Quasi-injectives and bounded comparisons for Ind categories

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A formal filtered presentation changes which objects are finite tests. Exactness of Hom against those tests is useful even when it does not give an injective object of the whole completion. We construct these relative injectives, specify the Ext groups their resolutions compute, and prove that bounded complexes with represented cohomology come from the original abelian category.

Fix universes $\mathbb U\in\mathbb V$ and choice. Let $\mathcal C$ be an abelian category with $\mathbb U$-small Hom sets; its objects need not form a $\mathbb U$-small set. Write $\mathcal A=\operatorname{Ind}_{\mathbb U}(\mathcal C)$ and $\iota\colon\mathcal C\to\mathcal A$. Every formal presentation is indexed by a $\mathbb U$-small filtered category. Choose a sufficiently large ambient universe in which the categories of complexes, their localizations and the derived Hom groups below are defined.

We use the complete proofs in Two cokernels, controlled rows and abelian completion, §§2–4, 6–7: $\mathcal A$ is abelian, $\iota$ is exact and fully faithful, small filtered colimits are exact, small coproducts exist, and $\begin{aligned} &\operatorname{Hom}_{\mathcal A}(\iota X,\mathop{\mathrm{colim}}_j A_j)\\ &\qquad=\mathop{\mathrm{colim}}_j\operatorname{Hom}_{\mathcal A}(\iota X,A_j) \end{aligned} \tag{1.1}$ for filtered diagrams. In particular, maps out of a constant into a formal presentation factor through one stage. The represented subcategory is closed under ambient kernels and cokernels of maps between represented objects, and under extensions. It need not be closed under arbitrary ambient subobjects. Small coproducts are exact: express a coproduct of short rows as the filtered colimit of its finite partial biproduct rows and use exact filtered colimits.

Lemma 1.1 (refining an epimorphism). If $E\twoheadrightarrow F$ is epic in $\mathcal A$ and $a\colon\iota X\to F$, there are an epimorphism $p\colon P\twoheadrightarrow X$ in $\mathcal C$ and $b\colon\iota P\to E$ with $b$ followed by the given epimorphism equal to $a\iota p$.

Proof. The complete stage-refinement theorem in the prerequisite, Theorem 4.4, applies without essential smallness of $\mathcal C$. Here is its construction. Present the epimorphism on a common small filtered index by epimorphisms $E_j\twoheadrightarrow F_j$ in $\mathcal C$, using the exact-row stage theorem. Factor $a$ through $\iota F_j$. In $\mathcal C$ take $P=X\times_{F_j}E_j$. The projection $p$ is epic because it is the pullback of an epimorphism; the other projection, followed by the stage map into $E$, is $b$. Their square gives the stated equality. All objects and maps used after the stage choice lie in the original category. $\square$

Throughout, $\operatorname{Ext}^n_{\mathcal A}(B,D)=\operatorname{Hom}_{D(\mathcal A)}(B,D[n])$, with cohomological shifts. Higher extensions, roofs and Yoneda composition, §§1–4, proves that every positive class has an exact-row representative, that a last epimorphism of that row kills the class by pullback, and that the normalized short-row boundary has the positive cone sign. These statements require no injectives or projectives.

2. The quasi-injective test and every higher Ext group

An object $Q\in\mathcal A$ is quasi-injective relative to $\mathcal C$ if $X\mapsto\operatorname{Hom}_{\mathcal A}(\iota X,Q)$ is exact on $\mathcal C^{\mathrm{op}}$. Since Hom is left exact, this asks precisely that every map $\iota X\to Q$ extend across $\iota X\hookrightarrow\iota Y$ for every monomorphism $X\hookrightarrow Y$ in $\mathcal C$. We denote this class by $\mathcal Q$. The tests are all objects of $\mathcal C$, including its large objects when it is not essentially small.

