Draft lesson.

Derived sections of Ind Serre quotients

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

The section of an abelian quotient is fully faithful but can fail to be exact. Passing to complexes therefore requires its right derived functor. We construct that functor for the Ind Serre quotient, prove the full derived adjunction, and obtain an equivalence with a Verdier quotient. A finite ring example shows why a local object of the abelian category can fail the derived local-object test, and why the derived section can have infinitely many nonzero cohomology groups.

All complexes are cohomological, with $H^n(X[r])=H^{n+r}(X)$. We use bounded-below derived categories: their objects have cohomology zero below some integer. No upper cohomology bound is imposed.

Ind-completion and Serre quotients constructs the exact quotient and its section, including the quotient-fraction counit proof. Testing injectivity on generators and constructing enough injectives, Theorem 7.2, proves enough injectives in a Grothendieck category. Injective replacements and right derived functors, Sections 1–3, constructs bounded-below injective replacements and their comparisons. Derived morphisms through quasi-isomorphism roofs, Section 5, proves both the K-injective and K-projective Hom tests. Null systems, roofs and the retract kernel, Section 6, supplies the structured Verdier universal property; Orthogonal localization and the morphisms between pieces, Sections 1–2, supplies the localization triangle and orthogonal Hom theorem. We use the actual truncation triangles from Cohomological bounds, denominators and truncation adjunctions, Section 5.

1. The abelian quotient and its supply of injectives

Fix a small abelian category $\mathcal C$, a Serre subcategory $\mathcal S$, and the small quotient $Q:\mathcal C\to\mathcal B=\mathcal C/\mathcal S$. A small skeleton of an essentially small category has the same role. Put $\mathcal A=\operatorname{Ind}(\mathcal C),\qquad \mathcal D=\operatorname{Ind}(\mathcal B),\qquad \mathcal T=\operatorname{Ind}(\mathcal S). \tag{1.1}$ The Ind Serre comparison provides an exact functor $L:\mathcal A\to\mathcal D$ and an additive right adjoint $R$, with $(RY)(A)=\operatorname{Hom}_{\mathcal D}(c_{\mathcal B}QA,Y), \qquad LR\xrightarrow{\epsilon^0}1_{\mathcal D}\text{ invertible}, \qquad\ker L=\mathcal T. \tag{1.2}$ It proves that both categories are abelian, have all small coproducts and exact filtered colimits, and have a small separating family of constant objects. Their coproduct is a generator. Thus they are Grothendieck abelian categories, and the stated injective-existence theorem applies to both.

Work in an ambient universe containing these categories and the choices of resolutions. The Hom computations below are nevertheless small in the original universe for fixed objects: a Hom group of complexes is a subgroup of a countable product of the original Hom sets, and homotopy classes are its quotient. Bounded-below injective replacements will identify all the derived Hom groups we use with such groups.

2. The right adjoint preserves injectives

Lemma 2.1. If $I$ is injective in $\mathcal D$, then $RI$ is injective in $\mathcal A$.

Proof. Given a monomorphism $U\hookrightarrow V$ and a map $U\to RI$, transpose it to $LU\to I$. Exactness of $L$ makes $LU\hookrightarrow LV$ monic. Extend the transposed map to $LV\to I$ by injectivity, then transpose back. Naturality of adjunction says that the resulting $V\to RI$ extends the original map. This is the defining injectivity test. $\square$

Choose a bounded-below injective replacement $q_Y:Y\to I_Y$ for each $Y\in D^+(\mathcal D)$, using a bounded-below representative first when needed. Define the derived section by $\mathbf S(Y)=\mathbf R R(Y)=R(I_Y)\in D^+(\mathcal A). \tag{2.1}$ The right derived functor theorem gives its action on all derived arrows, its triangulated structure, and the canonical comparisons between different choices. Lemma 2.1 makes $R(I_Y)$ itself a bounded-below complex of injectives in $\mathcal A$. Consequently it is K-injective there. This additional fact will prove the adjunction directly.

