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Draft lesson.

Pointwise abelian structure and natural splittings

Written by GPT-6.1 Sol (OpenAI), October 2026. Section 1 written by Claude Opus 5.5 (Anthropic), October 2026. Original text: CC0.

A diagram has objects and transition maps. A kernel must respect both. Taking a kernel at each object gives the correct underlying objects; its universal property then supplies the transition maps and proves their compatibility. This explains why diagrams in an abelian category form another abelian category. Splittings behave differently: unrelated splittings at individual objects need not assemble to a splitting of diagrams.

Work in a fixed universe with small indexing categories and locally small coefficient categories. We use the complete kernel, cokernel and exactness tests in Testing exactness with doubled objects, Sections 1 and 3, the compatible splitting maps in Extending maps, splittings and exact functors, Section 2, and the cohomology descriptions in Exact complexes, signs and lifting covers, Section 2.

1. Assemble the pointwise universal objects

Throughout this section, \(I\) and \(J\) are small categories, \(\mathcal C\) is a locally small category, and \(D:J\to\operatorname{Fun}(I,\mathcal C)\) is a diagram. Each object \(j\) of \(J\) gives a functor \(D(j):I\to\mathcal C\). Each arrow \(b:j\to j'\) gives a natural transformation \(D(b):D(j)\Rightarrow D(j')\), with components \(D(b)_i\). Fixing an object \(i\) of \(I\) produces the evaluated diagram \(\mathrm{ev}_iD:J\to\mathcal C\), which sends \(j\) to \(D(j)(i)\) and \(b\) to \(D(b)_i\).

A cone on \(D\) consists of a functor \(M:I\to\mathcal C\) and natural transformations \(\mu_j:M\Rightarrow D(j)\) with \(D(b)\mu_j=\mu_{j'}\) for every arrow \(b:j\to j'\). Transformations compose componentwise, so for each object \(i\) the arrows \((\mu_j)_i:M(i)\to D(j)(i)\) form a cone on \(\mathrm{ev}_iD\) with vertex \(M(i)\). Call \((M,\mu)\) pointwise universal if each of these evaluated cones is a limit cone in \(\mathcal C\).

Limit cones can be cancelled. Let \((\lambda_j:V\to E(j))_j\) be a limit cone on a diagram \(E:J\to\mathcal C\), and let \(w,w':T\to V\) satisfy \(\lambda_jw=\lambda_jw'\) for every \(j\). The arrows \(\lambda_jw\) form a cone with vertex \(T\), and \(w\) and \(w'\) both factor it through the limit cone; by uniqueness, \(w=w'\). For empty \(J\), a limit is a terminal object, and the rule says that two arrows into it agree.

Pointwise limit theorem. Let \(I,J\) be small, \(\mathcal C\) locally small, and \(D:J\to\operatorname{Fun}(I,\mathcal C)\).

(a) Every pointwise universal cone on \(D\) is a limit cone of \(D\) in \(\operatorname{Fun}(I,\mathcal C)\).

(b) Suppose that each evaluated diagram \(\mathrm{ev}_iD\) has a limit, and choose a limit cone \((\pi^i_j:L(i)\to D(j)(i))_j\) for every object \(i\) of \(I\). Then \(i\mapsto L(i)\) extends to a functor \(L:I\to\mathcal C\) for which every family \(\pi_j=(\pi^i_j)_i\) is a natural transformation \(L\Rightarrow D(j)\). The resulting cone \((L,\pi)\) is pointwise universal, hence a limit of \(D\).

(c) Under the hypothesis of (b), a cone on \(D\) is a limit cone if and only if it is pointwise universal. In particular, each evaluation functor \(\mathrm{ev}_i:\operatorname{Fun}(I,\mathcal C)\to\mathcal C\) carries every limit cone of \(D\) to a limit cone of \(\mathrm{ev}_iD\).

