Skyscraper sheaves cohomology

Question

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 $H^i(X,\mathcal{S})=0, \: \forall i>0$ ?

Answer 1

These notes probably have enough detail to give you what you want. I'll assume you mean $G$ to be an abelian group. Basically the argument goes like this: if $G$ is an abelian group, take an injective resolution $G \to I^\bullet$ of abelian groups. Then $\pi_*G \to \pi_*I^\bullet$ is an injective resolution, where $\pi:{\ast} \to X$ is the inclusion of a point. Note that $\pi_*G$ is your skyscraper sheaf. To compute $H^i(X,G)$ take global sections of $\pi_*G \to \pi_*I^\bullet$, which just gives back the resolution $G\to I^\bullet$, which shows that the higher cohomology of $\pi_*G$ vanishes.

Answer 2

A skyscraper sheaf is flasque, hence has no cohomology: Hartshorne Chap. III, Prop.2.5, page 208.

Answer 3

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.)