Draft lesson.

Thick abelian subcategories and derived comparison

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

An extension can have both endpoints in a small abelian category while its intermediate objects live in a larger one. In degree one, closure under extensions brings the middle object back into the smaller category. In higher degrees there are several middle objects, and endpoint membership alone does not control them. This lesson identifies the exact additional condition, proves it in both variances, and then reconstructs bounded complexes from their cohomology and connecting morphisms.

Conventions and prerequisites

Work in a fixed ambient universe in which the categories and derived Hom groups under discussion are defined. Complexes are cohomological, with \(X[r]^j=X^{j+r}\) and shifted differential \((-1)^r d_X\). The cone of \(f:X\to Y\) has terms \(Y^j\oplus X^{j+1}\), differential \((y,x)\mapsto(d_Yy+fx,-d_Xx)\), and positive projection to \(X[1]\). Neither abelian category is assumed to have enough injectives or projectives.

We use the complete proofs in Higher extensions, roofs and Yoneda composition, Sections 1–4: every positive derived class is represented by an exact extension, endpoint naturality holds, and normalized splicing equals composition. Short exact extensions and the Baer group, Sections 1–4 proves degree-one classification and the endpoint-preserving middle-isomorphism criterion. Triangles after quasi-isomorphism localization, Sections 2 and 6 proves exact Hom sequences and the natural triangle of a short exact row. Cohomological bounds and truncation adjunctions, Sections 3–5 supplies bounded representatives, negative Ext vanishing, heart full faithfulness and the natural truncation triangles. The projective computations in the exercises use Hereditary categories and cohomology splittings, Lemma 3.1, which proves that a bounded-above projective complex computes the actual derived Hom.

1. The precise subcategory and the comparison maps

Let \(\mathcal A\) be abelian. By a thick abelian subcategory \(\mathcal B\subset\mathcal A\) we mean a full replete subcategory containing zero, closed under ambient kernels and cokernels of arrows between its objects, and closed under extensions: if \(0\to B'\to E\to B''\to0\) is exact and \(B',B''\in\mathcal B\), then \(E\in\mathcal B\). This is also called a weak Serre subcategory. A Serre subcategory has the stronger property that every ambient subobject and quotient of each of its objects belongs to it. We use the weaker hypothesis throughout.

The stated closure properties make \(\mathcal B\) abelian and its inclusion exact. Indeed \(B'\oplus B''\) is an extension of its two summands, so finite biproducts belong to \(\mathcal B\). Addition and zero maps are inherited because the subcategory is full. The ambient kernel and cokernel have the same universal properties against objects of \(\mathcal B\). The ambient image and coimage belong to \(\mathcal B\), and their canonical isomorphism and its inverse are arrows in the full subcategory. Thus the abelian axioms and exactness of the inclusion follow.

A useful consequence is the five-term property. If

is exact and its four displayed \(B_i\) belong to \(\mathcal B\), then so does \(Z\). Let \(C=\operatorname{coker}(B_0\to B_1)\) and \(K=\ker(B_3\to B_4)\). Exactness identifies \(C\) with the image in \(Z\) and \(K\) with its quotient, giving \(0\to C\to Z\to K\to0\). Both ends lie in \(\mathcal B\), hence extension closure gives the assertion.

Define

as a full replete subcategory. It is triangulated. Shifts preserve the condition. If two vertices of a triangle have cohomology in \(\mathcal B\), the long exact cohomology sequence gives (1.1) with the remaining vertex's \(H^j\) in the middle; rotation covers every placement of that vertex. It is also closed under any existing direct summand: on cohomology a summand is the kernel of an idempotent complement endomorphism of an object of \(\mathcal B\). All complexes remain cohomologically bounded.

The exact inclusion preserves acyclic complexes and cones, so its functor on homotopy categories descends through inversion of quasi-isomorphisms to an exact functor

Kernels, images and quotients computing cohomology agree in the two categories, so \(H^j(FX)=H^j(X)\) as objects of \(\mathcal A\). For objects \(X,Y\in\mathcal B\), put

Here Ext means \(\operatorname{Hom}_D(X,Y[n])\). Bounded derived categories compute these groups by the bounded-denominator theorem in the prerequisite. The maps \(\phi^n\) are induced by \(F\). They preserve addition, composition and the exact Ext sequences of short rows: the inclusion preserves the actual cone, its positive boundary projection and its cone-to-quotient quasi-isomorphism.

