Derived Ind functors and filtered Ext comparisons

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

Passing to formal filtered systems makes every original object a finite test. This changes the filtered-colimit behavior of Ext. We construct the derived functor of an Ind extension, identify every higher value and its canonical maps, and explain exactly which filtered comparisons follow.

1. The categories and prerequisites

Fix universes and choice as in Quasi-injectives and bounded comparisons for Ind categories, §1. Let $\mathcal C,\mathcal D$ be locally small abelian categories in the same small universe; their object collections may be larger. Assume $\mathcal C$ has enough injectives. Write $\begin{gathered} \mathcal A=\operatorname{Ind}(\mathcal C),\quad \mathcal B=\operatorname{Ind}(\mathcal D),\\ \iota\colon\mathcal C\to\mathcal A,\qquad \jmath\colon\mathcal D\to\mathcal B. \end{gathered} \tag{1.1}$ All filtered diagrams and colimits called small use that small universe. Derived categories and their Hom groups use a sufficiently large ambient universe. Let $F\colon\mathcal C\to\mathcal D$ be additive and left exact, and put $\widehat F=\operatorname{Ind}(F)$.

Ind-objects through their elements, §3, proves that termwise extension is independent of presentation and preserves every small filtered colimit, with the specified identification $\widehat F\iota=\jmath F$. Two cokernels, controlled rows and abelian completion, §§4–6, proves left exactness of $\widehat F$, exactness of filtered colimits, and this important control: a short exact Ind row whose middle term is a filtered formal system of objects in a full additive class $\mathcal J\subset\mathcal C$ has a short exact stage presentation with middle stages in $\mathcal J$.

The quasi-injective lesson supplies a cogenerating full additive class $\mathcal Q\subset\mathcal A$: for $Q\in\mathcal Q$, the functor $X\mapsto\operatorname{Hom}_{\mathcal A}(\iota X,Q)$ is exact on $\mathcal C^{\mathrm{op}}$. It proves that $Q$ is a small filtered formal colimit of base injectives, that this class is closed under extensions and cokernels of monomorphisms between its objects, and that $\begin{gathered} \operatorname{Ext}^n_{\mathcal A}(\iota X,Q)=0\\ (X\in\mathcal C,\ n>0). \end{gathered} \tag{1.2}$ These are relative Hom tests; no genuine injective supply in $\mathcal A$ is assumed.

2. A lifting lemma and the adapted resolution class

Lemma 2.1. For a short exact row $0\to Q\to E\xrightarrow{e}T\to0$ in $\mathcal A$, with $Q\in\mathcal Q$, the row obtained by applying $\widehat F$ is short exact.

Proof. Present $T=\operatorname{colim}_i\iota T_i$. Pulling the row back along any stage map $\iota T_i\to T$ gives an extension of that constant by $Q$. It splits by (1.2) in degree one. A splitting supplies $b_i\colon\iota T_i\to E$ with $eb_i$ the stage map. No compatibility between these chosen lifts is needed. Applying $\widehat F$ shows that every structure map $\jmath F(T_i)\to\widehat F(T)$ factors through $\widehat F(e)$. Those structure maps are jointly epic: their target is their colimit. Hence $\widehat F(e)$ is epic. Left exactness supplies the remaining two exactness positions. $\square$

Theorem 2.2. The class $\mathcal Q$ computes a universal bounded-below right derived functor $R\widehat F\colon D^+(\mathcal A)\longrightarrow D^+(\mathcal B). \tag{2.1}$ Every outgoing quasi-isomorphism $K\to Q^\bullet$, with a bounded-below complex of quasi-injective terms, computes its value by $\widehat F(Q^\bullet)$.

Proof. The full same-lower-bound replacement construction in Relative replacements and finite resolution bounds, §§1–2, applies because $\mathcal Q$ is cogenerating. It gives every required replacement, preserving a specified lower degree bound.