Proposition 2.1. An object $Q$ belongs to $\mathcal Q$ if and only if $\operatorname{Ext}^1_{\mathcal A}(\iota X,Q)=0\qquad(X\in\mathcal C). \tag{2.1}$

Proof. Suppose first that $Q$ is quasi-injective. For $0\to Q\xrightarrow{i}E\xrightarrow{e}\iota X\to0$, Lemma 1.1 gives $P\twoheadrightarrow X$ and $b\colon\iota P\to E$ lifting it. Put $K=\ker(P\to X)$ in $\mathcal C$. The restriction of $b$ to $\iota K$ factors uniquely as $i c$. Quasi-injectivity extends $c$ to $r\colon\iota P\to Q$. Thus $b-ir$ vanishes on $\iota K$ and descends to $s\colon\iota X\to E$. The equality $es\iota p=\iota p$, with $\iota p$ epic, gives $es=1$. Every extension splits, proving (2.1).

Conversely a short exact row $0\to X\to Y\to Z\to0$ in $\mathcal C$ gives an exact Hom sequence ending in $\operatorname{Ext}^1_{\mathcal A}(\iota Z,Q)$. Its vanishing gives the required extension of every map out of $\iota X$. $\square$

Theorem 2.2. If $Q\in\mathcal Q$, then $\begin{gathered} \operatorname{Ext}^n_{\mathcal A}(\iota X,Q)=0\\ (X\in\mathcal C,\ n>0). \end{gathered} \tag{2.2}$

Proof. Degree one is Proposition 2.1. Suppose the assertion holds in degree $n-1>0$, simultaneously for every constant source. Represent $\xi\in\operatorname{Ext}^n(\iota X,Q)$ by an exact row. Pullback by its last epimorphism kills $\xi$. Lemma 1.1 refines that epimorphism by $\iota p\colon\iota P\twoheadrightarrow\iota X$, with $p$ epic in $\mathcal C$, so $(\iota p)^*\xi=0$. For $K=\ker p$, the contravariant Ext sequence of $0\to\iota K\to\iota P\to\iota X\to0$ contains $\begin{gathered} \operatorname{Ext}^{n-1}(\iota K,Q)\longrightarrow \operatorname{Ext}^n(\iota X,Q)\\ \longrightarrow \operatorname{Ext}^n(\iota P,Q). \end{gathered} \tag{2.3}$ Exactness puts $\xi$ in the image of the first group, which is zero by induction. This proves all degrees. $\square$

3. Closure and presentations by base injectives

Proposition 3.1. The class $\mathcal Q$ is closed under small filtered colimits, extensions, direct summands, and cokernels of monomorphisms between its objects.

Proof. For filtered colimits, apply (1.1) to all three objects of a short row in $\mathcal C$; exact filtered colimits of abelian groups preserve the exact Hom row. Direct summands preserve exactness of that row. For an extension of two quasi-injectives, the long exact sequence in the second variable makes $\operatorname{Ext}^1(\iota X,-)$ of the middle term zero. For $0\to Q'\to Q\to Q''\to0$, with $Q',Q\in\mathcal Q$, its segment $\operatorname{Ext}^1(\iota X,Q)\to\operatorname{Ext}^1(\iota X,Q'')\to\operatorname{Ext}^2(\iota X,Q')$ has zero ends by Theorem 2.2. Proposition 2.1 gives the last assertion. $\square$

Theorem 3.2. If $\mathcal C$ has enough injectives, then $Q\in\mathcal Q$ precisely when it is a small filtered formal colimit of objects injective in $\mathcal C$.

Proof. A represented base injective has exact Hom on $\mathcal C$, by full faithfulness of $\iota$. Proposition 3.1 proves the reverse implication for any filtered presentation by such objects. For the other direction, take a map $\iota X\to Q$, choose $X\hookrightarrow I$ with $I$ injective in $\mathcal C$, and extend the map to $\iota I\to Q$. Every represented test map has now factored through a base injective. The complete allowed-piece recognition theorem in Finite diagrams and comma objects, §4, says exactly that this factorization condition puts $Q$ in $\operatorname{Ind}_{\mathbb U}(\operatorname{Inj}\mathcal C)$. Its small cofinal extraction applies to a full subcategory with possibly large object class; no small skeleton of all injectives is assumed. $\square$

The injectivity in this presentation is injectivity in $\mathcal C$. It is not an assertion that filtered colimits of arbitrary injectives in $\mathcal A$ are injective. The additional noetherian hypotheses for that different assertion are treated in Noetherian generators, filtered injectives and Ext.