Proposition 1.1. Comparison is an isomorphism in degrees zero and one; both sides vanish in negative degrees.

Proof. Negative vanishing and degree-zero identification with ordinary Hom follow from the cohomological-bounds prerequisite. The latter comparison is an isomorphism because \(\mathcal B\) is full. A class in \(T^1(X,Y)\) is represented by \(0\to Y\to E\to X\to0\). Extension closure puts \(E\) in \(\mathcal B\), giving surjectivity. If two such rows have the same ambient class, the degree-one classification supplies an ambient middle isomorphism with identity endpoints. Fullness puts that map and its inverse in \(\mathcal B\), so the classes agree intrinsically. The zero row and addition are preserved by the exact functor, giving an isomorphism of groups. \(\square\)

2. Intrinsic effacement and the higher-degree theorem

An ambient class \(\alpha\in T^n(X,Y)\), \(n>0\), is effaced on the left end if there is a monomorphism \(i:Y\to J\) with \(J\in\mathcal B\) and \(i_*\alpha=0\). It is effaced on the right end if there is an epimorphism \(p:P\to X\) with \(P\in\mathcal B\) and \(p^*\alpha=0\). These are elementwise assertions: \(i\) or \(p\) may depend on the particular class and its degree. A single arrow killing a whole Ext group is a stronger condition.

Lemma 2.1. In any abelian category, every positive Ext class can be effaced in either end by an arrow in that category.

Proof. Realize the class by an exact row

For \(n=1\), its class is the boundary \(\delta:X\to Y[1]\) of the short row. The triangle equations give \(i[1]\delta=0\) and \(\delta p=0\).

For \(n\ge2\), put \(C=\operatorname{coker}i\). It is the image of the next map in the row. The left short row \(0\to Y\to E_{n-1}\to C\to0\) has class \(\eta:C\to Y[1]\), and the remaining row has class \(\gamma:X\to C[n-1]\). Normalized splicing gives \(\alpha=\eta[n-1]\gamma\). Consequently \(i[n]\alpha=0\), since \(i[1]\eta=0\). At the other end use \(K=\ker p\), the last short row's boundary \(\beta:X\to K[1]\), and the front row's class \(\theta:K\to Y[n-1]\). Then \(\alpha=\theta[1]\beta\) and \(\alpha p=0\). This also proves the two intrinsic factorizations that will be used below. Every term of (2.1) is in the category where the class was realized. \(\square\)

Theorem 2.2. Under the hypotheses of Section 1, the following conditions are equivalent.

1. Every \(\phi^n_{X,Y}\) is an isomorphism for all \(X,Y\in\mathcal B\) and all \(n>0\).
2. Every positive ambient class with endpoints in \(\mathcal B\) can be effaced on the left end by a monomorphism within \(\mathcal B\).
3. Every positive ambient class with endpoints in \(\mathcal B\) can be effaced on the right end by an epimorphism within \(\mathcal B\).
4. The functor (1.3) is fully faithful.
5. The functor (1.3) is an equivalence.

In particular requiring both effacements simultaneously gives the same condition, but either one alone already suffices. Sections 2–4 prove every implication.

Condition 1 implies 2 and 3: lift an ambient class to its intrinsic class and apply Lemma 2.1 in \(\mathcal B\). Naturality of \(\phi\) makes the corresponding ambient pushforward or pullback zero. The monomorphism and epimorphism are ambient ones too, because the inclusion is exact.

3. Both effacement inductions, including injectivity

First assume condition 2. We prove all comparison maps are isomorphisms by induction on \(n\), simultaneously for every pair of objects. Degree zero is Proposition 1.1. Suppose comparison in degree \(n-1\) is an isomorphism for every pair.

