Flabby sheaves are acyclic using Godement resolution

Question

Let $\mathcal{F}$ be a sheaf of abelian groups on a topological space $X$. Define the sheaf cohomology groups as $$H^\bullet(X,\mathcal{F}):=H^\bullet(G^\bullet(\mathcal{F})(X)), $$ where $G^\bullet(\mathcal{F})$ is the Godement resolution of $\mathcal{F}$. How do I prove that flasque sheaves are acyclic using this definition without using derived functors?

I know that in an exact sequence of sheaves $0 \to \mathcal{F}\to \mathcal{G} \to \mathcal{H}\to 0$, the induced sequence $$0 \to \mathcal{F}(X)\to \mathcal{G}(X) \to \mathcal{H}(X)\to 0$$ is exact, if $ \mathcal{F}$ is flasque.

Answer 1

First observe that if you have a short exact sequence $$0\rightarrow F \rightarrow G \rightarrow H \rightarrow 0,$$ with $F, G$ flasque, then $H$ is flasque. (To prove this, use the fact that taking sections over an open set still yields a short exact sequence since $F$ is flasque.)

Now, recall how you construct the Godement resolution: you look at the continuous map $f:X^{disc}\rightarrow X$ where $X^{disc}$ is the discrete space and $f$ is identity as a set map. Now $G^0(F)= f_*f^{-1}(F)$ and there is a canonical injective morphism of sheaves $\alpha: F\rightarrow G^{0}(F)$. But if $F$ is flabby, then $Coker(\alpha)= (G^0(F)/F)$ is again flabby. Now, $G^1(F)=G^0(G^0(F)/F)$ etc. and we get the complex $0\rightarrow G^0(F) \rightarrow G^1(F) \rightarrow \cdots$ .

Observe that since $F$ is flabby, we have $$0\rightarrow F(X) \rightarrow G^0(F)(X) \rightarrow (G^0(F)/F)(X)\rightarrow 0$$ to be exact. But $G^0(F)/F$ is also flabby, and we have a short exact sequence $$0\rightarrow G^0(F)/F\rightarrow G^1(F)\rightarrow G^1(F)/G^0(F)\rightarrow 0$$ so that $G^1(F)/G^0(F)$ is also flabby and we have short exact sequences $$0\rightarrow (G^0(F)/F)(X) \rightarrow G^1(F)(X) \rightarrow (G^1(F)/G^0(F))(X)\rightarrow 0$$ $$0\rightarrow G^1(F)/G^0(F)(X) \rightarrow G^2(F)(X) \rightarrow (G^2(F)/G^1(F))(X)\rightarrow 0.$$ From the above short exact sequences, we see that $$(G^0(F)/F)(X)=ker(G^1(F)(X) \rightarrow G^2(F)(X))$$ and $G^0(F)(X)$ surjects onto $(G^0(F)/F)(X)$, so that $H^1(X,F)= ker(G^1(F)(X) \rightarrow G^2(F)(X))/G^0(F)(X)=0$. A similar argument tell us that $G^{i+1}(F)/G^i(F)$ are flabby and $H^i(X,F)=0$ for all $i>0$.