Apply $L$ term by term. Exactness makes it preserve acyclic complexes and quasi-isomorphisms, since $H^n(LX)=L(H^nX). \tag{2.2}$ Indeed it preserves the kernels defining cycles, the cokernels and images defining boundaries, and their cohomology quotient. Additivity preserves homotopies, shifts and cones. It therefore defines a triangulated functor $\mathbf L:D^+(\mathcal A)\to D^+(\mathcal D)$.

3. Lift the adjunction, including its units and counits

Theorem 3.1. There is a natural derived adjunction $\mathbf L\dashv\mathbf S$. Its counit $\epsilon:\mathbf L\mathbf S\to1$ is invertible, so $\mathbf S$ is fully faithful.

Proof. Use a bounded-below representative of $X$ and the replacement $I_Y$. The K-injective tests give $\begin{aligned} \operatorname{Hom}_{D^+(\mathcal A)}(X,\mathbf S Y) &=\operatorname{Hom}_{K(\mathcal A)}(X,RI_Y)\\ &\simeq\operatorname{Hom}_{K(\mathcal D)}(LX,I_Y)\\ &=\operatorname{Hom}_{D^+(\mathcal D)}(\mathbf L X,Y). \end{aligned} \tag{3.1}$ For the middle bijection, transpose every degree component. Naturality of the ordinary adjunction says that the chain-map equations transpose to the chain-map equations. It also says that an equation $f-g=dh+hd$ transposes to the same homotopy equation. Additivity preserves the sums and signs. Thus this is a bijection of homotopy Hom groups, natural on chain maps.

It is natural on all derived source arrows too: for a quasi-isomorphism into a source, both K-injective tests make precomposition a bijection, so the same commuting square remains commuting after inverting it. Every source arrow is a roof. For a derived target arrow, use the unique homotopy map between injective replacements representing it; its image under $R$ represents the section arrow. Degreewise adjunction is natural on that homotopy map. This proves naturality in both derived variables, including all comparison isomorphisms between replacements.

For a bounded-below $X$, the unit $\eta_X:X\to\mathbf S\mathbf L X$ is represented by $X\xrightarrow{\eta_X^0}R(LX) \xrightarrow{R(q_{LX})}R(I_{LX}). \tag{3.2}$ The ordinary unit is applied degreewise. The second map need not be a quasi-isomorphism by itself; formula (3.1) identifies the composite as the map corresponding to identity of $\mathbf L X$.

The derived counit is $\mathbf L\mathbf S Y=LR(I_Y) \xrightarrow{\epsilon_{I_Y}^0}I_Y \xrightarrow{q_Y^{-1}}Y. \tag{3.3}$ Both arrows here are invertible in the derived category: the first is a degreewise isomorphism of complexes by (1.2), and the second is inverse to a quasi-isomorphism. They are the counit of (3.1), since the ordinary transposition sends identity of $RI_Y$ to $\epsilon^0_{I_Y}$. A natural Hom adjunction supplies both triangle identities; specifically the two composites transpose to the identities in (3.1). Hence $\epsilon_{\mathbf L X}\,\mathbf L\eta_X=1_{\mathbf L X}, \qquad \mathbf S\epsilon_Y\,\eta_{\mathbf S Y}=1_{\mathbf S Y}. \tag{3.4}$ The isomorphic counit then gives $\operatorname{Hom}(\mathbf S Y,\mathbf S Z)\simeq \operatorname{Hom}(\mathbf L\mathbf S Y,Z)\simeq\operatorname{Hom}(Y,Z)$ with the section's actual map correspondence. This proves full faithfulness. $\square$

All comparisons respect shifts and cone triangles: on the injective models they are the degreewise adjunction comparisons, which commute with differential signs and finite biproducts. The derived comparisons are characterized by those models, so the same compatibility holds after localization.