Proof. (a) Let \((M,\mu)\) be pointwise universal and let \((X,\xi)\) be any cone on \(D\). For each object \(i\), the arrows \((\xi_j)_i\) form a cone on \(\mathrm{ev}_iD\) with vertex \(X(i)\), so there is exactly one arrow \(u_i:X(i)\to M(i)\) with \((\mu_j)_iu_i=(\xi_j)_i\) for all \(j\). Let \(a:i\to i'\) be an arrow of \(I\). The arrows \(M(a)u_i\) and \(u_{i'}X(a)\) both go from \(X(i)\) to \(M(i')\). Naturality of \(\mu_j\) and of \(\xi_j\) at \(a\) gives, for every \(j\), \[ \begin{gathered} (\mu_j)_{i'}M(a)u_i=D(j)(a)(\mu_j)_iu_i=D(j)(a)(\xi_j)_i,\\ (\mu_j)_{i'}u_{i'}X(a)=(\xi_j)_{i'}X(a)=D(j)(a)(\xi_j)_i. \end{gathered} \tag{1.1} \] The two right-hand sides coincide, so cancelling the limit cone at \(i'\) gives \(M(a)u_i=u_{i'}X(a)\). Thus \(u=(u_i)_i\) is a natural transformation \(X\Rightarrow M\), and \(\mu_ju=\xi_j\) because both sides have the same components. If a natural transformation \(u'\) also satisfies \(\mu_ju'=\xi_j\) for every \(j\), then each \(u'_i\) satisfies the equations that determine \(u_i\), so \(u'=u\). Hence every cone on \(D\) factors through \((M,\mu)\) in exactly one way.

(b) Let \(a:i\to i'\) be an arrow of \(I\). For \(b:j\to j'\) in \(J\), naturality of \(D(b)\) at \(a\) and the cone condition at \(i\) give \[ D(b)_{i'}D(j)(a)\pi^i_j=D(j')(a)D(b)_i\pi^i_j=D(j')(a)\pi^i_{j'}. \] Hence the arrows \(D(j)(a)\pi^i_j\) form a cone on \(\mathrm{ev}_{i'}D\) with vertex \(L(i)\), and there is exactly one arrow \(L(a):L(i)\to L(i')\) with \[ \pi^{i'}_jL(a)=D(j)(a)\pi^i_j\qquad\text{for every object }j\text{ of }J. \tag{1.2} \] Since \(D(j)(1_i)\) is an identity, \(1_{L(i)}\) satisfies (1.2) for \(a=1_i\); so \(L(1_i)=1_{L(i)}\). For a second arrow \(a':i'\to i''\), equation (1.2) for \(a'\) and then for \(a\), together with functoriality of \(D(j)\), gives \[ \pi^{i''}_jL(a')L(a)=D(j)(a')D(j)(a)\pi^i_j=D(j)(a'a)\pi^i_j, \] which is also \(\pi^{i''}_jL(a'a)\) by (1.2) for \(a'a\). Cancelling the limit cone at \(i''\) gives \(L(a')L(a)=L(a'a)\), so \(L\) is a functor. Equation (1.2) says precisely that each \(\pi_j\) is natural at \(a\); by cancellation, no other arrows \(L(a)\) would make all the \(\pi_j\) natural. The equation \(D(b)\pi_j=\pi_{j'}\) holds because it holds at every component, by the cone condition at that object. Therefore \((L,\pi)\) is a cone on \(D\). Its evaluation at \(i\) is the chosen limit cone, so it is pointwise universal, and (a) makes it a limit of \(D\).

(c) Part (a) gives one implication. Conversely, let \((M,\mu)\) be any limit cone of \(D\), and let \((L,\pi)\) be the limit cone of (b). Each of the two cones factors through the other, by natural transformations \(\varphi:M\Rightarrow L\) and \(\psi:L\Rightarrow M\) with \(\pi_j\varphi=\mu_j\) and \(\mu_j\psi=\pi_j\) for all \(j\). Then \(\pi_j\varphi\psi=\pi_j\), so \(\varphi\psi\) and \(1_L\) both factor \((L,\pi)\) through itself, and \(\varphi\psi=1_L\). Likewise \(\psi\varphi=1_M\). In components, \(\varphi_i\) is invertible with inverse \(\psi_i\), and \(\pi^i_j\varphi_i=(\mu_j)_i\). Now let \((t_j:T\to D(j)(i))_j\) be a cone on \(\mathrm{ev}_iD\). It factors through the chosen limit cone by a unique arrow \(w\), and then \((\mu_j)_i\psi_iw=\pi^i_jw=t_j\). If also \((\mu_j)_iw'=t_j\) for all \(j\), then \(\pi^i_j\varphi_iw'=t_j\), so \(\varphi_iw'=w\) and \(w'=\psi_iw\). Thus the evaluation of \((M,\mu)\) at every object \(i\) is a limit cone. \(\square\)

The proof uses limits in \(\mathcal C\) only through the existence and uniqueness of factorizations. It therefore covers an empty \(J\), where a cone is just its vertex and the evaluated limits are terminal objects. It also covers an empty \(I\): then \(\operatorname{Fun}(I,\mathcal C)\) has exactly one object and one arrow, and every cone on \(D\) is both pointwise universal and a limit cone.

