# A characterization of AB5 and AB4 categories. Does this result show up in the literature?

EDIT NOTICE (10/6/24): Former Corollary 4 is now implication 1$$\Rightarrow$$5 of Proposition 1 (I realized that 5$$\Rightarrow$$1 is also true).

I am interested on finding the statement or the proof in the literature of a result about AB5 and AB4 categories. Before stating it, I'll give some background.

Let $$\mathcal{A}$$ be an AB5 (resp., AB4) category. Spelled out, this means that

1. $$\mathcal{A}$$ is abelian,
2. $$\mathcal{A}$$ is cocomplete. Equivalently, $$\mathcal{A}$$ has all coproducts (abelian categories have all coequalizers), and
3. taking filtered colimits (resp., taking arbitrary coproducts) in $$\mathcal{A}$$ is exact.

We give a precise description of condition 3. Let $$\mathcal{I}$$ be any category. If $$\mathcal{A}$$ is a category that is $$\mathsf{P}\in\{\text{preadditive, additive, preabelian, abelian}\}$$, then the category $$\operatorname{Fun}(\mathcal{I},\mathcal{A})$$ is also $$\mathsf{P}$$. Suppose now $$\mathcal{I}$$ is small. If $$\mathcal{A}$$ is an AB3 category (a cocomplete abelian category), then the functor $$\tag{1}\label{colim_fun} \operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$ is right exact, since colimits commute with colimits. An AB3 category is AB5 (resp., AB4) precisely when for all $$\mathcal{I}$$ small filtered categories (resp., for all $$\mathcal{I}$$ small discrete), the functor \eqref{colim_fun} is exact (equivalently, left exact; equivalently, it preserves monomorphisms).

Given an additive category $$\mathcal{A}$$, we denote $$\operatorname{Ch}(\mathcal{A})$$ to the category of cochain complexes with terms in $$\mathcal{A}$$. Recall that if $$\mathcal{A}$$ is (pre)abelian, then so is $$\operatorname{Ch}(\mathcal{A})$$. Remember as well that, by definition, a quasi-isomorphism is a morphism in $$\operatorname{Ch}(\mathcal{A})$$ that induces isomorphisms on all cohomology objects. The result I'm interested in is the following characterization of AB5 and AB4 categories.

Proposition 1. Let $$\mathcal{A}$$ be an AB3 category. The following are equivalent:

1. $$\mathcal{A}$$ is AB5 (resp., AB4),
2. $$\operatorname{Ch}(\mathcal{A})$$ is AB5 (resp., AB4),
3. for all filtered (resp., discrete) small categories $$\mathcal{I}$$, the functor $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ sends diagrams of acyclic complexes (i.e., a diagram $$F:\mathcal{I}\to\operatorname{Ch}(\mathcal{A})$$ such that $$F(i)$$ is acyclic for all $$i\in\mathcal{I}$$) to an acyclic complex. In other words, filtered colimits (resp., arbitrary coproducts) of acyclic complexes are acyclic,
4. for all filtered (resp., discrete) small categories $$\mathcal{I}$$, the functor $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ sends each natural transformation whose components are quasi-isomorphisms to a quasi-isomorphism, and
5. cohomology commutes with filtered colimits (resp., with arbitrary coproducts), i.e., for all filtered (resp., discrete) small categories $$\mathcal{I}$$ the square of functors $$\require{AMScd} \begin{CD} \operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A})) @>{H_*^n}>> \operatorname{Fun}(\mathcal{I},\mathcal{A})\\ @V{\text{colim}}VV @VV{\text{colim}}V \\ \operatorname{Ch}(\mathcal{A}) @>>H^n> \mathcal{A} \end{CD}$$ commutes (up to isomorphism of functors, say).

In condition 5, $$H^n$$ is the cohomology functor, and if $$\mathsf{C},\mathsf{D},\mathsf{E}$$ are categories, then a functor $$F:\mathsf{D}\to\mathsf{E}$$ induces a functor $$F_*:\operatorname{Fun}(\mathsf{C},\mathsf{D})\to\operatorname{Fun}(\mathsf{C},\mathsf{E})$$. In conditions 3 and 4 the colimit functor on makes sense: in general one has the result “if $$\mathcal{A}$$ is a cocomplete additive category, then so is $$\operatorname{Ch}(\mathcal{A})$$,” since colimits in $$\operatorname{Ch}(\mathcal{A})$$ may be computed degreewise.