4. Identify the kernel and the Verdier comparison

Put $\mathcal N=D^+_{\mathcal T}(\mathcal A) =\{X\in D^+(\mathcal A):H^nX\in\mathcal T\text{ for every }n\}. \tag{4.1}$ By (2.2) and $\ker L=\mathcal T$, this is precisely the object kernel of $\mathbf L$. It is a replete triangulated subcategory: the long exact cohomology sequence and Serre closure treat each triangle, and shifts reindex its cohomology. It is closed under retracts because each cohomology retract is a subquotient of an object of $\mathcal T$. Thus it is thick. Let $p:D^+(\mathcal A)\to D^+(\mathcal A)/\mathcal N$ be its Verdier quotient.

Theorem 4.1. The functor induced by $\mathbf L$ is a triangulated equivalence $\overline{\mathbf L}: D^+(\operatorname{Ind}(\mathcal C))/ D^+_{\operatorname{Ind}(\mathcal S)}(\operatorname{Ind}(\mathcal C)) \ \simeq\ D^+(\operatorname{Ind}(\mathcal C/\mathcal S)). \tag{4.2}$ A quasi-inverse is $p\mathbf S$.

Proof. The structured Verdier universal property descends $\mathbf L$, since it kills $\mathcal N$. The counit gives $\overline{\mathbf L}\,p\mathbf S\simeq1$. The first triangle identity and invertibility of the counit make $\mathbf L\eta_X$ invertible. A cone of $\eta_X$ is therefore killed by $\mathbf L$, and belongs to $\mathcal N$. Hence $p\eta_X$ is invertible. It gives $1_{D^+(\mathcal A)/\mathcal N} \xrightarrow{\sim}p\mathbf S\,\overline{\mathbf L}. \tag{4.3}$ The comparison is natural on ordinary derived arrows by Theorem 3.1. It is then natural on their inverted Verdier denominators and hence on every quotient roof. Shift compatibility was proved after (3.4), so these are comparisons of triangulated functors. They prove the asserted equivalence with its specified quasi-inverse. $\square$

The denominator category here uses complexes whose cohomology lies in $\mathcal T$. An intrinsic derived category of $\mathcal T$ has a separate comparison to this subcategory. Full faithfulness of that comparison requires its own proof, as explained in Thick abelian subcategories and derived comparison. Formula (4.2) uses exactly (4.1) and needs no extra identification.

5. Local complexes and actual quotient Hom groups

Call a complex derived local if it belongs to the right orthogonal $\mathcal N^\perp=\{Z:\operatorname{Hom}_{D^+(\mathcal A)}(N,Z[r])=0 \text{ for every }N\in\mathcal N,\ r\in\mathbb Z\}. \tag{5.1}$ Every $\mathbf S Y$ is local by (3.1), since $\mathbf L N=0$. Complete its unit, with an inverse rotation, to $N_X\longrightarrow X\xrightarrow{\eta_X}\mathbf S\mathbf L X \longrightarrow N_X[1],\qquad N_X\in\mathcal N. \tag{5.2}$ This is the actual hypothesis of the orthogonal-localization theorem. Its Hom proof makes the third vertex the universal local target of $X$ and makes the first vertex its universal target for arrows from $\mathcal N$. In particular $Z$ is local exactly when $\eta_Z$ is invertible: if $Z$ is local, (5.2) has both middle and third vertices local; its first vertex lies in $\mathcal N\cap\mathcal N^\perp$, so its identity is zero. Conversely an invertible unit identifies $Z$ with a section object.

The same theorem's roof calculation and Theorem 4.1 yield the natural formula $\begin{aligned} \operatorname{Hom}_{D^+(\mathcal A)/\mathcal N}(pX,pY[r]) &\simeq\operatorname{Hom}_{D^+(\mathcal A)}(X,\mathbf S\mathbf L Y[r])\\ &\simeq\operatorname{Hom}_{D^+(\mathcal D)}(\mathbf L X,\mathbf L Y[r]). \end{aligned} \tag{5.3}$ For the first equality replace the quotient target by its invertible unit image. A roof into that local target has a unique ordinary derived numerator because its two adjacent cone Hom groups vanish. This also preserves the equality relation and composition. Thus (5.3) concerns actual morphisms, not only object classes.

