# A characterization of AB5 categories. Does this result show up on the literature?

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

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

1. $$\mathcal{A}$$ is abelian,
2. $$\mathcal{A}$$ is cocomplete (all small colimits exist). Equivalently, $$\mathcal{A}$$ has all coproducts (abelian categories have all coequalizers), and
3. taking filtered colimits 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 precisely when for all $$\mathcal{I}$$ small filtered categories, the functor \eqref{colim_fun} is exact (equivalently, left exact; equivalently, it preserves monomorphisms). Equivalently, when for all directed sets $$\mathcal{I}$$ the functor \eqref{colim_fun} is exact (all small filtered categories contain a cofinal directed set, see here).

Given an additive category $$\mathcal{A}$$, we denote $$\operatorname{Ch}(\mathcal{A})$$ to the category of chain 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 homology objects. The result I'm interested in is the following characterization of AB5 categories.

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

1. $$\mathcal{A}$$ is AB5,
2. $$\operatorname{Ch}(\mathcal{A})$$ is AB5,
3. for all filtered 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 of acyclic complexes are acyclic, and
4. for all filtered 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.

The colimit functor on conditions 3 and 4 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 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 on 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. An interesting corollary to my proof of Proposition 1 is the result “in AB5 categories, homology commutes with filtered colimits.” Here's the precise statement, copied from my answer.

Corollary 4. If $$\mathcal{A}$$ is an AB5 category and $$\mathcal{I}$$ is small and filtered, then 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).

Where $$H^n$$ is the homology functor, and where if $$\mathsf{C},\mathsf{D},\mathsf{E}$$ are categories, then a functor $$F:\mathsf{D}\to\mathsf{E}$$ induces a functor $$F_*:\operatorname{Fun}(\mathsf{C},\mathsf{D})\to\operatorname{Fun}(\mathsf{C},\mathsf{E})$$.

This result has been pointed out on MSE before, but I think there aren't proofs here, so I'll include it on my answer as well.

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.

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 chain 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 chain 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 chain 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 chain complex of diagrams is the same thing as a diagram of chain complexes.”

Proof. For me, differentials will go up. We can realize the category of chain 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 chain complex with terms in $$\mathcal{A}$$ is the same thing as an additive functor $$\mathsf{Diff}\to\mathcal{A}$$, and a chain 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 chain map between the chain complexes of functors $$\eta^\bullet:C_{(-)}^\bullet\to D_{(-)}^\bullet$$. The chain 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 chain 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 chain 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 chain complexes $$C^\bullet_\alpha:C^\bullet_i\to C^\bullet_j$$. Commutativity of this front face means that $$C^\bullet_\alpha$$ is indeed a chain 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 chain 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 chain 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 chain 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 chain 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. If $$\mathcal{A}$$ is an AB3 category, then for a small category $$\mathcal{I}$$ we can consider the functor $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$. Suppose now $$\mathcal{I}$$ is filtered. By Lemma 1, the following conditions are equivalent:

1. The functor $$\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$ is exact ($$\mathcal{A}$$ is AB5),
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 chain 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 chain 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. $$\square$$

Corollary 4. In AB5 categories, homology commutes with filtered colimits. More precisely, if $$\mathcal{A}$$ is an AB5 category and $$\mathcal{I}$$ is small and filtered, then 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).

Where $$H^n$$ is the homology functor, and where if $$\mathsf{C},\mathsf{D},\mathsf{E}$$ are categories, then a functor $$F:\mathsf{D}\to\mathsf{E}$$ induces a functor $$F_*:\operatorname{Fun}(\mathsf{C},\mathsf{D})\to\operatorname{Fun}(\mathsf{C},\mathsf{E})$$.

Proof. 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 $$\mathcal{A}$$ is an AB5 category, we can apply this observation to the exact functor $$F=\operatorname{colim}:\operatorname{Fun}(\mathcal{I},\mathcal{A})\to\mathcal{A}$$ to 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 in the proof of Proposition 1, 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 Proposition 1 and interpreting which thing is which, and what thing is mapped to / induced by what. $$\square$$