The few stuff I've read on the literature about AB5 and AB4 categories don't talk about this characterization (nLab, wikipedia, Stacks Project and posts on MSE, although perhaps I haven't looked well enough on the SP; gosh, the SP is huge). I was wondering if this result has been pointed out before by somebody else or any book or reference. And in that case, if there is there a proof of it.

As an answer to this post, I will write a proof of Proposition 1 I've come up with myself. Apart from answering if Proposition 1 is found somewhere in the literature, other possible answers I would be grateful to see are other proof ideas for them, different than mine. I think the proof I came up with gets sometimes too technical.

• The other characterization result I know for the AB5 axiom is Theorem 1.1 from here, but it is different from Proposition 1. Commented Jun 10 at 14:51

Proposition 1 follows from the more general result. We obtain Proposition 1 from it by quantifying $$\mathcal{I}$$ over all filtered small categories or small discrete categories.

Proposition 1'. Let $$\mathcal{A}$$ be an AB3 category. Let $$\mathcal{I}$$ be a small category. The following are equivalent:

1. $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$ is exact,
2. $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ is exact,
3. $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ sends diagrams of acyclic complexes (i.e., a diagram $$F:\mathcal{I}\to\operatorname{Ch}(\mathcal{A})$$ such that $$F(i)$$ is acyclic for all $$i\in\mathcal{I}$$) to an acyclic complex. In other words, $$\mathcal{I}$$-shaped colimits of acyclic complexes are acyclic,
4. $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ sends each natural transformation whose components are quasi-isomorphisms to a quasi-isomorphism, and
5. cohomology commutes with $$\mathcal{I}$$-shaped colimits, i.e., the square of functors $$\require{AMScd} \begin{CD} \operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A})) @>{H_*^n}>> \operatorname{Fun}(\mathcal{I},\mathcal{A})\\ @V{\text{colim}}VV @VV{\text{colim}}V \\ \operatorname{Ch}(\mathcal{A}) @>>H^n> \mathcal{A} \end{CD}$$ commutes (up to isomorphism of functors, say).

We will deduce Proposition 1' from the upcoming two lemmas.

Lemma 2. Let $$F:\mathcal{A}\to\mathcal{B}$$ be an additive functor between preabelian categories. The following are equivalent:

1. $$F:\mathcal{A}\to\mathcal{B}$$ is exact,
2. $$\operatorname{Ch}(F):\operatorname{Ch}(\mathcal{A})\to\operatorname{Ch}(\mathcal{B})$$ is exact,
3. $$\operatorname{Ch}(F)$$ preserves acyclic complexes, and
4. $$\operatorname{Ch}(F)$$ preserves quasi-isomorphisms.

Proof. (1$$\Rightarrow$$2). Follows from the fact that exactness in the category of cochain complexes is equivalent to termwise exactness.

(2$$\Rightarrow$$1). Follows from the fact that $$\require{AMScd} \begin{CD} \mathcal{A} @>{F}>> \mathcal{B}\\ @V{[0]}VV @VV{[0]}V \\ \operatorname{Ch}(\mathcal{A}) @>>{\operatorname{Ch}(F)}> \operatorname{Ch}(\mathcal{B}) \end{CD}$$ is a commutative diagram of functors and $$[0]$$ is an exact full embedding of an abelian category into the associated category of cochain complexes.

(1$$\Rightarrow$$3). Easy.

(3$$\Rightarrow$$4). The functor $$\operatorname{Ch}(F)$$ preserves cones for $$F$$ is additive. Since a morphism in the category of cochain complexes is a quasi-isomorphism if and only if its cone is acyclic, the result follows.

(4$$\Rightarrow$$1). If $$0\to x\to y\to z\to 0$$ is a short exact sequence in $$\mathcal{A}$$, then $$\require{AMScd} \begin{CD} \cdots@>>>0@>>>x@>>>y@>>>z@>>>0@>>>\cdots\\ @.@VVV@VVV@VVV@VVV@VVV@.\\ \cdots@>>>0@>>>0@>>>0@>>>0@>>>0@>>>\cdots \end{CD}$$ is a quasi-isomorphism. Thus $$\require{AMScd} \begin{CD} \cdots@>>>0@>>>Fx@>>>Fy@>>>Fz@>>>0@>>>\cdots\\ @.@VVV@VVV@VVV@VVV@VVV@.\\ \cdots@>>>0@>>>0@>>>0@>>>0@>>>0@>>>\cdots \end{CD}$$ is a quasi-isomorphism and therefore $$0\to Fx\to Fy\to Fz\to 0$$ is short exact. $$\square$$