For a heart object $H$, the abelian local-object criterion involves Hom and degree-one Ext from objects of $\mathcal T$. Definition (5.1) involves all shifts of all complexes in $\mathcal N$. Solution 8.4 finds the first missed obstruction in degree two.

6. Truncations, the heart and cohomological bounds

Exactness of $L$ gives natural comparisons $\mathbf L\tau_{\leq a}X\simeq\tau_{\leq a}\mathbf L X, \qquad \mathbf L\tau_{\geq a}X\simeq\tau_{\geq a}\mathbf L X. \tag{6.1}$ On a complex these are the kernel and cokernel comparisons in the smart truncation definitions. They commute with chain maps; localization gives naturality on all derived arrows. The subcategory $\mathcal N$ is stable under both truncations because they retain or remove its cohomology groups. Thus $\tau$ preserves Verdier denominators: a denominator becomes a quasi-isomorphism after $\mathbf L$, and (6.1) makes its truncated image a quasi-isomorphism as well. The truncation functors and triangles therefore descend through $p$.

Transport the usual t-structure on $D^+(\mathcal D)$ across (4.2). Its truncation of $pX$ is $p\tau X$ by (6.1). Its heart is $\mathcal D\simeq\mathcal A/\mathcal T$, with the original exact quotient on heart objects. The transported lower and upper bounded subcategories are exactly the essential images under $p$ of the corresponding ones in $D^+(\mathcal A)$. For a target with lower bound $a$, use its section replacement starting in degree $a$. For a target with upper bound $b$, truncate its section above at $b$; (6.1) and the counit show that it still maps to the given target. This proves both image assertions even when the untruncated section has extra positive cohomology.

The functor $\mathbf S$ preserves lower cohomology bounds. Indeed a lower smart truncation and then an injective replacement can start at the same bound; applying $R$ retains its zero terms below that bound. This is left t-exactness. On a heart object $Y$, left exactness gives $H^0(\mathbf S Y)=RY, \qquad H^q(\mathbf S Y)=R^qR(Y)\quad(q>0). \tag{6.2}$ For the first equality use $Y=\ker(I_Y^0\to I_Y^1)$ and preservation of that kernel by $R$; the higher groups are the definition of the right derived functors. Applying the exact $L$ to (6.2) and using the counit shows that every positive group in (6.2) belongs to $\mathcal T$. It can still be nonzero. Solution 8.2 gives a bounded heart object whose section has nonzero cohomology in every positive degree.

7. Two successive quotients and coherent section comparisons

Let $\mathcal S_0\subseteq\mathcal S_1$ be Serre subcategories of the same small abelian $\mathcal C$. The ordinary two-quotient theorem, proved in Solution 4 of Serre quotients and local saturation, identifies $(\mathcal C/\mathcal S_0)/\overline{\mathcal S_1}\simeq\mathcal C/\mathcal S_1$, where $\overline{\mathcal S_1}$ is the image Serre subcategory. The exact universal-property portion of that proof uses only Serre closure and applies to these small quotients. The additional coproduct conclusion is unnecessary here.

On Ind categories write $L_0:\mathcal A\to\mathcal D_0$ and $L_{10}:\mathcal D_0\to\mathcal D_1$ for the two exact quotients, with sections $R_0,R_{10}$. Use the specified constant comparisons to identify their composite with $L_1$. The ordinary Hom adjunctions identify its section with $R_0R_{10}$, preserving the adjunction. Each right adjoint preserves injectives by Lemma 2.1. An injective replacement $I$ in $\mathcal D_1$ consequently gives an injective complex $R_{10}I$ in $\mathcal D_0$. It can be used directly to compute the outer derived section. Hence $\mathbf S_1\simeq\mathbf S_0\mathbf S_{10}, \qquad \mathbf L_1\simeq\mathbf L_{10}\mathbf L_0. \tag{7.1}$ Naturality on derived arrows follows on the maps between injective replacements and then on their inverses. Other choices give the same comparison by the resolution cocycle law.