If $E^\bullet$ is an acyclic complex of quasi-injectives, zero below $m$, its cycle $Z^m$ is zero. Inductively each short row $0\to Z^n\to E^n\to Z^{n+1}\to0$ shows $Z^{n+1}\in\mathcal Q$, by closure under cokernels of monomorphisms. Lemma 2.1 makes every such row exact after $\widehat F$. Their image factorizations identify the boundaries and cycles of $\widehat F(E^\bullet)$, so it is acyclic. Thus $\mathcal Q$ is an adapted class in the precise sense of Universal derived functors and compatible composition, §4. Its Proposition 4.1 and the class-localization proof supply (2.1), its universal unit, independence of all replacement choices, and naturality on actual derived morphisms. Additivity preserves shifts and the signed two-term cone formulas, so the descended functor is exact. $\square$

In particular every $Q\in\mathcal Q$ is $\widehat F$-acyclic: its degree-zero complex is itself an adapted replacement. This conclusion concerns this derived functor, and does not assert K-injectivity of $Q[0]$.

3. The full derived comparison on the original category

Enough base injectives similarly construct $RF\colon D^+(\mathcal C)\to D^+(\mathcal D)$. Exactness of $\iota,\jmath$ gives functors on derived categories, written with the same letters. There is a canonical isomorphism $\jmath RF\xrightarrow{\sim}R\widehat F\,\iota \quad\text{on }D^+(\mathcal C). \tag{3.1}$

Proof. Replace $K$ by a bounded-below base-injective complex $I^\bullet$. Its image $\iota I^\bullet$ has quasi-injective terms. The two proposed values are represented by the very same complex $\jmath F(I^\bullet)=\widehat F(\iota I^\bullet). \tag{3.2}$ The identifications include every differential. The class-localization equivalence in the prerequisites represents arrows by roofs between the chosen class complexes and proves their common-refinement equality test. Applying the identical termwise functors in (3.2) to those roofs proves naturality and independence of choices. The comparison on a base-injective complex is the identity of (3.2). For a general replacement, the normalizations of both derived units therefore give $\Gamma_K\,\jmath\eta^F_K=\eta^{\widehat F}_{\iota K}, \tag{3.3}$ under $\jmath F(K)=\widehat F(\iota K)$. Here $\Gamma$ denotes (3.1), and the equality takes place after passage to the relevant derived categories. This specifies the canonical comparison, including the unit. Since (3.2) respects shifts, cone inclusions and positive cone projections, so does $\Gamma$; in particular every short-row connecting square commutes. $\square$

We do not infer full faithfulness on all of $D^+(\mathcal C)$ from this assertion about derived functors. The bounded fully faithful inclusion was proved separately in the quasi-injective lesson.

4. Every higher derived value, with its actual comparison

For $n\ge0$, write $F_n=R^nF$ on base objects and $\widehat F_n=\operatorname{Ind}(F_n)$. These are additive functors. There are canonical natural isomorphisms $\theta^n\colon\widehat F_n\xrightarrow{\sim}R^n\widehat F. \tag{4.1}$

First define the map, without assuming filtered continuity of the right side. If $A=\operatorname{colim}_i\iota A_i$, take the maps $\begin{gathered} \jmath R^nF(A_i)\xrightarrow{H^n\Gamma_{A_i}} R^n\widehat F(\iota A_i)\\ \longrightarrow R^n\widehat F(A). \end{gathered} \tag{4.2}$ Their last arrows come from the structure maps into $A$. They form a cocone, hence induce $\theta^n_A$ from their colimit $\widehat F_n(A)$. Every constant-source map into an Ind-object factors through a stage. Consequently naturality on all those factorizations proves naturality for arbitrary Ind-arrows. It also proves independence of the presentation: the same maps (4.2), prescribed on every represented piece, determine the map from either colimit. In degree zero it is the specified identity comparison of $\widehat F$.

Proof of invertibility. Embed an arbitrary $A$ into $Q\in\mathcal Q$, with cokernel $B$. Choose a presentation of $Q$ by base injectives. The controlled-middle-stage theorem in §1 gives a presentation of the entire row by $\begin{gathered} 0\to X_i\to I_i\to Y_i\to0,\\ I_i\text{ injective in }\mathcal C. \end{gathered} \tag{4.3}$ Its formal colimit, with its maps, is $0\to A\to Q\to B\to0$. This specifies only the middle stages, exactly as the control theorem permits.

