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I'm trying to prove the snake lemma for sheaves of abelian groups, namely:

If we have the following commutative diagram with exact rows:

enter image description here

Then we have an exact sequence: $$0\to\text{ker}(f)\to\text{ker}(g)\to\text{ker}(h)\to\text{coker}(f)\to\text{coker}(g)\to\text{coker}(h)\to 0$$

The part $0\to\ker(f)\to\ker(g)\to\ker(h)$ is easily constructed from $i,j$ and it's easy to verify it is exact. Now let's call $\mathcal{F}'',\mathcal{G}'',\mathcal{H}''$ the presheaves defined by $U\mapsto \mathcal{F}'(U)/f(\mathcal{F}(U))$, $U\mapsto \mathcal{G}'(U)/g(\mathcal{G}(U))$ and $U\mapsto \mathcal{H}'(U)/h(\mathcal{H}(U))$, whose sheafifications are $\mathcal{F}''\to\text{coker}(f)$, $\mathcal{G}''\to\text{coker}(g)$ and $\mathcal{H}''\to\text{coker}(h)$ respectively. The presheaf morphisms $\mathcal{F}''\to\mathcal{G}''\to\mathcal{H}''$ are naturally induced by $i',j'$ and, using the universal property of sheafification, we get the following commutative diagram:

If I'm not mistaken, we can conclude $\text{coker}(g)\to\text{coker}(h)$ is an epimorphism directly by the fact that $\mathcal{G}'\to\mathcal{H}'$ is an epimorphism.

Now I'm stuck trying to proving that $\text{coker}(f)\to\text{coker}(g)\to\text{coker}(h)$ is exact and I have no idea how to build the connection map $\ker(h)\to\text{coker}(f)$.

Any ideas?

rmdmc89
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  • Sheaves of abelian groups form an abelian category. The linked duplicate explains how to prove the Snake Lemma in an arbitrary abelian category, and you can use the solution there to solve your problem. – KReiser May 18 '20 at 20:56
  • @KReiser, that's an interesting approach. But I was hoping for a more elementary solution. Isn't there a direct way to build the connection morphism, for example? Thank you! – rmdmc89 May 19 '20 at 00:16
  • That solution is elementary: the only things one uses are the definitions of kernels/cokernels/quotients and the like. I think this is a perfectly direct way to build the connection morphism - what do you mean when you say "direct"? – KReiser May 19 '20 at 00:32
  • @KReiser I think the issue is that a priori I don't know whether the category of sheaves of abelian groups is abelian, and it takes a considerable amount of work to prove it, right? So I thought there could be a way around it (in that sense, a more "direct" way) – rmdmc89 May 19 '20 at 00:56
  • It does not take much work to prove. An abelian category is a preadditive category where you have a zero object, binary biproducts, kernels and cokernels, and monos/epis are normal. All of these conditions except the last are trivial, and the final one is not particularly difficult. If you are going to attempt to learn algebraic geometry at any sort of modern level, you will need to see this stuff at least once - why not now? – KReiser May 19 '20 at 01:07
  • @KReiser fair enough – rmdmc89 May 19 '20 at 01:17

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