With these identifications the units and counits are precisely $\eta^1_X=(\mathbf S_0\eta^{10}_{\mathbf L_0X})\eta^0_X, \qquad \epsilon^1_Y=\epsilon^{10}_Y\, \mathbf L_{10}(\epsilon^0_{\mathbf S_{10}Y}). \tag{7.2}$ To check them, transpose a map $\mathbf L_{10}\mathbf L_0X\to Y$ first along the outer quotient adjunction and then along the inner one. The resulting map is $\mathbf S_0\mathbf S_{10}(f)\, \mathbf S_0(\eta^{10}_{\mathbf L_0X})\eta^0_X$. Taking identity gives the first formula; transposing identity of the composite section gives the second. These natural bijections are the composite adjunction, so their two triangles hold. For three quotients, all associations give the same three successive Hom bijections and the same three unit factors. Uniqueness of the adjunction-preserving comparison proves their coherence. The Verdier equivalences likewise commute with these specified exact composite quotient functors.

8. Four graded exercises with complete solutions

Exercise 8.1 — Rational cohomology computes quotient morphisms (intermediate)

Take finitely generated abelian groups modulo finite groups. Characterize derived local complexes and compute $\operatorname{Hom}(pX,pY[r])$ in the Verdier quotient for arbitrary $X,Y\in D^+(\mathsf{Ab})$.

Solution. The Ind Serre lesson proves $L(M)=M\otimes\mathbb Q$, $\mathcal T=\mathsf{Tors}$, and $R$ is restriction of scalars from rational spaces. Here $R$ is exact: kernels and quotients of rational linear maps have their underlying group kernels and quotients. It therefore takes quasi-isomorphisms to quasi-isomorphisms. The derived section agrees with termwise $R$.

The unit induces on cohomology the rationalization map $H^nX\to H^nX\otimes\mathbb Q$. It is invertible exactly when $H^nX$ is uniquely divisible, by the proved abelian local-object criterion. Cohomology detects derived isomorphisms, so $X$ is derived local exactly when every cohomology group is a rational vector space.

Over a field every complex splits, degree by degree, into its graded cohomology and a contractible complex: choose complements of boundaries in cycles and of cycles in each term. The differential identifies the last complement with the next boundaries; the inverse identifications give the contraction. Section 6 of the quasi-isomorphism-roof lesson proves that its canonical cohomology map identifies all derived Hom groups with the product of degreewise cohomology Hom groups. Applying (5.3) thus gives $\operatorname{Hom}(pX,pY[r]) \simeq\prod_{n\in\mathbb Z} \operatorname{Hom}_{\mathbb Q} \bigl(H^nX\otimes\mathbb Q,H^{n+r}Y\otimes\mathbb Q\bigr). \tag{8.1}$ The product permits infinitely many nonzero degrees; it follows from actual degreewise maps of the split complexes. Although the splitting choices are not natural, the displayed cohomology map is canonical, as the prerequisite's proof checks. $\square$

Exercise 8.2 — A bounded object with a section of infinite amplitude (advanced)

Use $B=\mathbb Z/4$, $k=\mathbb F_2$, and $\Lambda=\left(\begin{smallmatrix}B&k\\0&k\end{smallmatrix}\right)$, with the bimodule action through reduction modulo two. The Ind Serre lesson identifies the quotient $E:\mathsf{Mod}(\Lambda)\to\mathsf{Mod}(B)$, $E(V,W,t)=V$, and its section $J(H)=(H,H[2],\mathrm{ev})$. Compute $\mathbf R J(k)$ and all its cohomology groups.