Apply the base derived long sequences to (4.3), then $\jmath$, then the exact filtered colimit in $\mathcal B$. All positive groups of the $I_i$ vanish. The resulting sequence has its initial cokernel identification and its higher boundary isomorphisms $\begin{gathered} \operatorname{coker}\bigl(\widehat F(Q)\to\widehat F(B)\bigr) \simeq\widehat F_1(A),\\ \widehat F_{n-1}(B)\xrightarrow{\delta}\widehat F_n(A) \quad(n\ge2). \end{gathered} \tag{4.4}$ The derived long sequence of the global row has exactly the analogous formulas for $R^n\widehat F$, because $Q$ is acyclic. Each constant stage row maps to this global row. Equation (3.3), compatibility with its boundary, and the definition (4.2) make the comparison diagrams commute after taking the filtered colimit. The degree-zero identity therefore identifies the two cokernels for $n=1$. For $n\ge2$, the boundary square and the induction hypothesis for $B$ identify both sides of (4.4). Induction, simultaneously for every $A$, proves (4.1).

This argument also specifies the long-sequence comparisons for any short exact Ind row. Present that row by short exact base rows and take their derived long sequences and filtered colimit. Their maps to the global row give the same commuting boundary squares as above. Invertibility of $\theta$ identifies the resulting sequence with the canonical derived long sequence, proving presentation independence, every boundary square and exactness. $\square$

The comparisons respect a natural transformation $F\to G$ of left exact base functors: on constants this follows from the same injective replacement, and on any Ind-object (4.2) then forces the square. Identity and composition comparisons obey the corresponding laws for the same reason.

5. Filtered continuity and the formal meaning of Ext

Since every $\widehat F_n$ preserves small filtered colimits, (4.1) identifies the canonical maps as isomorphisms $\begin{gathered} \mathop{\mathrm{colim}}_j R^n\widehat F(A_j)\\ \xrightarrow{\sim}R^n\widehat F(\mathop{\mathrm{colim}}_j A_j)\\ (n\ge0) \end{gathered} \tag{5.1}$ for every small filtered diagram in $\mathcal A$, including diagrams whose terms are already Ind-objects. Indeed naturality of $\theta$ identifies its square with the actual structure-map cocone defining (5.1). We have proved a statement about $\operatorname{Ind}(\mathcal C)$, not about ordinary filtered colimits in a possibly large $\mathcal C$.

For a fixed $X\in\mathcal C$, take the base functor $H_X=\operatorname{Hom}_{\mathcal C}(X,-)$ with values in small abelian groups. The ordinary realization $\alpha\colon\operatorname{Ind}(\operatorname{Ab})\to\operatorname{Ab}$ is exact: a short exact formal row has a stagewise short exact presentation, and filtered colimits of groups are exact. The constant-source Hom formula identifies $\alpha\operatorname{Ind}(H_X)(A)=\operatorname{Hom}_{\mathcal A}(\iota X,A). \tag{5.2}$ The quasi-injective class is also adapted for this functor, either by applying exact $\alpha$ to §2 or by using the exact relative Hom tests directly. Its higher functors are a cohomological delta-functor with degree zero (5.2), and are effaced by embeddings into $\mathcal Q$. The delta-functor $\operatorname{Ext}^n_{\mathcal A}(\iota X,-)$, defined by derived morphisms, is likewise effaced by these embeddings, by (1.2). The complete effacement uniqueness theorem in Universal cohomology and acyclic resolutions, §§1–2, identifies these two functors by their degree-zero identity, including every boundary.

Thus if $A=\operatorname{colim}_i\iota A_i$, $\begin{gathered} \mathop{\mathrm{colim}}_i\operatorname{Ext}^n_{\mathcal C}(X,A_i)\\ \xrightarrow{\sim}\operatorname{Ext}^n_{\mathcal A}(\iota X,A),\\ n\ge0. \end{gathered} \tag{5.3}$ This is the map induced by the exact inclusion on each base Ext class and then by the actual stage map into $A$. To see that its normalization agrees with the one just constructed, restrict to $\mathcal C$: both comparisons extend the identity on ordinary Hom and preserve all short-row boundaries. Effacement uniqueness on $\mathcal C$ forces equality in every degree. Composing with each structure map proves the assertion for (5.3). It is natural in both endpoints and compatible with composition by derived arrows from base objects.