Colimits. A cocone on \(D\) consists of a functor \(Q:I\to\mathcal C\) and natural transformations \(\kappa_j:D(j)\Rightarrow Q\) with \(\kappa_{j'}D(b)=\kappa_j\) for every \(b:j\to j'\). It is pointwise universal if each evaluated cocone \(((\kappa_j)_i)_j\) is a colimit cocone of \(\mathrm{ev}_iD\). Statements (a)–(c) remain true when cones, limits and limit cones are replaced by cocones, colimits and colimit cocones. For chosen colimit cocones \((\kappa^i_j:D(j)(i)\to Q(i))_j\), the functor \(Q\) sends \(a:i\to i'\) to the unique arrow with \(Q(a)\kappa^i_j=\kappa^{i'}_jD(j)(a)\) for every \(j\).

These statements follow from the theorem by reversing arrows. A functor \(F:I\to\mathcal C\) is also a functor \(F^{op}:I^{op}\to\mathcal C^{op}\), with the same values on objects and arrows. Let \(\alpha:F\Rightarrow G\) be a natural transformation and \(a:i\to i'\) an arrow of \(I\). Read in \(\mathcal C^{op}\), the equation \(G(a)\alpha_i=\alpha_{i'}F(a)\) is the naturality equation, at the arrow \(i'\to i\) of \(I^{op}\) given by \(a\), for the same components regarded as a transformation \(\alpha^{op}:G^{op}\Rightarrow F^{op}\). Composites of transformations are reversed, identities are kept, and both assignments can be inverted. This gives an isomorphism of categories \[ \operatorname{Fun}(I,\mathcal C)^{op} \cong\operatorname{Fun}(I^{op},\mathcal C^{op}), \tag{1.3} \] which commutes with evaluation at each object \(i\). Composing \(D^{op}:J^{op}\to\operatorname{Fun}(I,\mathcal C)^{op}\) with (1.3) gives a diagram \(\bar D:J^{op}\to\operatorname{Fun}(I^{op},\mathcal C^{op})\). A cocone \((Q,\kappa)\) on \(D\) is the same data as a cone \((Q^{op},\kappa^{op})\) on \(\bar D\). Read in \(\operatorname{Fun}(I,\mathcal C)^{op}\), a colimit cocone of \(D\) is a limit cone of \(D^{op}\). An isomorphism of categories is bijective on objects and on Hom sets and respects composition, so it preserves and reflects limit cones. Hence \((Q,\kappa)\) is a colimit cocone of \(D\) exactly when \((Q^{op},\kappa^{op})\) is a limit cone of \(\bar D\). Moreover \(\mathrm{ev}_i\bar D=(\mathrm{ev}_iD)^{op}\), and a colimit cocone in \(\mathcal C\) is the same as a limit cone in \(\mathcal C^{op}\), so the pointwise universal cocones on \(D\) are the pointwise universal cones on \(\bar D\). Since \(I^{op}\) and \(J^{op}\) are small and \(\mathcal C^{op}\) is locally small, the theorem applies to \(\bar D\), and its three statements become the cocone statements. In particular, equation (1.2) for \(\bar D\), read in \(\mathcal C\), is the stated equation for \(Q(a)\).

The hypotheses of (b) and (c) concern only the evaluated diagrams of \(D\); \(\mathcal C\) need not have any other limits. If some \(\mathrm{ev}_iD\) has no limit, then no cone on \(D\) is pointwise universal, and the theorem makes no assertion about limits of \(D\) that \(\operatorname{Fun}(I,\mathcal C)\) may nevertheless have.

2. Why all diagrams form an abelian category

Let \(\mathcal A\) be abelian and \(I\) small. The objects of \(\mathcal A^I\) are all functors \(I\to\mathcal A\), and its maps are natural transformations. No additivity assumption is imposed on \(I\).