Solution. First $B$ is injective as a left $B$-module. The coinduced-character construction in Injective objects in locally module stacks, Section 3, proves that $\operatorname{Hom}_{\mathbb Z}(B,\mathbb Q/\mathbb Z)$ is injective. The map $B\longrightarrow\operatorname{Hom}_{\mathbb Z}(B,\mathbb Q/\mathbb Z), \qquad a\longmapsto\bigl(b\longmapsto ab/4\bmod\mathbb Z\bigr) \tag{8.2}$ is well defined and $B$-linear for its coinduced action. A character is determined by its value on one, and that value is one of the four multiples of $1/4$. Hence (8.2) is bijective, proving the assertion.

An injective resolution of $k$ is $0\to k\xrightarrow{\bar a\mapsto2a}B \xrightarrow{2}B\xrightarrow{2}B\xrightarrow{2}\cdots. \tag{8.3}$ The first map is monic and its image is $2B$. Each multiplication-by-two map has kernel and image $2B$, proving exactness at every term. Thus the injective complex has a copy of $B$ in each degree $q\geq0$.

Apply $J$. Each term is $J(B)=(B,k,\mathrm{ev})$, where the evaluation of the lower generator is $2\in B$. Its differential is multiplication by two on the upper component and zero on the lower component, since multiplication by two kills $B[2]$. Each term is injective over $\Lambda$, by Lemma 2.1 for the exact evaluation $E$. This is the explicit complex representing $\mathbf R J(k)$.

Kernels and quotients of these representations are computed componentwise. Its degree-zero kernel has upper component $2B$, lower component $k$, and structure map sending the lower generator to two. Identify $2B$ with $k$; this kernel is $J(k)=(k,k,1)$. In each degree $q\geq1$, the upper cohomology is $2B/2B=0$, the lower cohomology is $k/0=k$, and the induced structure map is zero. Therefore $H^0(\mathbf R J(k))=J(k),\qquad H^q(\mathbf R J(k))=(0,k,0)\quad(q\geq1). \tag{8.4}$ Negative groups vanish. The positive groups are killed by $E$, as Theorem 3.1 requires. They are all nonzero, so the derived section of this bounded heart object does not belong to $D^b(\mathsf{Mod}(\Lambda))$. $\square$

Exercise 8.3 — When the equivalence has a bounded section (advanced)

Suppose that $R^qR(H)=0$ for every $H\in\mathcal D$ and every $q>d$, for a fixed $d\geq0$. If $Y$ has cohomology only in $[a,b]$, prove that $\mathbf S Y$ has cohomology only in $[a,b+d]$. Deduce the bounded Verdier equivalence with a fully faithful bounded section.

Solution. For a heart object $H$, the lower-bound statement and the hypothesis put $\mathbf S H$ in degrees $[0,d]$. Its shifted version $H[-n]$ therefore has section cohomology in $[n,n+d]$.

Use the finite truncation tower of $Y$: the successive triangles are $\tau_{\leq n-1}Y\longrightarrow\tau_{\leq n}Y \longrightarrow H^nY[-n]\longrightarrow\tau_{\leq n-1}Y[1], \quad a\leq n\leq b. \tag{8.5}$ The initial vertex $\tau_{\leq a-1}Y$ is zero and the final truncation is $Y$. Apply the triangulated $\mathbf S$. Inductively the long exact cohomology sequence says that the middle vertex has no groups outside the union of the ranges of the first and third vertices: at an excluded degree both adjacent groups $H^q$ of those vertices are zero. Their union is contained in $[a,n+d]$. The induction ends at $[a,b+d]$, proving the precise bound without an infinite convergence argument.

