A skyscraper sheaf with group $G$ concentrated at the point $p\in X$ is a sheaf $\mathscr{F}$, such that $$\mathscr{F}(U)=\begin{cases}0, \: \text{if} \: \: p\not\in U\\ G, \: \text{if} \: \: p\in U. \end{cases}$$ It's well known that the following property $(S)$ $$\mathscr{F}_a=\begin{cases}0, \: \text{if} \: \: p\neq a\\ G, \: \text{if} \: \: p=a\end{cases}$$ holds for any $a\in X$. Is a sheaf with this property a skyscraper sheaf?
Characterization of skyscraper sheaves
0
$\begingroup$
sheaf-theory
1 Answers
2
Yes it is. Let $\mathcal{F}$ be a sheaf with your property $(S)$. Let $p:\{*\}\rightarrow X$ the inclusion of the point $p$, so that the functor $p^{-1}$ is just the stalk at $p$. Then by definition you have an isomorphism $p^{-1}\mathcal{F}\overset{\sim}\rightarrow G$. By adjunction you get a morphism $\mathcal{F}\rightarrow p_*G$. Now $p_*G$ is the skyscraper sheaf $G$ concentrated at the point $p$. It is easy to see that the morphism $\mathcal{F}\rightarrow p_*G$ is an isomorphism on stalks, hence an isomorphism.
By the way, the morphism $\mathcal{F}\rightarrow p_*G$ can be made explicit. Let $U$ be any open set. If $p\not\in U$, $\mathcal{F}(U)\rightarrow (p_*G)(U)$ is the zero morphism. If $p\in U$, then $\mathcal{F}(U)\rightarrow (p_*G)(U)=G$ is just the restriction map $\mathcal{F}(U)\rightarrow \mathcal{F}_p=G$. Check that this is indeed a morphism and an isomorphism on stalks.