The natural transformations from \(F\) to \(G\) form a subgroup of \[ \prod_{i\in\operatorname{Ob}I} \operatorname{Hom}_{\mathcal A}(F_i,G_i). \tag{2.1} \] Indeed, the equations \(G(a)\alpha_i=\alpha_{i'}F(a)\) are preserved under addition, negatives and zero because composition in \(\mathcal A\) is bilinear. Composition of transformations is bilinear componentwise. The constant zero diagram is a zero object. Finite biproducts are formed objectwise, with transition maps \(F(a)\oplus G(a)\); their injections and projections are natural and satisfy the biproduct identities. Alternatively, their product and coproduct universal properties are the two applications of §1. This gives the additive structure, including the empty biproduct.

Now let \(\alpha:F\to G\). Choose \(k_i:K_i\to F_i\) to be its component kernel. For \(a:i\to i'\), naturality gives \[ \alpha_{i'}F(a)k_i=G(a)\alpha_i k_i=0. \tag{2.2} \] The kernel property therefore defines a unique \(K(a):K_i\to K_{i'}\) with \(k_{i'}K(a)=F(a)k_i\). Testing against the monomorphism \(k\) proves the identity and composition equations. Thus \(K\) is a functor and \(k:K\to F\) natural.

For a natural \(t:T\to F\) with \(\alpha t=0\), each \(t_i\) factors uniquely as \(k_iv_i\). For \(a:i\to i'\), \[ \begin{aligned} k_{i'}K(a)v_i &=F(a)t_i\\ &=t_{i'}T(a)\\ &=k_{i'}v_{i'}T(a). \end{aligned} \tag{2.3} \] Cancel \(k_{i'}\) to obtain naturality of \(v\). Its componentwise uniqueness gives uniqueness as a natural transformation. Hence \(k\) is the kernel in \(\mathcal A^I\); this proves its full universal property.

Choose \(q_i:G_i\to Q_i\) as component cokernels. The equality \(q_{i'}G(a)\alpha_i=0\) defines a unique \(Q(a)\) with \[ Q(a)q_i=q_{i'}G(a). \tag{2.4} \] Cancel the epimorphisms \(q_i\) to prove identities and composition. A natural \(t:G\to T\) with \(t\alpha=0\) factors uniquely componentwise as \(v_iq_i\). Comparing \(T(a)v_iq_i=t_{i'}G(a)=v_{i'}Q(a)q_i\) and canceling \(q_i\) proves naturality. Thus \(q\) is the cokernel in the diagram category.

Consequently the functors \(\operatorname{Coim}\alpha=\operatorname{coker}(\ker\alpha)\) and \(\operatorname{Im}\alpha=\ker(\operatorname{coker}\alpha)\) are computed componentwise, and their canonical comparison has components \[ u_i:\operatorname{Coim}\alpha_i \longrightarrow\operatorname{Im}\alpha_i. \tag{2.5} \] Its naturality follows from the kernel and cokernel constructions just proved. Every \(u_i\) is invertible because \(\mathcal A\) is abelian. The inverses are natural: from \(\operatorname{Im}(a)u_i=u_{i'}\operatorname{Coim}(a)\), multiply by \(u_{i'}^{-1}\) on the left and \(u_i^{-1}\) on the right. Hence the comparison is an isomorphism of functors. This proves that \(\mathcal A^I\) is abelian.

Each evaluation \(\mathrm{ev}_i:\mathcal A^I\to\mathcal A\) is additive and preserves kernels and cokernels, so is exact by the retained complete tests. A complex of diagrams is exact at a position exactly when every evaluated complex is exact there: its kernel, image and their comparison are pointwise. Monomorphisms, epimorphisms and isomorphisms are likewise detected by all evaluations. For monomorphisms and epimorphisms use the zero-kernel and zero-cokernel criteria; for isomorphisms use the natural inverses above. These assertions include empty \(I\): its functor category has one object and one map and is the zero abelian category.

3. Products of categories and reversal of arrows

For a small family \((\mathcal A_s)_{s\in S}\) of abelian categories, an object of \(\prod_s\mathcal A_s\) is a tuple \((X_s)_s\); its maps are tuples of component maps. Hom groups, bilinear composition, zero objects and finite biproducts are obtained coordinatewise. The universal property of a kernel, or a cokernel, reduces to the separate universal properties in each coordinate: a tuple of maps annihilated by the given morphism has exactly one tuple of factors. The canonical coimage–image comparison is the tuple of the component comparisons, and its inverse is the tuple of their inverses. Therefore the product category is abelian, even when \(S\) is infinite. For empty \(S\), it is the category with one object and one map, again the zero abelian category.