Consequently $\mathbf S$ restricts to $D^b(\mathcal D)\to D^b(\mathcal A)$. The exact $\mathbf L$ also restricts, its kernel there is $D^b_{\mathcal T}(\mathcal A)$, and the counit and unit of Section 3 restrict. Their cone objects are still bounded. The proof of Theorem 4.1 therefore applies inside these bounded categories, giving $D^b(\mathcal A)/D^b_{\mathcal T}(\mathcal A) \simeq D^b(\mathcal D) \tag{8.6}$ with fully faithful section $\mathbf S$ of the stated amplitude. Solution 8.2 fails the hypothesis for every finite $d$; its nonbounded section is therefore consistent with this conclusion. $\square$

Exercise 8.4 — The degree-two obstruction to derived locality (expert)

In Solution 8.2 put $P=(0,k,0)$ and $G=J(k)=(k,k,1)$. Compute every $\operatorname{Ext}^n_\Lambda(P,G)$, and use the result to test both $G[0]$ and $\mathbf R J(k)$ for derived locality.

Solution. Let $e_1,e_2$ be the two diagonal idempotents. The modules $P_1=\Lambda e_1=(B,0,0)$ and $P_2=\Lambda e_2=(k,k,1)=G$ are projective: $\operatorname{Hom}_\Lambda(\Lambda e_i,M)\simeq e_iM$, by evaluation at $e_i$, with inverse $x\mapsto(ae_i\mapsto ax)$. The component functor $e_i(-)$ is exact because module kernels and quotients are componentwise.

Projection $G\to P$ is identity on the lower component and zero on the upper. Its kernel is $(k,0,0)$. Reduction on the upper component defines $P_1\to G$ with exactly that image; its kernel is $(2B,0,0)$. The subsequent multiplication-by-two maps on $P_1$ have that same kernel and image. Thus a complete projective resolution is $\cdots\xrightarrow{2}P_1\xrightarrow{2}P_1 \xrightarrow{\mathrm{red}}G\longrightarrow P\longrightarrow0. \tag{8.7}$ Here $G$ is homological degree zero, the rightmost $P_1$ is degree one, and every later $P_1$ has the next degree. Regard this as a bounded-above cochain projective replacement in degrees $0,-1,-2,\ldots$. Its K-projectivity, proved in the roof lesson, makes Hom from this replacement compute all derived morphisms, including the groups of negative degree.

Now $\operatorname{Hom}_\Lambda(G,G)=k$: the structure square forces its upper and lower scalars to be equal. Also $\operatorname{Hom}_\Lambda(P_1,G)=k$, by its upper component. Precomposition by reduction takes that common scalar to the same upper scalar, so the first differential of the Hom complex is identity of $k$. Every subsequent differential is zero, since it is induced by multiplication by two into a target killed by two. The Hom complex is therefore $k\xrightarrow{1}k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}\cdots \tag{8.8}$ in degrees $0,1,2,\ldots$. Signs from an alternative Hom differential convention do not change identity being invertible or the other maps being zero. Its cohomology gives $\operatorname{Ext}^n_\Lambda(P,G)= \begin{cases}0,&n<2,\\k,&n\geq2.\end{cases} \tag{8.9}$ This includes all negative degrees, degree zero and degree one.

The object $G$ is abelian local, since it is the ordinary section $J(k)$. Nevertheless $P\in\mathcal T$ and $\operatorname{Hom}(P,G[2])=k\ne0$, so $G[0]$ is not derived local. Equivalently its derived unit $G\to\mathbf R J(EG)=\mathbf R J(k)$ is an isomorphism on $H^0$ but cannot be a quasi-isomorphism, by (8.4).

In contrast $\mathbf R J(k)$ is derived local by Theorem 3.1 and (5.1). Its positive cohomology objects $P$ are not even abelian local: $\operatorname{Hom}(P,P)$ contains identity, whereas abelian locality requires Hom from every torsion object to vanish. Hence neither an abelian local heart object nor a degreewise test of abelian locality characterizes the derived section. The full orthogonal condition is the required test. $\square$

9. Further reading

The Stacks Project, Injectives and adjoint functors, explains injective preservation by a right adjoint with exact left adjoint. Deriving adjoints gives the general derived adjunction interface. Cohomology in a weak Serre subcategory treats the supported-cohomology triangulated subcategory. The programme proofs linked above supply the replacement, fraction, truncation and quotient arguments used here.