4. Enough quasi-injectives without a small base

Theorem 4.1. If $\mathcal C$ has enough injectives, every $F\in\mathcal A$ admits a monomorphism into an object of $\mathcal Q$.

Proof, small closure. Start with the small family of objects occurring in a presentation of $F$. Repeatedly adjoin zero, a chosen finite biproduct of every finite family, and chosen kernels and cokernels of every arrow between the objects already selected. For each selected $X$, also adjoin a chosen monomorphism $X\hookrightarrow I_X$ with $I_X$ injective in the ambient $\mathcal C$. At each step only a $\mathbb U$-small family of choices is needed: the current object set and each Hom set are $\mathbb U$-small. Iterate countably many times. Every finite family and arrow eventually occurs, and each selected object eventually receives its chosen injective embedding. The full subcategory $\mathcal C_0$ on the union is essentially $\mathbb U$-small, abelian, and its inclusion in $\mathcal C$ is exact. Indeed the chosen ambient kernels, cokernels and biproducts give the required universal objects in this full subcategory; its coimage-to-image map is the ambient isomorphism. We require neither closure under all ambient subobjects nor extension closure.

Ind-completion of this exact fully faithful inclusion is exact and fully faithful: the common-stage Hom formula proves full faithfulness, and the stage kernel/cokernel construction proves exactness. We may therefore work inside $\mathcal A_0=\operatorname{Ind}_{\mathbb U}(\mathcal C_0)$, containing $F$, and subsequently embed it exactly in $\mathcal A$.

Proof, successive extensions. Choose representatives of all monomorphisms $u\colon X_u\hookrightarrow Y_u$ in $\mathcal C_0$, a small set $S$. Put $F_0=F$. For each $n$, form the small set of pairs $(u,a)$, where $u\in S$ and $a\colon\iota X_u\to F_n$. In $\mathcal A_0$ form the pushout $\begin{array}{ccc} \displaystyle\bigoplus_{(u,a)}\iota X_u&\xrightarrow{\oplus\iota u}& \displaystyle\bigoplus_{(u,a)}\iota Y_u\\ \big\downarrow{\scriptstyle(a)}&&\big\downarrow\\ F_n&\xrightarrow{j_n}&F_{n+1}. \end{array} \tag{4.1}$ The top arrow is monic by exactness of small coproducts. Pushout of a monomorphism is monic, so $j_n$ is monic. Its universal square extends each $a$ over its specified $\iota Y_u$. The Hom sets indexing this construction are small by the Ind Hom formula, and all coproducts and pushouts are small. Thus each $F_n$ and $Q=\operatorname{colim}_{n\ge0}F_n$ remains in $\mathcal A_0$. Exact filtered colimits give a monomorphism $F\hookrightarrow Q$. By compactness, any map $\iota X_u\to Q$ factors through some $F_n$, and its extension is present in $F_{n+1}$. Consequently $Q$ is quasi-injective relative to $\mathcal C_0$.

Proof, all ambient tests. Let $A\hookrightarrow B$ be any monomorphism in $\mathcal C$, and $a\colon\iota A\to Q$. Since $Q$ has a small presentation in $\mathcal C_0$, (1.1) factors $a$ as $\iota A\xrightarrow{\iota f}\iota X\xrightarrow{v}Q$ with $X\in\mathcal C_0$. Its chosen embedding $X\hookrightarrow I_X$ lies in $\mathcal C_0$. Relative quasi-injectivity extends $v$ to $w\colon\iota I_X\to Q$. Ambient injectivity of $I_X$ extends $A\xrightarrow f X\hookrightarrow I_X$ over $B$, say by $g\colon B\to I_X$. Then $w\iota g$ extends $a$. This proves exact Hom on every short row in $\mathcal C$, so $Q\in\mathcal Q$. $\square$

The construction requires enough injectives in the base only to select the $I_X$ and extend the ambient test maps. It assumes no generator for the large category $\mathcal C$ and makes no claim that $\mathcal A$ has enough genuine injectives.

