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For a topological space X, the category of sheaves on X with its values in (Ab) will form an Abelian Category.

Q1: Is it difficult to prove this?

Next, for the short exact sequence 0 ---> F ---> G ---> H ---> 0, its exactness is usually stated in terms of their stalks at each point x on X.

However, somebody told me that this is a ``theorem" rather than definition. Namely, once I know that the category of sheaves on X with values in (Ab) (we call this (Sh)_X makes an Abelian category, automatically the notion of exact sequence exists.

Hence, it only turns out that the short exact sequence defined via the characteristic of (Sh)_X being Abelian category is equivalent to the exactness of the given short exact sequence after taking stalk at an arbitrary point x on X.

Q2. I cannot see at all what this will mean. Please explain more plainly.

I heartily wish somebody's explanation. Sincerely, Pierre MATSUMI

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This is not really research level. Maybe you should post it on math stackexchange. –  Oliver Straser Jun 27 '13 at 21:10
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closed as off-topic by Fernando Muro, Steven Landsburg, Sasha, Benjamin Steinberg, Andrey Rekalo Jun 27 '13 at 22:48

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1 Answer

$\DeclareMathOperator{\sh}{Sh}\DeclareMathOperator{\psh}{PSh}$ A1. Not especially. Essentially, one uses the fact that $\psh(X)$ is abelian (which is essentially trivial to prove) and then the sheafification functor to show that $\sh(X)$ admits all finite limits and colimits (which include kernels and cokernels). What remains is to show that every monic morphism $f:F\to G$ of sheaves is the kernel of its (categorical) cokernel, and likewise that every epimorphism $g:F\to G$ is the cokernel of its (categorical) kernel. Using the first case, one has to show that the following sequence is exact: $$ 0 \to F \xrightarrow f G \to \text{coker}(f) \to0 $$ is exact. Here is where the lemma on stalks comes in. One shows that taking kernels and cokernels behaves well with respect to taking stalks, and that exactness can be checked on the stalks. So, one only has to check that for all $x\in X$, $$ 0 \to F_x \xrightarrow{f_x} G_x \to \text{coker}(f_x) \to 0 $$ is exact. But this is trivial. So the "stalk lemma" simply makes it easier to prove that $\sh(X)$ is an abelian category by allowing you to "move" a critical part of the proof to the category of abelian groups.

I should note that the "stalk lemma" is not necessary. If $\mathcal{A}$ is any reflective subcategory of an abelian category $\mathcal{B}$ such that the left adjoint to the inclusion $\mathcal{A}\hookrightarrow \mathcal{B}$ preserves finite limits, then $\mathcal{A}$ is also abelian. Applying this to the inclusion $\sh(X)\to \psh(X)$ yields the result.

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Dear Professor Miller, great thanks!! –  Pierre MATSUMI Jun 28 '13 at 10:13
    
Dear Professor Miller, great thanks!! I really understand it. Pierre –  Pierre MATSUMI Jun 28 '13 at 10:13
    
I'm glad my answer helped. I should say though: I'm not a professor - I'm just a grad student at Cornell. –  Daniel Miller Jun 29 '13 at 17:53
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