Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $FH \rightarrow HF$).

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.