5. A functorial model for a module base

For a $\mathbb U$-small ring $R$, put $\mathcal C=\operatorname{Mod}_{\mathbb U}(R)$ and take the injective cogenerator $J$ supplied by Injective cogenerators and faithful module realizations, Theorem 1.1. Its Grothendieck hypotheses hold here: the free module $R$ is a generator because maps from it specify arbitrary elements; small colimits are formed by taking the direct sum modulo the diagram relations; and the filtered-colimit representative and eventual-equality tests prove exactness of filtered colimits of module rows. The cited theorem proves both injectivity and separation of all pairs of maps, not only a test for nonzero objects.

Define an ordinary functor and a natural map by $\begin{gathered} E(M)=\prod_{t\in\operatorname{Hom}_R(M,J)}J,\\ \pi_t\eta_M=t,\qquad\pi_t E(f)=\pi_{tf}\\ (t\colon N\to J,\ f\colon M\to N). \end{gathered} \tag{5.1}$ Each product is injective, since every coordinate map extends across a monomorphism and the chosen extensions combine into a map to the product. The cogenerator separates distinct maps, hence $\eta_M$ is monic. Coordinate identities give $E(1)=1$, $E(gf)=E(g)E(f)$ and $E(f)\eta_M=\eta_N f$. In particular this is a functor on an entire small diagram, even though no additivity of the assignment $f\mapsto E(f)$ is asserted.

If $F=\operatorname{colim}_j\iota M_j$, the natural stage embeddings give $F\hookrightarrow\mathop{\mathrm{colim}}_j\iota E(M_j). \tag{5.2}$ Exact filtered colimits make this monic; the target is quasi-injective by Theorem 3.2. Thus no simultaneous choice of incompatible injective embeddings of the individual stage objects is necessary. This construction, like Theorem 4.1, supplies quasi-injectives in $\operatorname{Ind}(\operatorname{Mod}_{\mathbb U}(R))$, not a replacement of that category by ordinary modules.

6. What a quasi-injective coresolution computes

Suppose $F$ admits a coresolution $0\to F\to Q^0\to Q^1\to\cdots$ with $Q^j\in\mathcal Q$. Such coresolutions exist under Theorem 4.1 by embedding each successive cokernel. For every $X\in\mathcal C$ there are canonical isomorphisms $\begin{gathered} H^n\operatorname{Hom}_{\mathcal A}(\iota X,Q^\bullet)\\ \simeq\operatorname{Ext}^n_{\mathcal A}(\iota X,F)\quad(n\ge0). \end{gathered} \tag{6.1}$

Proof. Set $Z^0=F$, and factor each differential through its image so that $0\to Z^j\to Q^j\to Z^{j+1}\to0$ is exact. The kernel in degree zero of the Hom complex is $\operatorname{Hom}(\iota X,F)$. For $n\ge1$, its cycles in degree $n$ are $\operatorname{Hom}(\iota X,Z^n)$, and the exact sequence for the previous short row identifies the quotient by boundaries with $\operatorname{Ext}^1(\iota X,Z^{n-1})$: the next term $\operatorname{Ext}^1(\iota X,Q^{n-1})$ vanishes. Every earlier short row gives an isomorphism $\begin{gathered} \operatorname{Ext}^{r}(\iota X,Z^{j+1})\\ \xrightarrow{\delta}\operatorname{Ext}^{r+1}(\iota X,Z^j)\\ (r\ge1) \end{gathered} \tag{6.2}$ because both adjacent positive Ext groups of $Q^j$ vanish by Theorem 2.2. Composing these positive boundary maps proves (6.1). Naturality of each exact sequence proves naturality in $X$ and in any morphism of augmented coresolutions. In particular two chosen coresolutions compute the same canonical Ext groups; no unproved lifting of a map between arbitrary coresolutions is needed for this assertion. $\square$

The restriction to the represented source $\iota X$ is part of (6.1). Quasi-injectivity alone does not say that $\operatorname{Hom}_{\mathcal A}(B,Q)$ is exact for arbitrary $B$, or that $Q[0]$ is K-injective. Exercise 2 gives an explicit obstruction.