Surjectivity in degree \(n\). Take \(\alpha\in T^n(X,Y)\). Choose \(i:Y\hookrightarrow J\) in \(\mathcal B\) killing it and let \(C=\operatorname{coker}i\), which lies in \(\mathcal B\). The exact row gives the commutative exact-sequence diagram

Since the last image of \(\alpha\) is zero, \(\alpha=\partial_T\gamma\) for some \(\gamma\in T^{n-1}(X,C)\). Lift \(\gamma\) to \(\widetilde\gamma\in S^{n-1}(X,C)\) by the inductive surjectivity. Then \(\partial_S\widetilde\gamma\) maps to \(\alpha\).

Injectivity in degree \(n\). Take \(\beta\in S^n(X,Y)\) with \(\phi^n\beta=0\). Lemma 2.1, applied intrinsically, gives a monomorphism \(Y\hookrightarrow J\) in \(\mathcal B\) killing \(\beta\). This arrow is chosen for \(\beta\), and need not be the arrow used in the surjectivity proof. With \(C\) its cokernel, exactness gives \(\beta=\partial_S\gamma\), where \(\gamma\in S^{n-1}(X,C)\). The image of \(\phi^{n-1}\gamma\) under \(\partial_T\) is zero. Thus it is the image of some \(\lambda\in T^{n-1}(X,J)\). Lift \(\lambda\) to \(\widetilde\lambda\in S^{n-1}(X,J)\). Subtract its image from \(\gamma\). The result maps to zero under \(\phi^{n-1}\), so inductive injectivity makes it zero. Hence \(\gamma\) is in the image of the first arrow in (3.1), and \(\beta=\partial_S\gamma=0\). This proves condition 1 from condition 2 in every degree, without an ambient injective object.

For completeness do the other variance explicitly. Assume condition 3. Choose \(p: P\twoheadrightarrow X\) inside \(\mathcal B\) and let \(K=\ker p\). The exact diagram is

For surjectivity, start with \(\alpha\in T^n(X,Y)\), choose \(p\) killing \(\alpha\), and write \(\alpha=\partial_T\gamma\). Lift \(\gamma\) using the degree-\((n-1)\) comparison for \((K,Y)\). Its intrinsic boundary lifts \(\alpha\).

For injectivity, start with \(\beta\in S^n(X,Y)\) mapping to zero, choose an intrinsic \(p\) killing \(\beta\) by Lemma 2.1, and write \(\beta=\partial_S\gamma\). Its image \(\phi^{n-1}\gamma\) is in the image of \(T^{n-1}(P,Y)\). Lift such a preimage by the degree-\((n-1)\) comparison for \((P,Y)\), subtract its intrinsic image, and use degree-\((n-1)\) injectivity for \((K,Y)\). This shows \(\gamma\) is an intrinsic image, so its boundary \(\beta\) vanishes. Starting at degree zero proves condition 1 from condition 3. Only kernels or cokernels of arrows between objects of \(\mathcal B\) were needed; arbitrary ambient subobjects were never required.

4. From object Ext to all bounded complexes

We give the finite induction rather than assuming that agreement on a heart automatically gives agreement on its derived category.

Five-term comparison lemma. In a commuting diagram of exact sequences of abelian groups \(G_1\to G_2\to G_3\to G_4\to G_5\) and \(G'_1\to G'_2\to G'_3\to G'_4\to G'_5\), if the four vertical maps outside the middle are isomorphisms, then the middle map is an isomorphism.

For injectivity, let \(x\in G_3\) map to zero. Its image in \(G_4\) is zero by the fourth vertical map's injectivity, so \(x\) comes from \(y\in G_2\). The image of \(y\) below is in the image of \(G'_1\); lift that preimage to \(G_1\) and subtract its image from \(y\). The result has zero image in \(G'_2\), so it is zero. Thus \(x=0\). For surjectivity, map \(x'\in G'_3\) to \(G'_4\), lift to \(y\in G_4\), and use injectivity of the fifth vertical map to see that \(y\) maps to zero in \(G_5\). Write \(y\) as the image of \(x\in G_3\). Then \(x'-f_3(x)\) comes from \(G'_2\); lift that element to \(G_2\) and add its image to \(x\). This gives a preimage of \(x'\), proving the lemma.

