Direct limits and quasi-isomorphism

Question

Suppose, for each $n>0$, I have two complexes of abelian groups $(A_{n}, d)$ and $(B_{n},d')$ and a quasi-isomorphism $$f_{n}: A_{n} \rightarrow B_{n}.$$ Furthermore, suppose I have maps of complexes $A_1 \rightarrow A_2 \rightarrow A_3 \dots$ and similarly for $B_n$, compatible with $f_n$.

Is it true that the map of direct limit complexes $$\lim_{n} f_{n}: \lim A \rightarrow \lim B$$ is also a quasi-isomorphism?

Complexes can mean unbounded or maybe just bounded above/below, depending on your preference.

Also: what if I replace the limit over natural numbers with a more general filtered colimit?

Answer 1

Yes indeed. You may assume that the complexes belong to a Grothendieck category if you wish, or just stick with abelian groups. (Grothendieck category: abelian category with exact filtered direct limits that possesses a generator). In the countable case, you have Milnor's exact sequence

$$ \oplus_{n} A_n \longrightarrow \oplus_{n} A_n \longrightarrow \lim_{n} A_n $$

The first arrow, usually denoted by $1-shift$ takes an element in $A_n$ and sends it to itself in $A_n$ and to the negative of its image by the transition map in $A_{n+1}$. It is an exercise that $1-shift$ is injective on abelian groups (and the proof can be adapted to the general situation).

You have a similar sequence for the $B_n$'s and a map of exact sequences given by $\oplus_{n} f_n$ in the first and second factor and the induced map in the third. When you interpret this diagram in the derived catdegory it is clear that $\oplus_{n} f_n$ is a quasi-isomorphism and your diagram of exact sequences becomes a diagram of triangles. This implies that the induced map is a quasi-isomorphism.

For certain more general systems there are some results in my joint paper with Jeremías & Souto, here or in the CJM site. You can find here an exposition of homotopy colimitis in the context of complexes. Hope this helps.

Answer 2

Doesn't this follow from the exactness of direct limits for modules?