Lemma 3. Let $$\mathcal{I}$$ be any category and let $$\mathcal{A}$$ be an additive category. There is an exact natural isomorphism of abelian categories $$\tag{2}\label{ch_goes_inside} \operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A}))\cong\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A})),$$ natural in $$\mathcal{I}$$ and in $$\mathcal{A}$$. Moreover this isomorphism induces the following one-to-one correspondences: \begin{align*} \begin{Bmatrix} \text{Acyclic complexes}\\ \text{in }\operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A})) \end{Bmatrix} &\longleftrightarrow \begin{Bmatrix} \text{Functors }F:\mathcal{I}\to\operatorname{Ch}(\mathcal{A})\text{ whose image consists of acyclic}\\\text{complexes, i.e., } F(i) \text{ is an acyclic complex for all }i\in\mathcal{I} \end{Bmatrix} \\ \\ \begin{Bmatrix} \text{quasi-isomorphisms}\\\text{in }\operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A})) \end{Bmatrix} &\longleftrightarrow \begin{Bmatrix} \text{Natural transformations }\eta:F\Rightarrow G\\\text{ between functors } F,G:\mathcal{I}\rightrightarrows\operatorname{Ch}(\mathcal{A}) \text{ whose }\\\text{components are qis, i.e., }\eta_i \text{ is a qis }\forall i\in\mathcal{I} \end{Bmatrix} \end{align*}

Slogan for the isomorphism \eqref{ch_goes_inside}: “A cochain complex of diagrams is the same thing as a diagram of cochain complexes.”

Proof. For me, differentials will go up. We can realize the category of cochain complexes as a functor category. Let's see how: Define the category $$\mathsf{Diff}$$ with objects $$\operatorname{Ob}(\mathsf{Diff})=\mathbb{Z}$$ and with hom-sets

• $$\operatorname{Hom}_\mathsf{Diff}(n,n+1)=\mathbb{Z}$$,
• $$\operatorname{Hom}_\mathsf{Diff}(n,n)=\mathbb{Z}$$ (the generator is $$1_n$$), and
• the rest of hom-sets are zero.

Denote $$\mathsf{ADD}$$ to the category of locally small additive categories with additive functors. There is a category isomorphism $$\tag{3}\label{Ch_as_fun_cat} \operatorname{Ch}(\mathcal{A})\cong\operatorname{Fun}_{\mathsf{ADD}}(\mathsf{Diff},\mathcal{A}),$$ where with the subscript $$\mathsf{ADD}$$ we mean additive functors. In other words, a cochain complex with terms in $$\mathcal{A}$$ is the same thing as an additive functor $$\mathsf{Diff}\to\mathcal{A}$$, and a cochain map is the same thing as a natural transformation between two of these functors. For more details regarding the isomorphism \eqref{Ch_as_fun_cat}, see this answer.

Therefore, we deduce \begin{align*} \operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A})) &\cong\operatorname{Fun}_\mathsf{ADD}(\mathsf{Diff},\operatorname{Fun}(\mathcal{I},\mathcal{A}))\\ &\cong\operatorname{Fun}_{\mathsf{ADD}\times\mathsf{CAT}}(\mathsf{Diff}\times\mathcal{I},\mathcal{A}) \end{align*} where we are uncurrying, and where with the subscript $$\mathsf{ADD}\times\mathsf{CAT}$$ ($$\mathsf{CAT}$$ denotes the category of locally small categories) we just mean that the functors $$F:\mathsf{Diff}\times\mathcal{I}\to\mathcal{A}$$ are additive on the first component, $$F(f+g,h)=F(f,h)+F(g,h)$$. Continuing: \begin{align*} \operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A})) &\cong\operatorname{Fun}_{\mathsf{CAT}\times\mathsf{ADD}}(\mathcal{I}\times\mathsf{Diff},\mathcal{A})\\ &\cong\operatorname{Fun}(\mathcal{I},\operatorname{Fun}_\mathsf{ADD}(\mathsf{Diff},\mathcal{A}))\\ &\cong\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A})). \end{align*}

Let me add here a commutative cube that contains the significance of the isomorphism \eqref{ch_goes_inside}.

