb Z/p$, together with $x: \mathbb Z\to\mathbb Z[1/p]$ sending $1$ to $1/p$, satisfies the actual equation $fx=y\iota(e)$. Its source and target are in the stated $\mathcal C$.

The identity kernel transitions and multiplication-by-$p$ middle transitions are different controls on the same row. We now prove how to make such choices in any abelian base. The two control theorems prescribe one term at a time; they do not promise simultaneous control by unrelated subcategories.

Theorem 4.1. Every short exact sequence of ind-objects has a small filtered presentation by short exact sequences in $\mathcal C$. If its middle term belongs to $\operatorname{Ind}(\mathcal J)$, for a full additive subcategory $\mathcal J\subset\mathcal C$, the middle stages may be chosen in $\mathcal J$.

Proof. Present the epimorphism $A\to A''$ by arrows $a_i: E_i\to F_i$. Put

K_i=\ker a_i,\qquad H_i=\operatorname{im}a_i. \tag{4.4}

Their transition maps give a functor of short exact rows

0\longrightarrow K_i\longrightarrow E_i\longrightarrow H_i\longrightarrow0. \tag{4.5}

The kernel formula identifies the formal colimit of the first term with $A'$. The cokernel formula, applied to $K_i\to E_i$, identifies the last formal term with $\operatorname{coker}(A'\to A)$, hence with $A''$. These identifications commute with the original maps. This is a presentation of the whole sequence, not merely of its three objects.

If $A$ has a presentation with stages in $\mathcal J$, use it as the specified source presentation of $A\to A''$. The finite arrow strictification theorem supplies a common index with a cofinal projection to that specified index. Its source stages remain in $\mathcal J$. Construction (4.4) retains those middle stages; it requires no kernel closure of $\mathcal J$. $\square$

There is a stronger statement about a prescribed source of a monomorphism.

Theorem 4.2. Let $m: A\to B$ be a monomorphism, with a specified presentation $A=\mathop{\mathrm{colim}}_{i\in I}^{\rm form}\iota\alpha(i)$. There are a small filtered category $K$, a cofinal functor $p: K\to I$, a functor $\beta: K\to\mathcal C$, and a natural transformation

\alpha p\longrightarrow\beta \tag{4.6}

whose components are monomorphisms and whose formal colimit is the given $m$.

Proof. Apply Theorem 4.1 to the short exact sequence of $m$. It first gives some stage monomorphisms $u_j: D_j\to E_j$ presenting $m$. Strictify the identity of $A$, viewed as a map from this kernel presentation $D$ to the specified presentation $\alpha$. The common-arrow theorem gives small filtered $K$, cofinal projections $q: K\to J$, $p: K\to I$, and a stage transformation $t: Dq\to\alpha p$ presenting that identity.

For each $k$, form the pushout in $\mathcal C$

\beta(k)=E(qk)\mathbin{\oplus_{D(qk)}}\alpha(pk). \tag{4.7}

The map $\alpha(pk)\to\beta(k)$ is monic. Here is a direct check of the abelian pushout fact: the pushout is the cokernel of $(u,-t): D\to E\oplus\alpha$. If a map $(0,x)$ into $E\oplus\alpha$ vanishes in that cokernel, it factors through $(u,-t)$, because $(u,-t)$ is monic and is its cokernel's kernel. Its $E$-component gives $uz=0$, hence $z=0$ and $x=0$. Thus the map from $\alpha$ is monic.

Every arrow in $K$ induces a unique map between the chosen pushouts; uniqueness proves functoriality and naturality of (4.6). Finite colimits in the ind-category agree with these stage constructions, and colimits commute with colimits. Hence the formal colimit of (4.7) is the pushout of

A\xrightarrow{m}B,\qquad A\xrightarrow{1_A}A. \tag{4.8}

It is $B$, with its given map from $A$. The projection $p$ is the actual cofinal functor supplied by strictification, so the specified source system has been retained by cofinal reindexing. $\square$

Corollary 4.3. If the kernel term $A'$ of a short exact ind-sequence lies in $\operatorname{Ind}(\mathcal J)$, its short exact stage presentation can be chosen with kernel stages in $\mathcal J$.

