In one of Vakil's exercises he asks to prove that filtered colimits commute with $\textbf{homology}$ in $Mod_A$. For this he suggests to use the following results:

$\bullet$ For a covariant exact functor $F\colon \mathscr{A} \to \mathscr{B}$ between abelian categories and a complex $C^\bullet$ in $\mathscr{A}$, there is an isomorphism $F(H^{i}(C^\bullet)) \cong H^{i}(F(C^\bullet))$.

$\bullet$ filtered colimits in $Mod_A$ are exact.

With this it is an immediate consequence that filtered colimits commute with $\textbf{co}$homology in $Mod_A$. It is not hard to see that an exact functor between abelian categories also commutes with homology. So, I was just wondering why he asked to just prove that filtered colimits in $Mod_A$ commute with homology and not also cohomology. Is this just for convenience or what do you think was his intention right there?


A complex only has either cohomology or homology, depending on which indexing you use-whether the differential goes up or down. The categories of complexes with increasing and decreasing differentials are, of course, isomorphic, even isomorphic under $\mathcal{A}$. So there is no difference between the claims.

  • $\begingroup$ Thanks for your answer. Would you mind explaining what it means to be isomorphic $\textit{under}$ $\mathscr{A}$? $\endgroup$ – user363120 Sep 14 '16 at 22:12
  • $\begingroup$ Sure, this just means the isomorphism commutes with the inclusions of $\mathcal{A}$. $\endgroup$ – Kevin Carlson Sep 15 '16 at 5:40

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