The same proof and (5.1) give filtered continuity of $\operatorname{Ext}^n_{\mathcal A}(\iota X,-)$ for every filtered diagram of Ind-objects. The test $X$ need not be finitely presented in an ordinary module category: its constant is finitely presented in the formal completion. Exercise 3 makes the difference visible in degree one.

6. Bounded sources and a common lower bound

Let $P\in D^b(\mathcal C)$. Let $(K_i^\bullet)$ be a small filtered diagram of actual complexes in $\mathcal C$, with actual chain maps, and with a common lower term bound $K_i^q=0$ for $q<m$. Put $K=\operatorname{colim}_i\iota K_i^\bullet$, termwise in $\mathcal A$. It is a bounded-below complex. For every integer $n$, the canonical map is an isomorphism $\begin{gathered} \mathop{\mathrm{colim}}_i\operatorname{Hom}_{D^+(\mathcal C)}(P,K_i[n])\\ \xrightarrow{\sim}\operatorname{Hom}_{D^+(\mathcal A)}(\iota P,K[n]). \end{gathered} \tag{6.1}$ In particular this includes a filtered diagram of base objects in degree zero. No unspecified colimit of objects of a triangulated category is used here.

Proof. Let $b$ be an upper cohomological bound for $P$. A smart truncation triangle and $\operatorname{Hom}(D^{\le b},D^{\ge b+1})=0$ show that maps from $P$ to a target $L$ are unchanged on replacing $L$ by $\tau_{\le b}L$. The other adjacent vanishing is $\operatorname{Hom}(P,(\tau_{\ge b+1}L)[-1])=0$, since this shifted target starts in degree $b+2$. These truncation and orthogonality facts, including the full bounded embeddings, are the prerequisites of §7 of the quasi-injective lesson.

Apply this to $L=K_i[n]$. Their truncated targets all have cohomology in the common finite interval $[m-n,b]$; if the interval is empty both sides vanish. Exact filtered colimits commute with kernels, cokernels and images, hence with cohomology and these finite smart truncations after applying $\iota$. For a shifted base object as source and a shifted base object as target, (6.1) is (5.3), with negative Ext groups zero. Build the uniformly truncated targets by their finite cohomology truncation towers. At each step the exact Hom sequences, exactness of filtered colimits of groups, and the comparison isomorphisms in the neighboring degrees give the middle comparison by the five lemma. This proves the assertion for a shifted base source and all these targets. Now apply the same finite induction to the truncation tower of the fixed bounded source $P$, using exact Hom sequences in the first variable. Every step uses the actual canonical comparison, so the resulting isomorphism is (6.1), with naturality in chain maps, in $P$, and in shifts. $\square$

The common lower bound supplies the finite interval in this argument. We have not identified an arbitrary filtered diagram in a derived category with a strict complex diagram, or asserted an unbounded replacement theorem.

7. Four graded exercises with complete solutions

7.1. Nonzero stage groups that disappear — introductory

Exercise 1. Take $\mathcal C=\operatorname{Ab}$, a prime $p$, and $F(M)=\operatorname{Hom}_{\mathbb Z}(\mathbb Z/p,M)$. Let $A_r=\mathbb Z/p^r$, $r\ge1$, with transition $1\mapsto p$, and let $A=\operatorname{colim}_r\iota A_r$ be the formal colimit. Compute every $R^n\widehat F(A)$, including its transition maps. Compare $A$ with the constant Prüfer group $\iota P$, where $P=\mathbb Z[1/p]/\mathbb Z$, and show that the quotient of their natural monomorphism is nonzero but has zero derived $\widehat F$-value.

Solution. The free resolution $0\to\mathbb Z\xrightarrow{p}\mathbb Z\to\mathbb Z/p\to0$ shows $\begin{gathered} F(M)=M[p],\\ R^1F(M)=M/pM,\\ R^nF(M)=0\quad(n\ge2). \end{gathered} \tag{7.1}$ Indeed its Hom complex is $M\xrightarrow{p}M$; it computes the derived Hom functor because a bounded complex of free modules is K-projective, or by the complete bounded resolution argument in the course prerequisites. On $A_r[p]$, choose generator $p^{r-1}$. Multiplication by $p$ sends it to $p^r$, the next generator, so the degree-zero transition is the identity of $\mathbb F_p$. On $A_r/pA_r$, the generator is the class of 1 and its image is $p$ modulo $pA_{r+1}$, hence zero. Thus (4.1) gives $\begin{gathered} R^0\widehat F(A)=\iota\mathbb F_p,\\ R^n\widehat F(A)=0\quad(n>0), \end{gathered} \tag{7.2}$ where $\iota$ now also denotes the constant embedding in $\operatorname{Ind}(\operatorname{Ab})$. A filtered system with all successive maps zero has zero colimit: every element of every test value becomes zero at the next stage. The positive answer is therefore not obtained by replacing the degree-one transitions with identities.

