Homology of acyclic complex and left exact functor

homological-algebra

Let $F$ be a left exact functor and $A^\bullet$ be a chain complexes:
$A^\bullet: \cdots \xrightarrow{f^{i-1}} A^i \xrightarrow{f^i} A^{i+1} \xrightarrow{f^{i+1}} A^{i+2} \cdots$.
Moreover, assume that each $A^i$ is $F$-acyclic.
Then, I guess $F(H(A^\bullet)) = H(F(A^\bullet))$. Especially, if $A^\bullet$ is exact than $F(A^\bullet)$ is also exact.

I tried considering the short exact sequence:
$0\rightarrow \mathrm{ker}f^i \rightarrow A^i \rightarrow \mathrm{im} f^i\rightarrow0$
$0\rightarrow\mathrm{im} f^{i-1} \rightarrow \mathrm{ker}f^i \rightarrow H^i(A^\bullet)\rightarrow0 $

Applying $F$:
$0\rightarrow F(\mathrm{ker}(f^i)) \rightarrow F(A^i) \rightarrow F(\mathrm{im} (f^i))\rightarrow$
$0 \rightarrow F(\mathrm{im} (f^{i-1}) \rightarrow F(\mathrm{ker}f^i) \rightarrow F(H^i(A^\bullet))\rightarrow $
If I know $\mathrm{ker}f^i$ is $F$-acyclic, it follows that $\mathrm{im} f^i$ is also $F$-acyclic, so both of these sequence are also right exact. Comparing with the short exact sequence obtained from $F(A^\bullet)$, I will get $F(H(A^\bullet)) = H(F(A^\bullet))$

But I cannot show $\mathrm{ker}f^i$ is $F$-acyclic. Is there some way to fix this?

Best Answer

No this is not true. In fact, let $X$ be any object in your abelian category and $A^\bullet$ a resolution of $X$ by $F$-acyclic (for instance injectives). In particular, you have $H^i(A^\bullet)=0$ for $i>0$ and $H^0(A^\bullet)=X$ (which is by the way $\ker f^0$ and may not be acyclic). On the other hand, $F(A^\bullet)$ is a complex with very interesting cohomology : $H^i(F(A^\bullet))=R^iF(X)$ and this might not be zero for $i>0$. Thus $F(H(A^\bullet))\neq H(F(A^\bullet))$.

On the other hand, if $A^\bullet$ is exact (which was not the case in my previous comment) and bounded below, then this is true and your proof should work fine with an induction (this is where you need bounded below).

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