7. The bounded comparison for every abelian base

No enough-injectives hypothesis is imposed in this section. For every $X,Y\in\mathcal C$, the natural maps are isomorphisms $\begin{gathered} \operatorname{Ext}^n_{\mathcal C}(X,Y) \xrightarrow{\sim}\operatorname{Ext}^n_{\mathcal A}(\iota X,\iota Y)\\ (n\in\mathbb Z). \end{gathered} \tag{7.1}$ Here the negative groups are zero. More generally the exact embedding induces an equivalence $\begin{gathered} D^b(\mathcal C)\xrightarrow{\sim}D^b_{\iota\mathcal C}(\mathcal A),\\ D^b_{\iota\mathcal C}(\mathcal A) =\{K\in D^b(\mathcal A):\\ H^jK\simeq\iota X_j\text{ for all }j\}. \end{gathered} \tag{7.2}$

Proof of object Ext comparison. We verify the intrinsic effacement criterion, with all its hypotheses, from Thick abelian subcategories and derived comparison, §§1–4, Theorem 2.2. The represented subcategory is full, exact, and thick abelian in the precise weaker sense of that lesson: kernels and cokernels between its objects and extensions remain represented. These are the facts recalled in §1; arbitrary subobjects are not required. For every ambient positive Ext class with constant endpoints, an exact-row representative has a last epimorphism killing the class by pullback. Lemma 1.1 supplies an epimorphism $P\twoheadrightarrow X$ inside $\mathcal C$ factoring through that last arrow. Thus each individual class is killed by an epimorphism within the represented subcategory. This is condition 3 of that theorem, and its complete surjectivity and injectivity inductions prove (7.1). The epimorphism may depend on the class; no single epimorphism killing an entire Ext group is being claimed. The maps are those of the exact embedding, so preserve the short-row boundary, addition, both endpoint naturalities and Yoneda composition.

For completeness, the mechanism of that criterion is particularly transparent here. If $K=\ker(P\to X)$, a class killed by $P\to X$ comes from $\operatorname{Ext}^{n-1}(K,Y)$, giving surjectivity by induction. For injectivity, first kill an intrinsic class by an intrinsic epimorphism, using its own exact-row representative. Lift the resulting degree-$n-1$ class on $K$; if its boundary is ambiently zero, exactness makes it an ambient restriction from $P$. Inductive comparison lifts that restriction intrinsically and subtracts it, leaving zero boundary. This proves injectivity in the same degree. The exact commuting diagrams and both endpoint inductions are written out in §3 of the cited lesson.

Proof for bounded complexes. Finite smart-truncation towers build each bounded complex from its shifted cohomology objects. The exact embedding commutes with these truncations. For a single shifted object against a single shifted object, comparison is (7.1). Apply the exact Hom sequences to a truncation triangle in one variable, with the comparison isomorphisms in neighboring degrees, and then in the other variable. Finite induction on the two numbers of nonzero cohomology groups proves full faithfulness on all bounded complexes, as in §4 of the same prerequisite. This uses the full exact sequences, not a splitting of the truncation triangles.

