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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.

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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}.

enter image description here

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

enter image description here

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

enter image description here

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. $\square$

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