Assume condition 1. If \(X=A[r]\), \(Y=B[s]\) are shifted objects of \(\mathcal B\), comparison for \(\operatorname{Hom}(X,Y)\) is comparison for \(\operatorname{Ext}^{s-r}(A,B)\). It is an isomorphism for positive degree by condition 1, zero degree by fullness, and negative degree because both groups vanish. Thus all Hom comparisons between shifted objects are isomorphisms.

First keep \(X\) a shifted object and allow \(Y\) to be any bounded derived object of \(\mathcal B\). Induct on the number of nonzero cohomology objects of \(Y\). Zero cohomology means the zero object; one cohomology object means a shifted heart object, by the bounded-representative theorem. Otherwise let \(b\) be the largest nonzero cohomology degree, set \(Y'=\tau_{\le b-1}Y\) and \(H=H^b(Y)[-b]\), and use the natural triangle

The inclusion commutes with these smart cuts because it preserves their kernels and quotients. Applying \(\operatorname{Hom}(X,-)\) gives the exact five-term window

The outer four comparison maps are isomorphisms by the shifted-object case and the induction hypothesis; shifts preserve the number of nonzero cohomology objects. The comparison lemma proves the middle assertion, simultaneously for all shifts of \(X\) and \(Y\).

Now fix any such \(Y\) and induct on the number of nonzero cohomology objects of \(X\). The base cases were just proved. Write \(X'\to X\to H\to X'[1]\) for its top truncation triangle. The contravariant Hom sequence gives

Again the outer four comparisons are isomorphisms, so the middle one is too. This proves full faithfulness for every pair of bounded complexes: condition 1 implies condition 4. The reverse implication restricts to \(X\) and \(Y[n]\) for heart objects.

Essential surjectivity. Suppose condition 4 holds and take \(Z\in D^b_{\mathcal B}(\mathcal A)\). Induct on the number of its nonzero cohomology objects. The zero and one-object cases are immediate. Otherwise its top truncation triangle is

By induction choose \(U\in D^b(\mathcal B)\) and an isomorphism \(FU\simeq L\); put \(V=H^b(Z)[-b]\) inside \(D^b(\mathcal B)\). Fullness lifts the arrow \(d:FV\to FU[1]\) to \(\widetilde d:V\to U[1]\). Complete \(-\widetilde d[-1]:V[-1]\to U\) to a triangle and rotate it once. The rotation rule makes its last arrow \(-(-\widetilde d[-1])[1]=\widetilde d\), so the triangle is

Apply \(F\). Rotate (4.5) and (4.4) to make their identified arrow \(d\) the first arrow; the triangle morphism axiom extends the identity square on that arrow to a morphism of triangles. Its third component is an isomorphism: the long exact cohomology sequences and the five-term comparison lemma show it is an isomorphism on all cohomology, and cohomology detects derived isomorphisms. Rotate back to obtain \(FW\simeq Z\). Hence condition 4 implies condition 5. An equivalence is fully faithful, so this completes Theorem 2.2.

The induction is finite because cohomology is bounded. It proves the bounded comparison theorem stated here; it supplies no conclusion about arbitrary unbounded complexes.

5. Four graded exercises with complete solutions

Exercise 1 (basic: the weaker hypothesis). Let \(\mathcal A\) be abelian groups and \(\mathcal B\) the full subcategory of uniquely divisible groups, equivalently rational vector spaces. Prove that it is thick abelian but is not Serre. Verify degree-one comparison directly. Explain why modules over \(k[t]/(t^2)\) annihilated by \(t\) provide a different example which does not meet the extension-closure hypothesis.

Solution. A uniquely divisible group has multiplication by every positive integer bijective. Define \((a/b)x\) to be the unique solution of \(by=ax\); unique divisibility verifies addition, multiplication and independence of the representative of \(a/b\), giving a rational vector space. A group homomorphism between such groups commutes with division by \(b\), since it takes the unique solution of \(by=x\) to the unique solution of the corresponding equation. It is therefore rational-linear. Kernels and cokernels of these arrows are rational vector spaces, and the subcategory is full and replete.