Proof. Choose a presentation of $A'$ in $\mathcal J$, apply Theorem 4.2 to $A'\to A$, and take each stage cokernel. Formula (2.6) identifies their formal colimit with $A''$. No closure of $\mathcal J$ under cokernels is needed. $\square$

The middle-term and kernel-term controls are separate conclusions. They do not prescribe both terms simultaneously under arbitrary unrelated choices of subcategories.

The failed torsion lift in the worked row suggests the correct test: allow a cover of the represented test object. The construction is a pullback, so the lifted square is an actual square of maps.

Theorem 4.4. A morphism $f: A\to B$ of ind-objects is epic precisely when, for every $Y\in\mathcal C$ and every $y: \iota Y\to B$, there exist $X\in\mathcal C$, an epimorphism $e: X\to Y$ in $\mathcal C$, and $x: \iota X\to A$ such that

f x=y\,\iota(e). \tag{4.9}

Proof. If $f$ is epic, Theorem 4.1 presents it by stage epimorphisms $E_i\to H_i$. Finite presentability of $\iota Y$ factors $y$ through some $\iota H_i$, by an actual $Y\to H_i$. In $\mathcal C$, form $X=Y\times_{H_i}E_i$. Its map to $Y$ is epic, and the other projection followed by $\iota E_i\to A$ is the desired lift. We used the abelian pullback fact: a pullback of an epimorphism is epic. It also follows by dualizing the pushout-of-monomorphism proof in Section 4 inside the opposite abelian category.

Conversely present $B$ by $(Y_i)$ and form $P_i=A\times_B\iota Y_i$ in the ind-category. The assumed factorization for its structure map $\iota Y_i\to B$ gives a map $\iota X_i\to P_i$ whose composite to $\iota Y_i$ is epic. Thus $P_i\to\iota Y_i$ is epic. The commutation of filtered colimits with finite limits gives

\mathop{\mathrm{colim}}_i P_i\simeq A\times_B B\simeq A. \tag{4.10}

Colimits preserve cokernels, so the colimit of this natural transformation of epimorphisms is epic. Under (4.10) that colimit map is $f$. $\square$

Corollary 4.5. A complex $A\xrightarrow{f}B\xrightarrow{g}D$, with $gf=0$, is exact at $B$ precisely when every $y: \iota Y\to B$ killed by $g$ admits a lift as in (4.9) after an epimorphism $X\to Y$ in $\mathcal C$.

Proof. Such $y$ factors uniquely through $\ker g$. The complex is exact at $B$ exactly when the induced $A\to\ker g$ is epic. Apply Theorem 4.4 to that map. $\square$

The worked row proves that the cover is essential: it need not be possible to take $X=Y$ and $e=1_Y$. Its torsion test map already exhibits this obstruction.

5. Test one-sided functors on that same row

The controlled row is also a test of which exactness property a functor preserves. On its finitely generated abelian-group category, set $\begin{gathered} H(M)=\operatorname{Hom}_{\mathbb Z}(\mathbb Z/p,M),\\ T(M)=M\otimes_{\mathbb Z}\mathbb F_p. \end{gathered} \tag{5.1}$ First prove the extension result using kernels and cokernels; then compute every stage map for these two functors.

Let $F: \mathcal C\to\mathcal C'$ be additive between abelian categories with the same universe convention. Applying $F$ to a formal system defines $\operatorname{Ind}(F)$ on objects and maps; the Hom formula and common stage models give its independence of representatives.

Proposition 5.1. $\operatorname{Ind}(F)$ is additive. If $F$ is left exact, so is $\operatorname{Ind}(F)$. If $F$ is right exact, so is $\operatorname{Ind}(F)$.

Proof. Addition of two maps is computed after strictifying the finite diagram of their sources and targets. Additivity of $F$ respects that addition and zero, and hence the Hom group structures. If $F$ preserves kernels, a common arrow model for $f$ gives natural isomorphisms

\begin{aligned} \operatorname{Ind}(F)(\ker f) &\simeq\mathop{\mathrm{colim}}_i^{\rm form}\iota F(\ker f_i)\\ &\simeq\mathop{\mathrm{colim}}_i^{\rm form}\iota\ker F(f_i)\\ &\simeq\ker\bigl(\operatorname{Ind}(F)(f)\bigr). \end{aligned} \tag{5.2}

Theorem 2.2 in the two categories proves these identifications and identifies their kernel maps. Together with additivity this is left exactness. If $F$ preserves cokernels, the identical argument using the second formula in (2.6) proves right exactness. Both conclusions hold for all ind-arrows, without assuming that a stage presentation already has short exact rows. $\square$

the worked calculation below shows that either one-sided hypothesis may hold without the other.

