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?
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