For an exact row \(0\to U\to E\to V\to0\) with both ends uniquely divisible, multiplication by \(n>0\) on \(E\) is injective: an element killed by \(n\) maps to zero in \(V\), belongs to \(U\), and is zero by injectivity there. It is surjective: for \(e\in E\), divide its image in \(V\) by \(n\), lift that quotient to \(e'\in E\), then \(e-ne'\in U\); dividing this difference in \(U\) gives \(u\) with \(n(e'+u)=e\). Thus \(E\) is uniquely divisible, proving extension closure. However \(\mathbb Z\subset\mathbb Q\) is an ambient subgroup which is not uniquely divisible, so the subcategory is not Serre.

Every ambient extension with rational endpoints is now a row of rational vector spaces. Choose a rational basis of the quotient, lift its basis vectors, and extend rational-linearly to a section. Consequently both degree-one Ext groups vanish, and comparison is the isomorphism between zero groups.

For the last assertion put \(R=k[t]/(t^2)\) and \(S=R/(t)\). The full category of modules annihilated by \(t\) is the category of \(k\)-vector spaces, but

is an ambient extension whose middle object is not annihilated by \(t\). It is nonsplit: an \(R\)-linear section would send \(1\in S\) to \(1+bt\), which is not killed by \(t\). Intrinsic Ext in degree one is zero by vector-space splitting, whereas this ambient degree-one class is nonzero. Extension closure is essential even before the higher comparison theorem begins.

Exercise 2 (intermediate: finite dual-number modules). Let \(R=k[t]/(t^2)\), let \(\mathcal A=R\text{-}\mathrm{Mod}\), and let \(\mathcal B\) consist of its finite-dimensional \(k\)-modules. Prove the comparison theorem holds for this inclusion. Compute \(\operatorname{Ext}^n_{\mathcal B}(S,S)\) for every \(n\ge0\), where \(S=R/(t)\).

Solution. Component dimensions in an exact row add, so the finite-dimensional modules form a Serre subcategory. We construct an ambient injective target inside it for each finite module, proving the left-end effacement condition.

Let \(\ell:R\to k\) extract the coefficient of \(t\). The pairing \((a,b)\mapsto\ell(ab)\) is nondegenerate: in the basis \(1,t\) its matrix is \(\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\). It gives an \(R\)-linear isomorphism \(R\simeq\operatorname{Hom}_k(R,k)\), where \((r\lambda)(a)=\lambda(ra)\). For every \(R\)-module \(M\), finite or not, there is a natural bijection

Its inverse sends \(g\) to \(m\mapsto(a\mapsto g(am))\); these formulas verify both compositions and \(R\)-linearity. A linear functional on a subspace extends by extending a basis, so the right-hand functor carries a monomorphism to a surjection on Hom. Thus \(R\), and every finite direct sum \(R^m\), is injective in the ambient module category.

For finite-dimensional \(Y\), its dual \(D(Y)=\operatorname{Hom}_k(Y,k)\) is an \(R\)-module with \((r\lambda)(y)=\lambda(ry)\). Choose finitely many \(R\)-generators to obtain \(R^m\twoheadrightarrow D(Y)\); a \(k\)-basis already gives such a finite generating set. Dualizing gives a monomorphism \(D(D(Y))\hookrightarrow D(R^m)\). Evaluation identifies \(Y\simeq D(D(Y))\), and the pairing above identifies \(D(R^m)\simeq R^m\). Hence \(Y\hookrightarrow R^m\) is an ambient monomorphism inside \(\mathcal B\).

Every positive ambient Ext group with second endpoint \(R^m\) is zero. Indeed represent a class by an extension row; its first short row splits because its left endpoint is injective. Lemma 2.1's factorization then makes the whole class zero. Therefore this embedding effaces every class with endpoint \(Y\), and Theorem 2.2 proves all Ext comparisons and the bounded derived equivalence.

