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.
Best Answer
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 $H^i(S)=H^i(\Gamma(I_\bullet))$.