Worked calculation.

$H$ is additive and left exact, since Hom into a kernel is its kernel of Hom maps. $T$ is additive and right exact: the tensor universal property identifies maps from a tensor of a cokernel with bilinear maps vanishing on its image, which is the cokernel property after tensoring. Finitely generated groups give finitely generated, hence finite dimensional, values of both functors. The ind-category of finite dimensional vector spaces is the category of all small vector spaces by the finite-basis case of the same compact-presentation theorem.

For $H$, both copies of $\mathbb Z$ have value zero. On $\mathbb Z/p^n$, the value is a one dimensional vector space with chosen generator the map

\overline1\longmapsto p^{n-1}\pmod {p^n}. \tag{5.3}

Postcomposition with the quotient transition sends this generator to $p^n$ modulo $p^{n+1}$, the next chosen generator. Thus the transition is the identity in these coordinates, and the quotient's ind-value is $\mathbb F_p$. Applying the extension to (4.3) gives the three terms

0\longrightarrow0\longrightarrow\mathbb F_p. \tag{5.4}

The first map has the correct kernel, while the second is not epic. Thus left exactness survives, and right exactness fails.

For $T$, the kernel stages form the constant system $\mathbb F_p$ with identity transitions. Each middle stage is $\mathbb F_p$, but its transition is multiplication by $p$, hence zero. Its filtered colimit is zero: every element is killed at the next stage. Each quotient stage is also $\mathbb F_p$; the same multiplication-by-$p$ transition is zero, so its colimit is zero too. The first stage arrow $p^n$ tensors to zero for every $n\geq1$. The three resulting terms and maps are

\mathbb F_p\xrightarrow{0}0\longrightarrow0. \tag{5.5}

This is right exact and the first arrow is not monic. Thus right exactness survives, and left exactness fails. These computations concern the actual induced functors on the ind-presentations; they do not assume a functor is exact merely because each original row was short exact.

6. Completion with its size conditions

Theorem 6.1 (completion). Under the locally small abelian hypotheses of Theorem 3.2, the ind-category has all small colimits and its small filtered colimits are exact. If $\mathcal C$ has all small limits, so does its ind-category. If $\mathcal C$ is essentially small, the ind-category has a small generating family, hence a generator, and is Grothendieck abelian. The first two conclusions do not require essential smallness.

Proof. Theorem 3.1 of the cited formal limits and colimits lesson gives all small colimits because $\mathcal C$ has finite colimits. Theorem 4.1 there gives commutation of filtered colimits with finite limits. Filtered colimit, as a left adjoint to the constant-diagram functor, also preserves cokernels. It thus preserves short exact sequences in the abelian ind-category: this is AB5. If $\mathcal C$ has small limits, Corollary 2.3 of that lesson gives small limits in its ind-category, with its stated universe convention.

Finally suppose $\mathcal C$ is essentially small. Choose a small representative family $(C_s)_{s\in S}$. Its represented objects jointly detect maps: two different transformations between ind-presheaves differ at some $C_s$ and some element there, which is a map from $\iota C_s$. The formal coproduct (2.5) is a single generator. Indeed, a detecting map from one summand extends to the whole coproduct by taking zero maps on the other summands. AB5, all small colimits and this generator give the asserted Grothendieck property. We have used essential smallness only for this final conclusion. $\square$

In particular (2.5) is the actual coproduct of the represented family. Small coproducts of arbitrary ind-objects are supplied by the general colimit theorem. The theorem does not assert a small generating family for an arbitrary large $\mathcal C$.

7. Extension closure and the subobjects it does not cover

For this lesson a full subcategory of an abelian category is thick when it is closed, up to isomorphism, under kernels and cokernels of maps between its own objects and under extensions. Closure under every subobject or quotient of an object in the subcategory is the stronger Serre condition.

Final application. Let $k$ be a field, $R=\prod_{r\in\mathbb N}k$, and, for this application, let $\mathcal C$ be the full category of finitely generated projective $R$-modules. It is abelian and its ind-completion is all modules. The ideal of finite-support sequences is a subobject of a represented object but is not itself represented. We verify the category and the actual subobject, then prove the closure theorem it does satisfy.

