Let $\mathcal{A}$ be an $\mathbf{AB5}$ category with enough projectives and let $F:\mathcal{A}\rightarrow\mathcal{B}$ be a right exact functor into abelian category that commutes with filtered colimits. What reasonable assumptions should one imposed on $\mathcal{A}$ and $F$ to obtain that letf derived functors of $F$ commute with filtered colimits?
When $\mathcal{B}$ is also AB5, this is the case iff filtered colimits of projectives are $F$acyclic:
"$\Rightarrow$" If $\{X_i\}$ is a filtered system of projectives, then $$(L^* F)(\mathrm{colim}_i X_i) = \mathrm{colim}_i (L^* F)(X_i) = 0,$$ i.e. $\mathrm{colim}_i X_i$ is $F$acyclic.
"$\Leftarrow$" If $\{X_i\}$ is some filtered system, choose for each $X_i$ some projective resolution $(P_{\bullet})_i \to X_i$. By assumption and since $\mathcal{A}$ is AB5, $\mathrm{colim} (P_{\bullet})_i$ is a $F$acylic resolution of $\mathrm{colim}_i X_i$, so that we can use it to compute the Left derived functors. It follows (since $\mathcal{B}$ is AB5)
$(L^* F)(\mathrm{colim}_i X_i) = H_*(\mathrm{colim}_i (P_{\bullet})_i) = \mathrm{colim}_i H_*(P_{\bullet})_i) = \mathrm{colim}_i (L^* F)(X_i)$
and this is induced by the canonical morphism $\leftarrow$. $\square$
For example, the tensor product $M \otimes_R  : \mathrm{Mod}(R) \to \mathrm{Ab}$ satisfies this assumption, since filtered colimits of projectives are flat and therefore acyclic with respect to the tensor product. It follows that $\mathrm{Tor}_*(M,)$ commutes with filtered colimits.

Martin thank you for your answer. Do you know any usefull criteria when acyclic objects are closed under taking filtered colimits? – Anonymous May 22 '12 at 17:09

I don't, but I think without any specific choice of $F$ there is no general criterion. – Martin Brandenburg May 22 '12 at 17:47

I wish know a reference about how the morphisms of the filtred system have a (coherent) lifting to a filtred system of projective resolutions. – Buschi Sergio May 22 '12 at 20:18


@Brandenburg You deliver the colimit $C_n:=colimi_{i\in I} P_{n, i}$, I dont understand (my ignorance probabily) how are defined the morphism in the $I$diagram $(P_{n,i})_{i\in I}$ , your proof builds only the objects of this diagram. – Buschi Sergio May 23 '12 at 11:58
This is only a comment on some aspect of the Brandenburg answere. I wish elucidate the aspect about the $I$naturality of projective resolutions:
considerind a example of a single morphism $f: A\to B$ and two projective resolution $P_\ast\to A$ and $Q_\ast\to B$ then is a well knowed fundamental lemma of homological algebrathat exist (homotopically unique) a extentions of $f$ to the (augmented) chain complexes. In the basic homological algebra text there is also the example of a resolution of exat sequences.
But what if insted a single morphism (or a exat sequence) we consider a small diagram?
From Weibel, "An introduction to homological algebra", 2.3.13 on p.43 (I see it from link text) follow that $\mathcal{A}^I$ has enought projectives, but isnt clear if for a projective $P\in \mathcal{A}^I$ each $P(i)\in \mathcal{A}$ is projective as we need in the context of the above Brandenburg proof
(this is true if for any $i\in I$ the right Kan extention of the $i$valutation $v(i): \mathcal{A}^I\to \mathcal{A}$ is exact, I dont know if this follow from the "$I$ is filtrant" hypothesis).
From the book "Theory of Categories (BArry Mitchell) cor.7.6 p. 138, let $T_i: \mathcal{A}^I\to \mathcal{A}$ ($i\in I$) the $i$valuation, and $S_i$ its left adjuction (the left Kan extention), now for a projective $P\in \mathcal{A}$ the object $S_i(P)(j),\ j\in I$ is a sum of copies of $P$ (see the Weibel reference above) then is projective. THe above corollary assert that projectives of $\mathcal{A}^I$ are objects of the form $\bigoplus_{i\in I}S_i(P_i)$ (where $P_i\in \mathcal{A}$ is a prjective) and all its retracts.
This is enought for ensure the existence a projective resolution of a $(X_i)_i\in \mathcal{A}^I$ with projective arguments.