This product of categories does not require an infinite product of objects inside any \(\mathcal A_s\). Exercise 3 separates the two constructions.

The opposite \(\mathcal A^{op}\) is also abelian. Its Hom groups have the same addition, composition remains bilinear after reversal, and the zero object remains zero. A biproduct has its injections and projections exchanged. The defining kernel and cokernel universal properties exchange under reversal. For \(f:X\to Y\), this gives \[ \begin{gathered} \ker(f^{op})=(\operatorname{coker}f)^{op},\\ \operatorname{coker}(f^{op})=(\ker f)^{op},\\ \operatorname{Coim}(f^{op})=(\operatorname{Im}f)^{op},\\ \operatorname{Im}(f^{op})=(\operatorname{Coim}f)^{op}. \end{gathered} \tag{3.1} \] Here equality means the canonical identifications given by those universal properties. The canonical coimage–image comparison for \(f^{op}\) is the opposite of the comparison for \(f\), with these identifications. It is invertible. This checks every abelian axiom and every variance in (3.1).

4. Pointwise splittings need an extra compatibility

A short exact sequence of diagrams is short exact at every object. A splitting is a natural section, or equivalently a natural retraction with the compatible maps of the retained splitting theorem. It therefore gives a splitting at every object. The converse requires the chosen component sections to commute with all diagram arrows.

For example, a one-object category associated to a group \(H\) gives \(H\)-actions in the coefficient category. A section of underlying vector spaces is a splitting in \(\operatorname{Fun}(H,\operatorname{Vect}_k)\) only if it commutes with every group operator. Exercise 2 exhibits the obstruction and proves the precise averaging repair when \(|H|\) is invertible in \(k\).

5. Four graded exercises with full solutions

Exercise 1 (introductory: an idempotent diagram). Let \(I\) have one object and two endomorphisms \(1,e\), with \(e^2=e\). Describe \(\operatorname{Fun}(I,\operatorname{Vect}_k)\), including its maps, and prove it is equivalent as an abelian category to \(\operatorname{Vect}_k\times\operatorname{Vect}_k\).

Solution. A functor is a vector space \(V\) with an idempotent \(p:V\to V\). Each vector decomposes as \(v=pv+(v-pv)\), where the terms lie in \(\operatorname{Im}p\) and \(\ker p\), respectively. Their intersection is zero because \(p\) is the identity on its image and zero on its kernel. Thus \(V=\operatorname{Im}p\oplus\ker p\). A natural map \((V,p)\to(W,q)\) is a linear map \(a\) with \(ap=qa\); this sends each summand to the corresponding summand. Conversely a pair of maps on those summands has a unique direct-sum map and satisfies that equation. This defines a fully faithful functor to the product category. The pair \((U,W)\) is obtained from \(U\oplus W\) with projector onto \(U\), proving essential surjectivity. The decompositions commute with all natural maps, so provide the natural inverse equivalences. Kernels and cokernels are those of the two summand maps, either by §2 or directly by this block description. Hence the equivalence is exact.

Exercise 2 (intermediate: a component section may fail). Suppose \(k\) has characteristic \(p>0\) and \(H=C_p\). On \(V=ke_1\oplus ke_2\), let a generator act by \(T=1+N\), where \(Ne_2=e_1\) and \(Ne_1=0\). Show that the short exact sequence of \(H\)-representations \[ 0\to ke_1\to V\to V/ke_1\to0 \tag{5.1} \] splits after evaluation but does not split naturally. Then prove that every underlying split short exact sequence of representations of a finite group \(H\) splits naturally if \(|H|\) is invertible in \(k\).

Solution. Since \(N^2=0\), \((1+N)^p=1+pN=1\); thus this is an \(H\)-action, also for \(p=2\). Both end terms have trivial action. The underlying vector-space section sends the quotient basis to \(e_2\). Every possible section sends it to \(e_2+ae_1\), but \[ \begin{aligned} T(e_2+ae_1)&=e_2+(a+1)e_1\\ &\ne e_2+ae_1. \end{aligned} \tag{5.2} \] There is no invariant lift and hence no natural section. A natural retraction would give a natural section by the retained compatible-splitting theorem, so none exists either.

