Pointwise abelian structure and natural splittings

Adapted and extended by GPT-6.1 Sol (OpenAI), October 2026; self-checked by the writing AI at Ultra. This lesson, including the adaptation and added teaching, is licensed under CC BY-NC-SA 4.0.

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

This section adapts Tom Leinster's Basic Category Theory, Theorem 6.2.5 and its dual, arXiv version 2, 26 August 2025, pp. 148–150. Source copyright © Tom Leinster 2014, 2016. The AI changes notation and expands the functor and naturality checks. The source and this adaptation retain CC BY-NC-SA 4.0, including its warranty disclaimer. No endorsement is implied.

Pointwise limit theorem. Let \(I,J\) be small, \(\mathcal C\) locally small, and \(D:J\to\operatorname{Fun}(I,\mathcal C)\). Chosen limits \(L_i=\lim_jD_j(i)\), with projections \(p_{j,i}\), assemble to a limit of \(D\). Every cone that evaluates to these limits is universal.

Proof. Write \(D_b:D_j\to D_{j'}\) for an arrow \(b:j\to j'\) of the diagram. For \(a:i\to i'\), naturality gives \(D_{b,i'}D_j(a)p_{j,i}=D_{j'}(a)D_{b,i}p_{j,i}=D_{j'}(a)p_{j',i}\). Thus \(D_j(a)p_{j,i}\) is a cone, and determines a unique \(L(a)\) satisfying \[ p_{j,i'}L(a)=D_j(a)p_{j,i}. \tag{1.1} \] For an identity arrow, both \(L(1_i)\) and \(1_{L_i}\) have projection \(p_{j,i}\) for every \(j\), so uniqueness makes them equal. For composable \(a:i\to i'\) and \(a':i'\to i''\), the projection of \(L(a')L(a)\) is \(D_j(a')D_j(a)p_{j,i}=D_j(a'a)p_{j,i}\), which is also the projection of \(L(a'a)\). Uniqueness proves the composition equation. Equation (1.1) makes each \(p_j\) natural. The pointwise equations \(D_{b,i}p_{j,i}=p_{j',i}\) say precisely that \((p_j)_j\) is a cone of natural transformations. These arguments include empty \(J\): its limit objects are terminal, and uniqueness still supplies all the equations.

For another cone \(q_j:X\to D_j\), let \(v_i\) be its unique pointwise factor. Naturality follows by testing against every projection: \[ \begin{aligned} p_{j,i'}L(a)v_i &=D_j(a)q_{j,i}\\ &=q_{j,i'}X(a)\\ &=p_{j,i'}v_{i'}X(a). \end{aligned} \tag{1.2} \] Thus \(v\) is natural and \(p_jv=q_j\). Componentwise uniqueness proves uniqueness of \(v\), and also the final assertion. \(\square\)

For colimits apply this argument to \(J^{op}\) and \(\mathcal C^{op}\), using \[ \operatorname{Fun}(I,\mathcal C)^{op} \simeq\operatorname{Fun}(I^{op},\mathcal C^{op}). \tag{1.3} \] This equivalence reverses natural transformations and both source and target arrows. Hence chosen pointwise colimits are colimits of diagrams of functors too.

The hypothesis concerns the evaluated diagrams actually being used. It does not assert that an arbitrary coefficient category has all small limits, or that every existing limit in a functor category must be pointwise when the requisite coefficient limits are absent.

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 and adaptation notice