Every object of $\mathcal C$ is the image of an idempotent finite matrix: write $P=eR^m$ and $Q=fR^n$. A map $A: P\to Q$ extends to a finite matrix $A: R^m\to R^n$ with $A=fAe$. At coordinate $r$, it is a map from $\operatorname{im}e_r$ to $\operatorname{im}f_r$, both finite dimensional vector spaces.

Choose a projection $a_r$ onto the kernel of this restricted map inside $\operatorname{im}e_r$, extending it by zero on $\ker e_r$. Then $a_r^2=a_r$, $a_re_r=e_ra_r=a_r$. Similarly choose $b_r$ projecting $\operatorname{im}f_r$ onto the image of the restricted map, extended by zero on $\ker f_r$. Choose a section $t_r$ of the map onto its image, composed with $b_r$, with values in $\operatorname{im}e_r$. Assemble all their entries into matrices over $R$. This is possible because a finite matrix over $R$ is exactly a sequence of finite matrices over $k$, with no boundedness condition on its coordinate entries. We obtain

\begin{gathered} a^2=a,\quad ae=ea=a,\\ b^2=b,\quad bf=fb=b,\\ At=b,\qquad et=t. \end{gathered} \tag{7.1}

As modules, $eR^m$ identifies with $\prod_r\operatorname{im}e_r$: choose the coordinate vectors freely and group their finitely many components into elements of $R$. Consequently the kernel is $aR^m$. The image is $bR^n$: the inclusion in it is coordinatewise, and $At=b$ gives the opposite inclusion with an actual preimage in $P$. Since $f-b$ is idempotent and complementary to $b$ on $Q$, the cokernel is $(f-b)R^n$. All three are finitely generated projective. Finite biproducts and zero stay in $\mathcal C$, and it is full. Thus it is abelian, with these actual kernels and cokernels and the ordinary coimage-to-image isomorphism.

For a map between finite free modules take $e,f$ to be identities. The same argument shows that every cokernel of a finite matrix is finitely generated projective. Every finitely presented module is such a cokernel, so it belongs to $\mathcal C$. Conversely a finitely generated projective module is a retract of a finite free module; the summand argument in Theorem 4.1 of the compact-presentation lesson proves it finitely presented. Hence

\begin{gathered} \mathcal C=\mathsf{Mod}_{\mathrm{fp}}(R),\\ \operatorname{Ind}(\mathcal C)\simeq\mathsf{Mod}(R). \end{gathered} \tag{7.2}

The equality is of full subcategories of modules, and the equivalence takes a formal filtered system to its module colimit.

Let $e_r\in R$ be the coordinate idempotents and let $I$ be the ideal of sequences with finite support. For a finite $F\subset\mathbb N$, put $e_F=\sum_{r\in F}e_r$. Then

I=\bigcup_{F\subset\mathbb N,\ F\text{ finite}} e_FR. \tag{7.3}

Each $e_FR$ is a represented finitely generated projective module. The displayed directed union presents $I$ as an ind-object, and its inclusion into $R$ is monic under (7.2). But $I$ is not finitely generated: finitely many elements have supports contained in one finite $F$, and their $R$-linear combinations cannot produce $e_r$ for $r\notin F$. Thus $I$ cannot be isomorphic to an object of $\mathcal C$. Proposition 7.1 only closes kernels and cokernels of maps between represented objects and extensions with represented ends. The map $I\to R$ has an unrepresented source, so it witnesses failure of the stronger Serre subobject condition without contradicting thickness.

Proposition 7.1. The represented copy of $\mathcal C$ is thick in $\operatorname{Ind}(\mathcal C)$.

Proof. Its kernels and cokernels stay represented by exactness of $\iota$. For extension closure take a short exact sequence with ends $\iota X,\iota Y$. Apply Theorem 4.2 to its kernel inclusion using the constant presentation of $X$. Complete it by stage cokernels. We get a small filtered short exact presentation

0\longrightarrow X\longrightarrow E_i\longrightarrow Y_i\longrightarrow0, \tag{7.4}

with identity transitions on the first term. The identity of $\iota Y$ factors through some quotient stage $\iota Y_i$. This factorization comes from a map $s: Y\to Y_i$ in $\mathcal C$, by full faithfulness. Pull back (7.4) along $s$. Its middle term $E$ lies in $\mathcal C$, and the resulting row maps to the original ind-row with identity end maps.