Finally the bounded-above free resolution of \(S\) has one copy of \(R\) in every degree at most zero, all differentials in negative degrees multiplication by \(t\), and augmentation \(R\to S\). It is exact in negative degrees because the image and kernel of multiplication by \(t\) are both \((t)\). Applying the projective-Hom prerequisite, Hom to \(S\) has one copy of \(k\) in every nonnegative cochain degree and zero differentials, since \(tS=0\). Thus

The comparison theorem permits nonzero Ext in every positive degree; it does not assert vanishing.

Exercise 3 (advanced: a genuine Serre obstruction in degree two). Let \(\mathcal A\) be finite-dimensional representations of the quiver \(1\xrightarrow{a}2\xrightarrow{b}3\), with relation \(ba=0\). Let \(\mathcal B\) consist of representations with zero space at vertex 2. Show it is Serre and degree-one comparison holds, but degree-two comparison and both effacement conditions fail. Construct an object of \(D^b_{\mathcal B}(\mathcal A)\) outside the essential image.

Solution. Kernels, cokernels and exact sequences of these representations are computed at each vertex: the induced arrows on kernels and quotients satisfy \(ba=0\), so they are representations. Zero space at vertex 2 is preserved by subobjects, quotients and extensions. Thus \(\mathcal B\) is Serre. It is equivalent to pairs of finite-dimensional \(k\)-vector spaces at vertices 1 and 3, with zero arrows, so every short exact row in \(\mathcal B\) splits. By Proposition 1.1 all its object degree-one Ext comparisons are isomorphisms, with both sides zero.

Write \(S_i\) for the one-dimensional simple at vertex \(i\), and consider

For any representation \(V\), a map from \(P_i\) is determined uniquely by the image of its vector at vertex \(i\), giving \(\operatorname{Hom}(P_i,V)\simeq V_i\). For \(P_1\) the relation \(ba=0\) ensures the required zero at vertex 3; the other two cases follow directly from the indicated identity arrow. Vertex evaluation is exact, so all three \(P_i\) are projective. The exact resolution is

Here \(P_3\to P_2\) is identity at vertex 3; \(P_2\to P_1\) is identity at vertex 2 and zero elsewhere; \(P_1\to S_1\) is identity at vertex 1. Vertexwise inspection checks every kernel and image. Hom of this projective resolution to \(S_3\) has zero in cochain degrees zero and one and \(k\) in degree two. Therefore \(T^2(S_1,S_3)=k\). In \(\mathcal B\), every positive class factors through a split short row, so \(S^2(S_1,S_3)=0\). Comparison fails in degree two.

Take any nonzero \(\alpha:S_1\to S_3[2]\). Every monomorphism \(S_3\to J\) in \(\mathcal B\) splits by vector-space splitting; a retraction shows its map on ambient Ext has a left inverse and cannot kill \(\alpha\). Every epimorphism \(P\to S_1\) in \(\mathcal B\) also splits; its section gives a left inverse to pullback on Ext. Thus neither end permits effacement.

Let \(U=S_3[1]\), \(V=S_1\), and complete \(-\alpha[-1]:V[-1]\to U\) to a triangle and rotate once. We obtain

Its cohomology exact sequence gives \(H^{-1}(Z)=S_3\), \(H^0(Z)=S_1\), and zero otherwise. Hence \(Z\) belongs to \(D^b_{\mathcal B}(\mathcal A)\). The triangle is its natural truncation triangle up to the displayed endpoint identifications: the cohomology bounds and truncation adjunctions identify its first and third vertices with \(\tau_{\le-1}Z\) and \(\tau_{\ge0}Z\).

Every bounded complex in \(\mathcal B\) is isomorphic to the direct sum of its shifted cohomology objects. To see this without a formality assertion, choose complements to the boundaries in each cycle space, at each of the two vertices. Their inclusions give a chain map from the zero-differential cohomology complex which induces the identity on every cohomology space. It is a quasi-isomorphism. If \(Z\) were in the essential image, it would therefore be isomorphic to \(S_3[1]\oplus S_1\). That direct sum's natural truncation triangle has zero connecting arrow, since its projection has the inclusion of \(S_1\) as a section. Naturality would make \(\alpha=0\), a contradiction. Thus even a Serre subcategory can fail essential surjectivity of the bounded comparison.