To check the essential image, let $K$ have represented cohomology in a finite interval. Induct on its length. At its highest nonzero degree $b$, the truncation triangle has vertices $U=\tau_{\le b-1}K$, $K$, and $V=H^b(K)[-b]$, and a connecting map $\delta\colon V\to U[1]$. The first and third vertices have preimages by induction. Full faithfulness lifts $\delta$ to those preimages. Form the cone of $-\delta[-1]\colon V[-1]\to U$. Rotation of its cone triangle gives $U\to\operatorname{Cone}(-\delta[-1])\to V\xrightarrow{\delta}U[1]$: the minus sign cancels the sign in rotation. Its image is isomorphic to the original triangle by the triangulated morphism axiom and the two isomorphisms at the ends; exact Hom sequences show the middle map is an isomorphism. The one-cohomology case is the usual smart-truncation quasi-isomorphism. This constructs a bounded preimage of $K$ and proves (7.2). $\square$

The argument concerns bounded objects and their actual connecting maps. It does not identify $K$ with the direct sum of its cohomology, extend (7.2) to unbounded categories, or require the quasi-injective existence theorem of §4.

8. Four graded exercises with complete solutions

8.1. Formal localization in a large base — introductory

Exercise 1. Take $\mathcal C=\operatorname{Ab}_{\mathbb U}$ and $F=\operatorname{colim}_{n\ge0}\iota(\mathbb Z\xrightarrow{2}\mathbb Z\xrightarrow{2}\cdots)$. Compute Hom from $\iota\mathbb Z$ and from $\iota\mathbb Z[1/2]$. Compare $F$ with the constant ordinary colimit. Construct a two-term quasi-injective coresolution and compute Ext from $\iota\mathbb Z$.

Solution. The constant-source formula gives $\operatorname{Hom}(\iota\mathbb Z,F)=\mathbb Z[1/2]$, the stage integer $a$ at stage $n$ corresponding to $a/2^n$. Every map $\mathbb Z[1/2]\to\mathbb Z$ is zero: its value on 1 must be divisible by every power of 2, and then its values on $1/2^n$ are also zero. Therefore $\operatorname{Hom}(\iota\mathbb Z[1/2],F)=0$. In contrast, $\operatorname{Hom}(\iota\mathbb Z[1/2],\iota\mathbb Z[1/2])$ contains the nonzero identity, so $F$ is not that constant object.

At each stage use $0\to\mathbb Z\to\mathbb Q\to\mathbb Q/\mathbb Z\to0$, with multiplication by 2 on all three terms. Exact filtered colimits give $\begin{gathered} 0\to F\to\iota\mathbb Q\to Q^1\to0,\\ Q^1=\mathop{\mathrm{colim}}_{n\ge0}\iota D_n,\\ D_n=\mathbb Q/\mathbb Z,\qquad D_n\xrightarrow{2}D_{n+1}. \end{gathered} \tag{8.1}$ The identification of the middle formal colimit with $\iota\mathbb Q$ sends stage $q$ at $n$ to $q/2^n$; all its transition maps are isomorphisms. In particular the embedding of $F$ sends $a$ at $n$ to $a/2^n$ and is monic. The groups $\mathbb Q$ and $\mathbb Q/\mathbb Z$ are divisible, by division of rational representatives, hence injective by the following extension argument.

Given a map $H\to D$ from a subgroup of $G$ into a divisible group, choose a maximal extension $h\colon H'\to D$ by Zorn's lemma; unions of chains are compatible extensions. If $x\notin H'$, the integers $m$ satisfying $mx\in H'$ form $d\mathbb Z$, $d\ge0$. For $d>0$ choose $y$ with $dy=h(dx)$; for $d=0$ choose $y=0$. The rule $h'(a+nx)=h(a)+ny$ extends $h$ to $H'+\mathbb Zx$. It is well-defined because any two expressions differ by an integer multiple of the relation $dx\in H'$, or, when $d=0$, have identical integer coefficients. This contradicts maximality. Thus $H'=G$, proving injectivity. Theorem 3.2 now makes both terms of (8.1) quasi-injective.

Applying Hom from $\iota\mathbb Z$ gives $\mathbb Q\to\mathbb Q/\mathbb Z[1/2]$, the ordinary quotient map. Indeed the stage class $q+\mathbb Z$ at $n$ maps to $q/2^n+\mathbb Z[1/2]$. This is surjective; a class is zero precisely when $2^kq$ is integral for some $k$, precisely the condition for eventual zero in that filtered system. Its kernel is $\mathbb Z[1/2]$. Formula (6.1) now gives Ext in degree zero equal to $\mathbb Z[1/2]$ and zero in every positive degree. These statements concern the constant test $\iota\mathbb Z$; they do not replace $F$ by its ordinary realization.