The compatible embeddings $A_r\to P$, $1\mapsto1/p^r$, give $A\hookrightarrow\iota P$. This is monic by the stage kernel formula, since each embedding is monic. It is not an isomorphism. Every map $P\to A_r$ is zero: its image is divisible, while multiplication by $p^r$ is zero on the target and can be surjective on that image only when the image is zero. Hence $\operatorname{Hom}(\iota P,A)=0$, while $\operatorname{Hom}(\iota P,\iota P)$ contains the nonzero identity. Its cokernel $T$ is consequently nonzero.

The group $P$ is divisible, hence injective as proved by the full subgroup-extension argument in the quasi-injective lesson, Exercise 1. Thus $RF(P)=F(P)[0]=\mathbb F_p[0]$. Under (3.1) the map from (7.2) to $R\widehat F(\iota P)$ is the identity in degree zero: its stage generator $p^{r-1}$ maps to $1/p$. It is an isomorphism in every cohomological degree and therefore in the derived category. The derived triangle for $0\to A\to\iota P\to T\to0$ now gives $R\widehat F(T)=0$. This does not make $T$ zero; a derived functor need not detect every nonzero object.

7.2. Extensions in every degree — intermediate

Exercise 2. Let $k$ be a field, $R=k[\epsilon]/(\epsilon^2)$, and let $\mathcal C$ be finite dimensional left $R$-modules. Verify the enough-injectives hypothesis and $\operatorname{Ind}(\mathcal C)\simeq\operatorname{Mod}(R)$. For $X=k=R/(\epsilon)$, let $W_r=k^r$, with trivial $\epsilon$-action and coordinate inclusions. Compute (5.3) in every degree, retaining the maps.

Solution. The coefficient of $\epsilon$ in $ab$ is a nondegenerate bilinear pairing on $R$: in basis $1,\epsilon$ its matrix is $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$. It gives an $R$-linear isomorphism $R\simeq\operatorname{Hom}_k(R,k)$, with $(r\lambda)(s)=\lambda(sr)$. For finite dimensional $M$, the adjunction $\begin{gathered} \operatorname{Hom}_R(M,\operatorname{Hom}_k(R,k))\\ \simeq\operatorname{Hom}_k(M,k) \end{gathered} \tag{7.3}$ sends a map to its value at 1, with inverse $\lambda\mapsto(m\mapsto[r\mapsto\lambda(rm)])$. The right side is exact, by extension of a linear functional from a subspace along a basis. Thus the dual regular module is injective in $\mathcal C$. A basis of $\operatorname{Hom}_k(M,k)$ gives a finite evaluation embedding of $M$ into a finite product of copies of that module, because a nonzero vector is separated by a linear functional. This proves enough injectives in the finite category.

Every $R$-module is the filtered union of its finitely generated submodules, which are finite dimensional since $R$ is finite dimensional. Each finite module has a finite presentation: a finite free surjection has a finite dimensional kernel, which has finitely many generators. The Ind Hom formula then proves full faithfulness of ordinary realization; the unions prove essential surjectivity. Exactness follows from the common-stage kernels and cokernels.