The induced middle map $\iota E\to A$ is an isomorphism. For completeness, a map into its kernel has zero quotient component, hence factors through $\iota X$; the identity on that end forces it to vanish. A map from its cokernel is a map $A\to T$ vanishing on $\iota E$. It vanishes on $\iota X$, so factors through $\iota Y$. The epimorphism $\iota E\to\iota Y$ forces that factor to vanish too. Thus the middle map has zero kernel and cokernel and is an isomorphism in the abelian ind-category. $\square$

Theorem 7.2. If a full additive $\mathcal J\subset\mathcal C$ is closed under extensions in $\mathcal C$, then $\operatorname{Ind}(\mathcal J)$ is closed under extensions in $\operatorname{Ind}(\mathcal C)$.

Proof. Take $0\to A'\to A\to A''\to0$ with both ends in $\operatorname{Ind}(\mathcal J)$. The allowed-piece criterion in Section 4 of the cited finite diagrams lesson says that it suffices to factor every map $v: \iota X\to A$, for $X\in\mathcal C$, through a represented object of $\mathcal J$.

Factor its quotient component through $\iota Y''$, with $Y''\in\mathcal J$. Write its first factor as the actual $a: X\to Y''$. By Theorem 4.4 applied to $A\to A''$ and the map $\iota Y''\to A''$, there is an epimorphism $e: X_1\to Y''$ in $\mathcal C$ with a lift $\iota X_1\to A$. Form

\begin{gathered} d=[a,e]\colon X\oplus X_1\longrightarrow Y'',\\ N=\ker d. \end{gathered} \tag{7.5}

The map $d$ is epic because $e$ is. The map from $\iota(X\oplus X_1)$ to $A$, given by $v$ and the chosen lift, has on $\iota N$ a map into $A'$. Factor the latter through $\iota Y'$, with $Y'\in\mathcal J$, using the same allowed-piece criterion. Its first factor is an actual map $N\to Y'$. Push out

0\to N\to X\oplus X_1\to Y''\to0. \tag{7.6}

along that map. The resulting short exact row has ends $Y',Y''$ and middle

P=Y'\mathbin{\oplus_N}(X\oplus X_1). \tag{7.7}

Extension closure puts $P$, up to isomorphism, in $\mathcal J$. The two maps into $A$ agree on $\iota N$, by construction. Since $\iota$ preserves this pushout, they induce $\iota P\to A$. Composing with $X\to X\oplus X_1\to P$ recovers $v$. Every represented test map now factors through $\mathcal J$, proving the claim. Fullness makes all the stage maps used in the allowed-piece criterion morphisms of the stated subcategory; no smallness or kernel closure of $\mathcal J$ was assumed. $\square$

8. Graded exercises with full solutions

Exercise 1 (foundation: detect failure of left exactness). Replace $2$ in the first problem by an integer $m\geq2$. Evaluate both cokernels at $\mathbb Z$ and $\mathbb Z/m$, and prove directly that the pointwise cokernel is not left exact.

Solution. The ind-cokernel is $h_{\mathbb Z/m}$, by the universal stage formula. The pointwise cokernel is $Q_m(Z)=\operatorname{Hom}(Z,\mathbb Z)/m\operatorname{Hom}(Z,\mathbb Z)$. Both values at $\mathbb Z$ are $\mathbb Z/m$, but $Q_m(\mathbb Z/m)=0$, while $h_{\mathbb Z/m}(\mathbb Z/m)=\mathbb Z/m$. Apply $Q_m$ contravariantly to $\mathbb Z\xrightarrow{m}\mathbb Z\to\mathbb Z/m\to0$. Left exactness would require $0\to0\to\mathbb Z/m\xrightarrow{m}\mathbb Z/m$ to be exact. The last map is zero, so its kernel is the nonzero whole group and the preceding image is zero. This proves the failure explicitly.

Exercise 2 (intermediate: a composite denominator). Build the controlled row for any integer $q\geq2$, using $q^n$ in place of $p^n$. Identify the transition maps and colimit, and give a test object whose map to the quotient needs a nonidentity cover.