For the positive assertion, write the quotient as \(q:V\to W\) and choose an underlying section \(s\). Use the two specified actions to define \[ \bar s=\frac1{|H|}\sum_{h\in H}h_V s h_W^{-1}. \tag{5.3} \] Equivariance of \(q\) gives \(q h_Vs h_W^{-1}=h_Wqsh_W^{-1}=1_W\), so \(q\bar s=1_W\). For \(g\in H\), conjugating each summand sends its index \(h\) to \(gh\). This permutes the finite sum, giving \(g_V\bar s g_W^{-1}=\bar s\). Thus \(\bar s\) is natural and is a section. The two hypotheses are used exactly in forming the finite sum and dividing by \(|H|\). No finiteness of the vector-space dimensions is needed.

Exercise 3 (intermediate: two meanings of product). Let \(\mathcal A_n=\operatorname{Vect}^{fin}_k\) for each \(n\ge0\). Explain why \(\prod_n\mathcal A_n\) is abelian and contains the object \((k)_n\), although a countable product of copies of \(k\) does not exist in \(\operatorname{Vect}^{fin}_k\).

Solution. Each finite-dimensional vector-space category is abelian. Its finite direct sums and zero space give the additive structure. A kernel is a subspace of a finite-dimensional domain, hence finite-dimensional; the images of a finite spanning family of the target span any quotient, so a cokernel is finite-dimensional too. For a linear map \(f\), the comparison \(V/\ker f\to\operatorname{Im}f\) sends the class of \(v\) to \(f(v)\): it is well-defined, surjective by the definition of image, and injective by the definition of kernel. Thus it is invertible. Section 3 applies to the small family; the tuple \((k)_n\) is an object by its definition.

If an internal product \(P\) existed in \(\operatorname{Vect}^{fin}_k\), its projections would give a bijection \[ \operatorname{Hom}(k,P)\longrightarrow \prod_{n\ge0}\operatorname{Hom}(k,k). \tag{5.4} \] The map is linear because each projection is linear, so its inverse is linear as well. The right side has the linearly independent coordinate vectors \(\delta_n\): each finite linear relation is zero only when all coefficients vanish, as its coordinates show. The left side is isomorphic to \(P\), hence finite-dimensional. This contradiction applies to finite and infinite fields alike. Coordinatewise structure in a product of categories and an internal product in one coefficient category have different universal problems.

Exercise 4 (hard: exact evaluation and cohomology). Let \(I\) be small and let \(F^\bullet\) be a complex in \(\mathcal A^I\). Prove that \[ \mathrm{ev}_i H^n(F^\bullet) \simeq H^n(\mathrm{ev}_i F^\bullet) \tag{5.5} \] canonically and compatibly with arrows of \(I\) and maps of complexes. Show by an explicit degreewise short exact sequence of complexes of vector spaces that \(H^0\) itself need not be exact.

Solution. The cohomology construction retained in Exact complexes, signs and lifting covers, §2, uses the kernel of \(d^n\) modulo the image of \(d^{n-1}\), with its specified universal comparison maps. Section 2 here proves that these kernels, images and the final cokernel in \(\mathcal A^I\) are all pointwise. Evaluation therefore identifies exactly the same objects and structure maps with those of the evaluated complex, proving (5.5). The transitions of \(F^\bullet\) and a map of complexes give commuting squares for the differentials. The uniqueness in the retained kernel and quotient constructions identifies their induced maps too; hence these comparisons are natural in \(i\) and in the complex, including empty \(I\).

For the counterexample use degrees \(0,1\). Let \(B^\bullet\) be \(k\xrightarrow{1}k\), let \(A^\bullet\) be \(0\to k\), and let \(C^\bullet\) be \(k\to0\). Include \(A^\bullet\) in \(B^\bullet\) by zero in degree 0 and identity in degree 1. Quotient to \(C^\bullet\) by identity in degree 0 and zero in degree 1. These are maps of complexes and \[ 0\longrightarrow A^\bullet\longrightarrow B^\bullet \longrightarrow C^\bullet\longrightarrow0 \tag{5.6} \] is short exact in each degree. But \(H^0(A)=0\), \(H^0(B)=0\), and \(H^0(C)=k\). The resulting sequence ends in \(0\to k\to0\), which is not exact at \(k\). Exactness of evaluation does not turn the cohomology functor into an exact functor on complexes.

6. References