The augmented periodic free resolution of $k$ has one copy of $R$ in each degree $0,-1,-2,\ldots$ and differential multiplication by $\epsilon$. It is exact in negative degrees because $\ker\epsilon=\epsilon R=\operatorname{im}\epsilon$, with cokernel $k$ in degree zero. Applying Hom into a module $W$ killed by $\epsilon$ gives one $W$ in every nonnegative degree and zero differentials. Therefore $\operatorname{Ext}^n_R(k,W)=W\qquad(n\ge0). \tag{7.4}$ The same resolution computes the ambient groups; alternatively all finite endpoint comparisons follow from the bounded comparison in the quasi-injective lesson. For $W=W_r$, the Ext transition in every degree is exactly the coordinate inclusion $k^r\hookrightarrow k^{r+1}$, because the Hom maps are componentwise those inclusions. At the colimit $W=\bigoplus_{j\ge1}k$, the same resolution still gives $W$ in every degree and zero differential. Thus (5.3) is, in every degree, the identity identification of the union of finite coordinate spaces with $\bigoplus_{j\ge1}k$. There is no finite homological-dimension bound in this example, despite the asserted continuity in every fixed degree.

7.3. A 2-adic Ext class and two different colimits — advanced

Exercise 3. Put $X=\mathbb Z[1/2]$ in the large base $\mathcal C=\operatorname{Ab}$, and set $A=\operatorname{colim}_r\iota(\mathbb Z\xrightarrow{2}\mathbb Z\xrightarrow{2}\cdots)$. Prove $\begin{gathered} \operatorname{Ext}^1_{\mathcal A}(\iota X,A) \simeq\mathbb Z_2/\mathbb Z\ne0,\\ \operatorname{Ext}^1_{\mathbb Z}(X,\mathbb Z[1/2])=0, \end{gathered} \tag{7.5}$ and compute the stage transition in the first group. Here $\mathbb Z_2=\lim_m\mathbb Z/2^m$.

Solution. Let $L=\bigoplus_{r\ge0}\mathbb Ze_r$, and define $d(e_r)=e_r-2e_{r+1}$ and $q(e_r)=2^{-r}$. Then $0\longrightarrow L\xrightarrow{d}L\xrightarrow{q}X\longrightarrow0 \tag{7.6}$ is a free resolution. To see injectivity of $d$, compare the coefficient at the least index of a nonzero finite sum. For the cokernel, its relations give $e_r=2e_{r+1}$, and identify every finite sum with an integer multiple of a chosen last $e_N$. If its rational value under $q$ is zero that integer is zero, so the kernel is precisely the relation subgroup. Every dyadic rational occurs, proving exactness.

Hom into $M$ gives the two-term product complex with $\begin{gathered} \Delta_M\colon\prod_{r\ge0}M\longrightarrow\prod_{r\ge0}M,\\ \Delta_M(a)_r=a_r-2a_{r+1}. \end{gathered} \tag{7.7}$ For $M=\mathbb Z$, define $\Phi(b)=\sum_{r\ge0}2^r b_r$ in $\mathbb Z_2$: modulo $2^m$ only the first $m$ summands matter, so this defines a compatible inverse-limit element for every integer sequence. Telescoping gives $\Phi(\Delta a)=a_0$, since $2^{m}a_m$ vanishes modulo every fixed power of 2 for large $m$. Thus $\Phi$ induces a map from $\operatorname{coker}\Delta_{\mathbb Z}$ to $\mathbb Z_2/\mathbb Z$. It is onto, because every compatible residue sequence has a binary digit expansion: successively choose the digit representing the next difference divided by its current power of 2.

If $\Phi(b)=z\in\mathbb Z$, set $a_0=z$ and $a_{r+1}=(a_r-b_r)/2$. These are integers: equivalently $2^r a_r=z-\sum_{j<r}2^j b_j, \tag{7.8}$ whose right side is divisible by $2^r$ by the defining residues of $\Phi(b)$. The recurrence gives $\Delta a=b$. This proves the desired cokernel isomorphism $\operatorname{Ext}^1_{\mathbb Z}(X,\mathbb Z)=\mathbb Z_2/\mathbb Z$.

Postcomposition by $2\colon\mathbb Z\to\mathbb Z$ multiplies every sequence coordinate by 2, hence acts on this quotient by multiplication by 2. This is an automorphism. If $2x$ is an ordinary integer for $x\in\mathbb Z_2$, that integer is even, since its residue modulo 2 is zero; division makes $x$ an integer, proving injectivity on the quotient. For surjectivity, add either 0 or 1 to a representative to make it even in $\mathbb Z_2$, then divide by 2. Consequently the formal filtered system of these Ext groups has ordinary colimit $\mathbb Z_2/\mathbb Z$. The explicit colimit identification sends a class $u$ at stage $r$ to $2^{-r}u$, where the inverse is taken in this quotient automorphism. Formula (5.3) proves the first equality of (7.5).

