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Statement: let $F$ be a right exact functor. Describe a map $FH \rightarrow HF$. (from Vakil's notes 1.6H)

attempt of a solution:

Let $K$ be the kernel of $FA^i \rightarrow FA^{i+1}$. Since F is right exact and the map $A^i \rightarrow A^{i+1}$ factors uniquely through the $\def\im{\rm{im}\,}\im$$d^i$, we have that $F d^i$ also factors through $F\im$$d^i$ (actually, how do we prove that right exact functors preserve "map factorizations")?

Hence the composition $F \ker d^i \rightarrow FA^i \rightarrow FA^{i+1}$ exists and is equal to the zero map (because a right exact functor, being additive preserves zero morphisms). It follows that we get a map $F \ker d^i \rightarrow K$.

Now comes the delicate point. In the category of modules I can prove that the last map is epi, but I think that I need mono (since I'm probably wrong and I am looking for a proof using only concepts of abelian categories I won't write down my reasoning here). Anyways, assuming it is monic, and that the map $F \im$ $d^i \rightarrow \im$ $Fd^i$ is epic (the existence of which map is easy to prove by a similar arguement as mine above; but I'm still stuck in showing that it is epic), we the map that we wanted $FH \rightarrow HF$ by the five lemma:

applying $F$ to:

$0\rightarrow \im$$d^{i-1}\rightarrow \ker d^i \rightarrow H^i\rightarrow 0$

we get

$F\im$$d^{i-1}\rightarrow F\ker d^i \rightarrow FH^i\rightarrow 0$ and

$\im$$Fd^{i-1}\rightarrow \ker Fd^i \rightarrow HF^i\rightarrow 0$ (I'm not sure of how to the first map here and commutativity with the vertical maps. But if all I said so far is true we get a mono $FH \rightarrow HF$).

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