Solution. Use $0\to\mathbb Z\xrightarrow{q^n}\mathbb Z\to\mathbb Z/q^n\to0$ with identity, multiplication by $q$, and $\overline z\mapsto\overline{qz}$ as the three transitions. The first square commutes since $qq^n=q^{n+1}$; the quotient square commutes by the same equality. Maps from the middle stages send $z$ to $z/q^n$, and maps from the quotient stages send $\overline z$ to $z/q^n+\mathbb Z$. They are compatible and injective, with unions $\mathbb Z[1/q]$ and $\mathbb Z[1/q]/\mathbb Z$. The kernel arrow sends $z$ to $q^nz/q^n=z$, giving the stated exact row. The quotient has the nonzero element $1/q+\mathbb Z$, of exact order $q$. Thus $y:\mathbb Z/q\to\mathbb Z[1/q]/\mathbb Z$, $\overline1\mapsto1/q+\mathbb Z$, has no lift to the torsion-free middle group. The quotient cover $\mathbb Z\to\mathbb Z/q$ and the map $1\mapsto1/q$ lift it and satisfy the required square. Primality was needed to name a Prüfer group in the first example, not for the controlled-stage construction.

Exercise 3 (advanced: separate stage maps from their colimits). For the prime-$p$ row, compute all arrows after $H=\operatorname{Hom}(\mathbb Z/p,-)$ and after $T=-\otimes\mathbb F_p$ at a fixed stage $n$. Then compute each transition to stage $n+r$, for $r\geq1$, and compare with the formal colimit maps in Section 5.

Solution. At every stage the $H$-terms are $0,0,\mathbb F_p$, so both row arrows are zero. The last group's generator is the map $\overline1\mapsto p^{n-1}$; the transition to $n+r$ multiplies its image by $p^r$, taking it to the chosen generator $p^{n+r-1}$. Thus this coordinate transition is the identity. Its colimit is $\mathbb F_p$, while the other two colimits vanish.

The $T$-terms at a fixed stage are all $\mathbb F_p$. The first arrow $p^n$ becomes zero, and the quotient arrow becomes the identity on the residue of $1$. Their transitions are respectively the identity on the kernel term and zero on both other terms, since $p^r=0$ in $\mathbb F_p$. These transitions commute with both row arrows: for the first arrow both composites are zero, and for the second both are zero. The colimit row is consequently $\mathbb F_p\to0\to0$, with zero first arrow. The middle and quotient stage terms were nonzero; their transition maps, rather than their dimensions, cause their colimits to vanish. The $H$-row retains left exactness and loses epimorphicity; the $T$-row retains right exactness and loses monicity.

Exercise 4 (expert: the quotient also escapes). For the product ring of Section 7, prove that $R/I$, where $I$ is the finite-support ideal, is nonzero and unrepresented in $\operatorname{Ind}(\mathcal C)$. Explain precisely why thickness permits this quotient.

Solution. The sequence constantly equal to $1$ is not in $I$, so $R/I\ne0$. If $R/I$ were represented, it would be finitely generated projective by the module identification in Section 7. The module epimorphism $R\to R/I$ would split. Its kernel $I$ would then be a direct summand of the cyclic module $R$, hence finitely generated: apply its summand projection to the generator $1$. But finitely many finite-support elements generate only elements supported on the union of their supports, whereas $I$ contains every coordinate idempotent. This contradiction proves that $R/I$ is unrepresented. Thickness controls cokernels of arrows whose source and target are both represented. The source of $I\to R$ is unrepresented, so that closure property makes no claim about this quotient. Both subobject and quotient Serre conditions fail in this model.

9. References

Ind-objects of an abelian category again form an abelian category; see Masaki Kashiwara and Pierre Schapira, Ind-sheaves, Astérisque 271 (2001). Ind-objects in general are treated in Pierre Schapira, An Introduction to Categories and Homological Algebra, lecture notes, version of 1 March 2026, Section 2.7.

Daniel Schäppi, Ind-abelian categories and quasi-coherent sheaves, arXiv:1211.3678v1, Theorem 1.3, gives a broader finitely cocomplete setting. Its abelian case is a comparison; no Tannakian result is a prerequisite here.

The primary Mathlib sources preadditive indization, abelian indization, and Grothendieck axioms for indization provide comparisons. Their small-base instances do not replace the large-base reduction of Theorem 3.2. This lesson has not been compiler verified.

The filtered-limit interface uses Stacks, Tag 04AX; ordinary colimits are discussed in Stacks, Tag 002P. The exact internal formal-object providers linked above retain the proof dependencies. All results here concern ordinary abelian categories and ordinary ind-objects.