8.2. Every object quasi-injective, yet a nonsplit extension — intermediate

Exercise 2. Let $R=\prod_{n\ge0}\mathbb F_2$, let $e_n$ be its coordinate idempotents, and let $\mathcal C$ be the category of finitely generated projective $R$-modules. Use the finite-matrix proof in §7 of the abelian-completion prerequisite that $\mathcal C$ is abelian, all its exact rows split, and $\operatorname{Ind}(\mathcal C)\simeq\operatorname{Mod}(R)$. Put $I=\bigoplus_{n\ge0}\mathbb F_2 e_n\subset R$. Is $I$ quasi-injective? Is it injective? What would fail if (6.1) were applied to an arbitrary Ind source?

Solution. Every short row in $\mathcal C$ splits because its quotient is projective as an $R$-module. Hom into any $R$-module preserves split exact rows, so every object of $\operatorname{Mod}(R)$, in particular $I$, is quasi-injective relative to this $\mathcal C$. The prerequisite's coordinate matrix argument proves that finite-presentation modules are precisely these finite projectives and gives the asserted exact Ind equivalence; no noetherian assumption on $R$ is needed.

The row $0\to I\to R\to R/I\to0$ does not split. A retraction $r\colon R\to I$ would be determined by $e=r(1)$, which has finite support. But for every $n$, $r(e_n)=e_n e$ would have to equal $e_n$, forcing every coordinate of $e$ to be 1. This contradicts finite support. Thus $I$ is not injective: injectivity would extend $1_I$ across $I\hookrightarrow R$, producing that retraction. The nonsplit row is a nonzero class in $\operatorname{Ext}^1_R(R/I,I)$.

The augmented complex $I[0]$ is already a quasi-injective coresolution of $I$. Its Hom complex from $R/I$ has no positive cohomology and misses this nonzero Ext class. In fact if $I[0]$ were K-injective, the derived Hom test for a K-injective target would compute that Ext by this very Hom complex, a contradiction. There is no contradiction with (6.1): the source $R/I$ is not represented by an object of $\mathcal C$. To see this directly, a finitely presented $R/I$ would have finitely generated relation ideal $I$; a finite set of elements of $I$ has support contained in one finite set and cannot generate all the $e_n$. This also illustrates why the finiteness hypotheses in filtered-injective theorems cannot simply be dropped.

8.3. A base with no nonzero injectives — advanced

Exercise 3. Fix a prime $p$, let $\mathcal C$ be finite abelian $p$-groups, and identify its Ind category. Show that $\mathcal C$ has no nonzero injective objects but its Ind category has the useful quasi-injective $P=\mathbb Z[1/p]/\mathbb Z$. For $A_r=\mathbb Z/p^r$, $r\ge1$, compute $\operatorname{Ext}^n(A_s,A_r)$ in both categories and identify a positive boundary class by an exact row.

Solution. Every $p$-primary torsion group is the filtered union of its finite subgroups: finitely many torsion elements generate a finite $p$-group, and sums give a common upper subgroup. A finite abelian group is finitely presented over $\mathbb Z$. Consequently Hom from each finite test commutes with filtered module colimits. The Ind Hom formula now proves full faithfulness of ordinary realization, and the stated unions prove essential surjectivity onto $p$-primary torsion groups. Stage kernel/cokernel computations show this equivalence is exact.

For a nonzero finite group $E$, choose $x\in E$ of maximal order $p^r$. The map $\mathbb Z/p^r\to E$ sending 1 to $x$ cannot extend across $\mathbb Z/p^r\hookrightarrow\mathbb Z/p^{r+1}$, $1\mapsto p$. Such an extension would give $y\in E$ with $py=x$, and $y$ would have order $p^{r+1}$, contrary to maximality. Thus no nonzero object is injective in $\mathcal C$.