The quotient is nonzero: 3 is invertible modulo every $2^m$, with compatible inverses, so $1/3$ defines a 2-adic integer. It is not an ordinary integer, because the embedding $\mathbb Z\to\mathbb Z_2$ is injective and no integer solves $3z=1$. Its class is nonzero.

For $M=\mathbb Z[1/2]$, the map $\Delta_M$ is onto: for any sequence $b$, choose $a_0=0$ and recursively $a_{r+1}=(a_r-b_r)/2$, now always an element of $M$. Its cokernel, the second Ext group in (7.5), is zero. Ordinary realization of $A$ is exactly $\mathbb Z[1/2]$. The two answers in (7.5) therefore concern two different targets in two different categories; replacing a formal filtered object by its ordinary realization loses the displayed extension class.

7.4. A Hom complex with signed transition maps — advanced

Exercise 4. Again take $\mathcal C=\operatorname{Ab}$ and a prime $p$. Let $P$ be the free complex $\mathbb Z\xrightarrow{p}\mathbb Z$ in degrees $-1,0$, so $P\simeq\mathbb Z/p$. Put $K_r=(A_r\xrightarrow{p}A_r)$ in degrees $0,1$, with $A_r=\mathbb Z/p^r$, and let the transition multiply both terms by $p$. Compute every group and transition in (6.1) by an actual Hom complex. Identify the termwise formal colimit of the targets.

Solution. In $\operatorname{Hom}^\bullet(P,K_r)$, put the two degree-one components in the order $P^{-1}\to K_r^0$, $P^0\to K_r^1$. The Hom differential $d(f)=d_Kf-(-1)^n f d_P$ gives $\begin{gathered} A_r\xrightarrow{d^0}A_r^2\xrightarrow{d^1}A_r,\\ d^0(a)=(-pa,pa),\\ d^1(b,c)=p(b+c). \end{gathered} \tag{7.9}$ It squares to zero, since $p(-pa+pa)=0$. These are degrees $0,1,2$. A bounded free source computes derived Hom into every target here, so the cohomology is the left Hom group of (6.1).

Degree zero is $A_r[p]$, and degree two is $A_r/pA_r$. In degree one set $u=b$, $v=b+c$. The cycle condition is $pv=0$, and a boundary has $u=-pa$, $v=0$. Thus the map $[(b,c)]\longmapsto(b\bmod pA_r,\ b+c) \tag{7.10}$ is an isomorphism $H^1\simeq(A_r/pA_r)\oplus A_r[p]$. It is onto by the representative $(b,v-b)$; its kernel consists precisely of the boundaries just displayed. Every transition multiplies $a,b,c$ by $p$. On the quotient factor of (7.10) this is zero; on the torsion factor, with generator $p^{r-1}$, it is the identity in the next generator. The same calculation makes the degree-zero transition the identity and the degree-two transition zero. The filtered colimits of the three cohomology groups are therefore $\mathbb F_p,\mathbb F_p,0$, with all other degrees zero.

The termwise formal colimit is $K=(A\xrightarrow{p}A)$, with the formal $A$ of Exercise 1. Its kernel is $\iota\mathbb F_p$, by the identity transitions on $A_r[p]$, and its cokernel is zero, by the zero transitions on $A_r/pA_r$. The inclusion of that kernel in degree zero is a quasi-isomorphism $\iota\mathbb F_p[0]\to K$. Thus the right side of (6.1) is the Ext from $\iota\mathbb Z/p$ into $\iota\mathbb F_p$, equal to $\mathbb F_p$ in degrees zero and one and zero otherwise. The exact constant inclusion and the free resolution identify these with ordinary Ext over $\mathbb Z$. The canonical comparison sends the surviving degree-zero torsion generator to $1\in\mathbb F_p$; in degree one the surviving coordinate is $v=b+c$, while the quotient coordinate becomes zero at the next stage. This checks the actual comparison and its differential signs, rather than only matching dimensions. All targets have the common lower term bound zero required by §6.