Where $$\alpha:i\to j$$ is a morphism in $$\mathcal{I}$$. This commutative cube can be interpreted in two ways (I will be thinking of the face with the $$C$$'s as the front face and the face with the $$D$$'s as the back face):

• From the perspective of the LHS of \eqref{ch_goes_inside}: the front face of this cube interprets as complex $$C_{(-)}^\bullet$$ of functors $$\mathcal{I}\to\mathcal{A}$$. That is, on degree $$n$$, we have a functor $$C_{(-)}^n:\mathcal{I}\to\mathcal{A}$$, and the differential $$d_{C_{(-)}^\bullet}$$, on degree $$n$$, is a natural transformation $$d_{C_{(-)}^\bullet}^n:C_{(-)}^n\to C_{(-)}^{n+1}$$ whose component at $$i\in\mathcal{I}$$ equals $$d_{C_i^n}$$. Naturality of $$d_{C_{(-)}^\bullet}^n$$ amounts to commutativity of this front face. Similarly, the back face of the cube defines a complex of functors $$D^\bullet_{(-)}:\mathcal{I}\to\mathcal{A}$$. The edges that point to the back define a cochain map between the cochain complexes of functors $$\eta^\bullet:C_{(-)}^\bullet\to D_{(-)}^\bullet$$. The cochain map on degree $$n$$ is given by a natural transformation $$\eta^n:C_{(-)}^n\to D_{(-)}^n$$ whose component at $$i\in\mathcal{I}$$ equals $$\eta^n_i$$. Naturality of $$\eta^n$$ and of $$\eta^{n+1}$$ amounts to commutativity of the left and right sides of the cube, respectively. The fact that $$\eta^\bullet$$ defines a cochain map means that $$d^n_{D_{(-)}^\bullet}\circ\eta^n=\eta^{n+1}\circ d^n_{C_{(-)}^\bullet}$$, i.e., the top and bottom faces commute.

• From the perspective of the RHS of \eqref{ch_goes_inside}: The front face defines a functor $$C^\bullet_{(-)}:\mathcal{I}\to\operatorname{Ch}(\mathcal{A})$$, that is, for each $$i\in\mathcal{I}$$, we get a cochain complex $$(C^\bullet_i,d_{C^\bullet_i})$$, and for each morphism $$\alpha:i\to j\in\mathcal{I}$$, we get an induced morphism of cochain complexes $$C^\bullet_\alpha:C^\bullet_i\to C^\bullet_j$$. Commutativity of this front face means that $$C^\bullet_\alpha$$ is indeed a cochain map. Similarly, the back face defines a functor $$D^\bullet_{(-)}:\mathcal{I}\to\operatorname{Ch}(\mathcal{A})$$. On the other hand, the morphisms that go to the back represent a natural transformation of functors $$\eta:C^\bullet_{(-)}\to D^\bullet_{(-)}$$ whose component at $$i\in\mathcal{I}$$ equals the cochain map $$\eta_i^\bullet:C^\bullet_{i}\to D^\bullet_{i}$$. Commutativity of the top and bottom faces means that $$\eta_i^\bullet$$ is indeed a cochain map. Naturality of $$\eta$$ means that for every morphism $$\alpha:i\to j\in\mathcal{I}$$, we have $$D_\alpha^\bullet\circ\eta^\bullet_i=\eta^\bullet_j\circ C_\alpha^\bullet$$, i.e., that the left and right squares commute.

Using this analysis of the isomorphism \eqref{ch_goes_inside} as this double interpretation of the commutative cube, one can see that the claimed one-to-one correspondences hold.

It remains to argue why the isomorphism \eqref{ch_goes_inside} is exact. Exactness of a sequence of cochain complexes living on the LHS of this isomorphism amounts to degreewise exactness, i.e., it amounts to exactness of a sequence of functors $$\mathcal{I}\to\mathcal{A}$$; but this in turn amounts to exactness objectwise. Similarly, exactness of a sequence of functors $$\mathcal{I}\to\operatorname{Ch}(\mathcal{A})$$ living on the RHS of \eqref{ch_goes_inside} amounts to objectwise exactness of a sequence of cochain complexes, which in turn amounts to exactness degreewise. In other words, with the notations of the commutative cube, a sequence $$C_{(-)}^\bullet\to D_{(-)}^\bullet\to E_{(-)}^\bullet$$ living on either side of \eqref{ch_goes_inside} is exact if and only if the sequences $$C_i^n\to D_i^n\to E_i^n$$ are exact for all $$i\in\mathcal{I}$$ and $$n\in\mathbb{Z}$$. $$\square$$