Multiplication by $p$ is surjective on $P$; multiplication by any integer prime to $p$ is invertible on it, by the elementary inverse modulo each $p^m$. Hence $P$ is divisible and injective as an abelian group by the extension argument in Exercise 1. An extension into $P$ of a map from a subgroup of a $p$-primary group remains an arrow in the full $p$-primary subcategory, so $P$ is injective there and in particular quasi-injective. The row $\begin{gathered} 0\to A_r\xrightarrow{j}P\xrightarrow{p^r}P\to0,\\ j(1)=1/p^r+\mathbb Z, \end{gathered} \tag{8.2}$ is exact. Its Hom complex from $A_s$ is $P[p^s]\xrightarrow{p^r}P[p^s]$, identified with multiplication by $p^r$ on $\mathbb Z/p^s$ via $1\mapsto1/p^s$. Its kernel and cokernel are cyclic of order $p^{\min(r,s)}$. Therefore Ext in degrees zero and one has that group, and Ext in higher degrees is zero. These are the ambient groups by (6.1), and the same intrinsic groups by (7.1), which does not require enough base injectives.

The positive boundary of $A_s\to P$, $1\mapsto1/p^s$, is the pullback of (8.2). Its middle group consists of pairs $(x,a)$ with $p^rx=a/p^s$ in $P$. The element $g=(1/p^{r+s},1)$ has order $p^{r+s}$ and satisfies $p^s g=(1/p^r,0)$. Every pair differs from an integer multiple of $g$ by such a kernel element, itself a multiple of $g$. Thus the pullback is exactly $0\to A_r\xrightarrow{\,p^s\,}\mathbb Z/p^{r+s} \to A_s\to0. \tag{8.3}$ The quotient sends 1 to 1, and the kernel generator has the indicated positive sign. It represents the class of 1 in $\operatorname{coker}(p^r\colon\mathbb Z/p^s\to\mathbb Z/p^s)$, a generator. This verifies the boundary convention as well as the group orders.

8.4. Infinite terms with finite bounded cohomology — advanced

Exercise 4. Let $\mathcal C$ be finite dimensional vector spaces over a field $k$, so that $\operatorname{Ind}(\mathcal C)\simeq\operatorname{Vect}_{\mathbb U}(k)$. On $V=\bigoplus_{n\ge0}k e_n$ let $S e_n=e_{n+1}$. Give an explicit bounded finite dimensional model for the complex $V\xrightarrow{S}V$ in degrees 0 and 1. Explain why finite dimensional cohomology in each degree alone does not suffice for a bounded conclusion.

Solution. The equivalence with vector spaces follows by writing every vector space as the filtered union of finite dimensional subspaces and using Hom from a finite basis to prove full faithfulness. The shift $S$ is injective, and its cokernel is the one dimensional span of the class of $e_0$. Let $K$ denote the displayed complex. Define a chain map $p\colon K\to k[-1]$ by taking the $e_0$-coefficient in degree 1 and zero in degree 0. Define $i\colon k[-1]\to K$ by $i^1(1)=e_0$. Then $pi=1$. Put $h\colon K^1\to K^0$ with $h(e_0)=0$ and $h(e_{n+1})=e_n$, extended linearly on finite sums. On degree 0, $hS=1$; on degree 1, $Sh=1-ip$. Thus $dh+hd=1-ip$, an explicit chain homotopy. This proves $K\simeq k[-1]$, with actual inverse maps in the homotopy category, even though its two terms are infinite dimensional.

In contrast, the complex with one copy of $k$ in every degree $n\ge0$ and zero differentials has $H^n=k$ for every such $n$. All these groups are represented by finite dimensional objects, but they are not confined to a finite interval. Any bounded complex has zero cohomology outside a finite interval, and a quasi-isomorphism preserves all cohomology groups. Hence this complex cannot be quasi-isomorphic to a bounded one. This is precisely the boundedness restriction in (7.2), rather than a claim about a larger derived comparison.