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This is Proposition 2.5 on page 208 of Hartshone's Algebraic Geometry:

If $\mathcal F$ is a flasque sheaf on a topological space $X$, then $H^i(X, \mathcal F) = 0$ for all $i>0$.

The proof considers this exact sequence $0 \rightarrow \mathcal F \rightarrow \mathcal J \rightarrow \mathcal G \rightarrow 0$, in which $\mathcal J$ is an injective sheaf containing $\mathcal F$, and $\mathcal G$ is the quotient sheaf. As $\mathcal F$ and $\mathcal J$ are flasque, so is $\mathcal G$.

Now since $\mathcal F$ is flasque, we have an exact sequence $0 \rightarrow \Gamma(X, \mathcal F) \rightarrow \Gamma(X, \mathcal J) \rightarrow \Gamma(X, \mathcal G) \rightarrow 0$. On the other hand, since $\mathcal J$ is injective, we have $H^i(X, \mathcal J) =0$ for $i>0$. Thus from the long exact sequence of cohomology, we get $H^1(X, \mathcal F) =0$ and $H^i(X, \mathcal F) \cong H^{i-1}(X, \mathcal G)$ for each $i \geq 2$. But $\mathcal G$ is also flasque, so by induction on $i$ we get the result.

What I don't understand is why we can discuss the cohomology group of $\mathcal F$ in the exact sequence $0 \rightarrow \mathcal F \rightarrow \mathcal J \rightarrow \mathcal G \rightarrow 0$. By definition, the cohomology functor $H^i(X,-)$ is the right derived functor of $\Gamma(X,-)$. So I think $H^i (X, \mathcal F)$ is related to the injective resolution of $\mathcal F$. But, in the above sequence, is $\mathcal G$ injective? If it is, why? If not, why is the proof valid?

Thanks to everyone.

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up vote 5 down vote accepted

The general fact is, given a left exact functor $F:A \rightarrow B$ where $A$ and $B$ are abelian categories and $A$ has enough injectives (that is, every object of $A$ admits an embedding into an injective object), every short exact sequence of objects of $A$ gives rise to a long exact sequence involving the derived functors of $F$ (these exist because $A$ has enough injective objects to define them). In your case, with $A$ equal to the category of sheaves on $X$, $B$ equal to the category of abelian groups, and $F=\Gamma$ equal to the global sections functor, every short exact sequence $0 \rightarrow S \rightarrow T \rightarrow U \rightarrow 0$ of sheaves gives rise to a long exact sequence $$\cdots \rightarrow H^{i-1}(U) \rightarrow H^i(S) \rightarrow H^i(T) \rightarrow H^i(U) \rightarrow \cdots $$ of cohomology groups.

There is no requirement that the objects in the SES be injective, though it is true that calculating the derived functors requires, a priori, an injective resolution: given an injective resolution $$ 0 \rightarrow S \rightarrow I_0 \rightarrow I_1 \rightarrow \cdots$$ one obtains $H^i(S)=H^i(\Gamma(I_\bullet))$.

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Thanks very much for the answer. Dose this mean I can take an arbitray short exact sequence of the sheaf to compute the cohomology group? Why is it wrong to take $0 \rightarrow \mathcal F \rightarrow \mathcal F \rightarrow 0 \rightarrow 0$? (Then for any sheaf $\mathcal F$, even if it is not flasque or injective, the cohomology group would be $0$ for $i>0$.) – ShinyaSakai Mar 2 '12 at 14:12
"Dose this mean I can take an arbitray short exact sequence of the sheaf to compute the cohomology group?" Only if you have some control over the cohomology groups of the other two sheaves in the sequences. – Stephen Mar 2 '12 at 16:34
"Why is it wrong to take 0→F→F→0→0?" It's not wrong. You get a long exact sequence. Write it out and you will see that it gives no information---just that the cohomology of $F$ is isomorphic to itself. So, "(Then for any sheaf F, even if it is not flasque or injective, the cohomology group would be 0 for i>0.) " is wrong. This doesn't follow. – Stephen Mar 2 '12 at 16:34
Thank you very much. Now I understand :) Thanks a lot. – ShinyaSakai Mar 3 '12 at 5:45

This is a bit of an old chain, but here's an additional answer. What's often very useful (because injectives are nice theoretical constructs, but hard to come by in practical situations, is that you can use any resolution by acyclic object to compute cohomology.

That is, if you have a resolution $0\rightarrow\mathcal{F}\rightarrow\mathcal{A}^{\cdot}$, where, for each k,

$H^{i}(\mathcal{A}^{k})=0$ for all $i>0$ (this is the acyclic part),

then the cohomology of $\mathcal{F}$ is given by the complex $0\rightarrow H^{0}(\mathcal{A}^{0})\rightarrow H^{0}(\mathcal{A}^{1})\rightarrow\cdot\cdot\cdot$

So, depending of where you're working, you can use fine sheaves (on smooth manifolds) or flasque (flabby) sheaves (on varieties or schemes), etc.

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