Proof of Proposition 1'. First we show 1$$\Leftrightarrow$$2$$\Leftrightarrow$$3$$\Leftrightarrow$$4. If $$\mathcal{A}$$ is an AB3 category, then for a small category $$\mathcal{I}$$ we can consider the functor $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$. By Lemma 2, the following conditions are equivalent:

1. The functor $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$ is exact,
2. The functor $$\operatorname{Ch}(\operatorname{colim}):\operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ is exact,
3. $$\operatorname{Ch}(\operatorname{colim})$$ preserves acyclic complexes, and
4. $$\operatorname{Ch}(\operatorname{colim})$$ preserves quasi-isomorphisms.

We want to transform the last three conditions of this list to the last three conditions of the statement of Proposition 1'. For this, it suffices to show that the functor $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ makes the diagram

commutative, since by exactness of the isomorphism \eqref{ch_goes_inside} we have that $$\operatorname{Ch}(\operatorname{colim}):\operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ will be exact if and only if $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A}))\to\operatorname{Ch}(\mathcal{A})$$ is exact; and also by the one-to-one correspondences of Lemma 2 we can transform the conditions 3 and 4 from the last list to the corresponding ones from Proposition 1'.

Why does the triangle commute? Well, commutativity of this diagram amounts to asserting that a colimit of cochain complexes can be computed degreewise, i.e., if we have a diagram $$F:\mathcal{I}\to\operatorname{Ch}(\mathcal{A})$$ then $$(\operatorname{colim}_{i\in\mathcal{I}}F(i))^n=\operatorname{colim}_{i\in\mathcal{I}}F(i)^n$$ and if we have a natural transformation between diagrams of cochain complexes

then, defining the diagram $$F^n:\mathcal{I}\to\mathcal{A}$$, $$i\in\mathcal{I}\mapsto F(i)^n$$ and the natural transformation $$\eta^n:F^n\to G^n$$ whose component at $$i\in\mathcal{I}$$ equals $$\eta^n_i$$, we have $$(\operatorname{colim}\eta)^n=\operatorname{colim}\eta^n$$. From the point of view of the isomorphism \eqref{Ch_as_fun_cat}, this is a particular case of the general categorical result which says that the colimit of a diagram of functors exists if it exists objectwise, and may be computed in an objectwise fashion.

5$$\Rightarrow$$1. Suppose $$A_i^\bullet$$ is an $$\mathcal{I}$$-shaped diagram of acyclic cochain complexes. Then $$H^n(\operatorname{colim}_iA_i^\bullet)=\operatorname{colim}_iH^n(A_i^\bullet)=\operatorname{colim}_i0=0$$.

1+2+3+4 $$\Rightarrow$$ 5. If $$F:\mathcal{A}\to\mathcal{B}$$ is an exact functor between preabelian categories, then the square $$\require{AMScd} \begin{CD} \operatorname{Ch}(\mathcal{A})@>{H^n}>>\mathcal{A}\\ @V{\operatorname{Ch}(F)}VV@VV{F}V\\ \operatorname{Ch}(\mathcal{B})@>>{H^n}>\mathcal{B} \end{CD}$$ commutes, since exact functors commute with finite limits and therefore in particular commute with quotients (aka cokernels of injective maps), kernels and images. In particular, if $$F=\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$ is exact, then we deduce that the square $$\require{AMScd} \begin{CD} \operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A}))@>{H^n}>>\operatorname{Fun}(\mathcal{I},\mathcal{A})\\ @V{\operatorname{Ch}(\operatorname{colim})}VV@VV{\operatorname{colim}}V\\ \operatorname{Ch}(\mathcal{A})@>>{H^n}>\mathcal{A} \end{CD}$$ commutes.

Thus, by the commutative triangle of functors from several paragraphs above, it then suffices to show that the composite $$\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A})) \xrightarrow{\cong}\operatorname{Ch}(\operatorname{Fun}(\mathcal{I},\mathcal{A})) \xrightarrow{H^n}\operatorname{Fun}(\mathcal{I},\mathcal{A})$$ equals $$H_*^n:\operatorname{Fun}(\mathcal{I},\operatorname{Ch}(\mathcal{A})) \to\operatorname{Fun}(\mathcal{I},\mathcal{A})$$.

This is a matter of looking at the commutative cube from proof of Lemma 3 and interpreting which thing is which, and what thing is mapped to / induced by what. $$\square$$