Exercise 4 (advanced: an image complex need not split). In Exercise 2's finite-module category, put \(K=(R\xrightarrow{t}R)\) in degrees \(-1,0\). Compute its cohomology and \(\operatorname{Hom}_D(K,S[n])\) for every integer \(n\). Identify the sign of its cohomology connecting class relative to the normalized two-extension. Prove that the derived comparison equivalence does not make \(K\) a direct sum of its cohomology.

Solution. The kernel and image of multiplication by \(t\) are \((t)\). Identify \(S\) with \((t)\) by \(\bar a\mapsto at\). Then \(H^{-1}(K)=S\), \(H^0(K)=S\), and all other cohomology is zero. Both terms of \(K\) are projective, so the projective-Hom lemma computes its actual derived Hom. The Hom cochain complex to \(S\) has \(k\) in degrees zero and one and zero elsewhere. Its differential is precomposition with \(t\), possibly with the Hom sign, and is zero because \(tS=0\). It follows that

These computations hold intrinsically and ambiently by Exercise 2's comparison.

Let \(j:S[1]\to K\) be the cycle inclusion, with degree-minus-one component \(\bar a\mapsto at\). The cone \(C_j\) has terms \(S,R,R\) in degrees \(-2,-1,0\), and differentials \(j\) and multiplication by \(t\). Its augmentation \(q:C_j\to S[0]\) is reduction at degree zero, a quasi-isomorphism. The positive last cone projection \(f:C_j\to S[2]\) is identity at degree \(-2\). The natural truncation triangle therefore has connecting arrow

For the row \(e:0\to S\xrightarrow{j}R\xrightarrow{t}R\to S\to0\), the higher-extension normalization is \(\delta(e)=\varepsilon_2 Q(f)Q(q)^{-1}\) with \(\varepsilon_2=-1\). Thus \(\kappa=-\delta(e)\), including in characteristic two where the minus sign equals plus. This computes the sign from the actual cone rather than suppressing it.

The class is nonzero. Let \(P\to S\) be Exercise 2's periodic projective resolution. There is a chain map \(P\to C_j\) with identity components at degrees zero and minus one, reduction \(R\to S\) at degree minus two, and zero below. The degree-minus-two equation is \(j\circ\mathrm{reduction}=t\), and the degree-minus-three equation is \(\mathrm{reduction}\circ t=0\); the other equations are identities. Its composite with \(q\) is the augmentation \(P\to S\), so it represents the same roof class. The composite with \(f\) is reduction in degree minus two, the nonzero coordinate \(1\in k\) in the Hom calculation of (5.3). Hence \(\kappa\ne0\).

One can also see nonsplitting directly from Hom groups. If \(K\simeq S[1]\oplus S\), its Hom to \(S[2]\) would be \(\operatorname{Ext}^1(S,S)\oplus\operatorname{Ext}^2(S,S)\simeq k\oplus k\), whereas (5.7) makes that Hom group zero. Thus it cannot split. Unlike Exercise 3, this nonsplit complex already has both terms in \(\mathcal B\). Full comparison preserves its nonzero cohomology connecting class; comparison equivalence and splitting into cohomology are different properties.

Freely accessible references

• The Stacks Project, Serre subcategories, Tag 02MN, for the distinction between Serre and weak Serre hypotheses.
• The Stacks Project, Triangulated subcategories of the derived category, Tag 06UP, for the cohomology-defined subcategory and its comparison functor.
• The Stacks Project, Ext groups, Tag 06XP, for derived Ext, exact sequences and extension classes in an arbitrary abelian category.
• The linked course prerequisites contain the complete proofs of the extension, truncation, triangle and projective-Hom interfaces used here.

The exposition and worked solutions in this lesson are released under CC0-1.0. They were prepared with AI assistance. External references retain their own licences.