Skyscraper sheaves cohomology Let $X$ be a topological space and $G$ an abelian group. Denote by $\mathcal{S}$ the skyscraper sheaf with  group $G$ at the point $x\in X$. How I can prove that $\mathcal{S}$ has not cohomology, i.e $H^i(X,\mathcal{S})=0, \: \forall i>0$ ?

You can also see this using Čech cohomology:
Consider an open cover $\mathfrak{U}=(U_i)_{i\in I}$ of your space $X$. You can always refine this cover so that only one of the sets $U_i$ contains the point $x$: Pick a set $U_0$ containing $x$ and consider the cover $\mathfrak{U}'$ consisting of $U_0$ and $U_i\setminus\{x\}$ for all $i\in I$. Then $\mathfrak{U}'$ is a refinement of $\mathfrak{U}$ and $x$ is only contained in the set $U_0$.
In particular, $x$ is not contained in any intersection of two or more distinct sets in $\mathfrak{U}'$, so the scyscraper sheaf has no sections over these. Hence all higher Čech-cocylces are $0$ and all higher cohomology groups vanish.
(I believe this approach can also be found in Forster's book on Riemann Surfaces.)
