This is my second attempt to solve exercise 1.6H from Vakil: Assume $F$ is a covariant right-exact functor and show that we get a map $HF^i \rightarrow H^iF$.
Attempt to a solution:
Apply $F$ to the exact sequence $ A^i \overset{d^i}\rightarrow A^{i+1} \rightarrow \def\coker{\operatorname{coker}\,}\coker d^i\rightarrow 0$ and simultaneously take the cokernel of the map $Fd^i: FA^i \rightarrow FA^{i+1}$. Then it follows from five lemma that $F \coker d^i \cong \coker Fd^i$.
Now consider the two sequences $$F\def\im{\operatorname{im}\,}\im d^i \rightarrow F A^{i+1} \rightarrow F \coker d^i \rightarrow 0$$
$$(0 \rightarrow) \ker F \coker d^i \rightarrow F A^{i+1} \rightarrow \coker Fd^i \rightarrow 0$$
where the first one is $F$ applied to the "fundamental sequence of chain complexes" and the second one is obtained from the first one by the isomorphism just described on the third term and substituting the first term with the kernel of the following map.
Since, again $F \coker d^i \cong \coker Fd^i$, the first term is actually $\ker \coker Fd^i$, which, by this post we can write as $\im Fd^i$. From the isomorphism of the fourth term and the universal property of the kernel we get a map $\kappa: F \im d^i \rightarrow \im F d^i$.
a) How do we show (without diagram-chasing) that this map is epi?
b) And why do we need it to be epi?
Now, we consider the next two exact sequences: $$FH^i \rightarrow F\coker d^{i-1} \rightarrow F\im d^i \rightarrow 0$$ $$(0 \rightarrow )H^iF \rightarrow \coker Fd^{i-1} \rightarrow \im Fd^i \rightarrow 0,$$ where the first one is $F$ applied to the "dual definition" of cohomology, and the second one is the cohomology of $Fd^i$. We have an isomorphism of the second term and a vertical downward map on the last term. Assuming that these make the diagram commute we get a map $HF^i \rightarrow H^iF$ by the universal property of the cokernel. Two more questions:
c) how do we show the diagram commutes? I think it must follow from the construction of $\kappa$, but there we had indices $i$ and $i+1$. Here we have indices $i$ and $i-1$ and the maps are in opposite order. How do I connect them?
d) Can we actually show that $HF^i \rightarrow H^iF$ is mono?
Best Answer
a) Use the fact "if $fg$ is epi then $f$ is epi" with
$f$:$F$[im $d^i$] $\rightarrow$ im$F$[$d^i$],
$g$:$F[A^i] \rightarrow F[$coker $d^i] \rightarrow F[$im $d^i]$
b) I think we don't need to.
c) Compare two sequences:
$F[A^i] \rightarrow F[$coker $d^{i-1}] \rightarrow F[$im $d^i] \rightarrow F[A^{i+1}]$
$F[A^i] \rightarrow $coker $F[d^{i-1}] \rightarrow $im $F[d^i] \rightarrow F[A^{i+1}]$
There is a vertical map on the third term and isomorphism on the other terms. So the rectangle with the second term and fourth term commutes. Also, from the construction of the vertical map, the right small square commutes. So each way from $F[$coker $d^{i-1}]$ to $F[A^{i+1}]$ give the same morphism. Finally using that "im $F[d^i] \rightarrow F[A^{i+1}]$" is monomorphism, two ways from $F[$coker $d^{i-1}]$ to $F[A^{i+1}]$ must be same.
d) I have no idea.