Take the 2-minute tour ×  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?  share|improve this question add comment  2
Answers  up vote 5 down vote accepted  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 \