Is $\mathscr{H\!}\mathit{om}_{\mathscr{O}_X}(\mathscr{F},\mathscr{G})_x\cong\operatorname{Hom}_{\mathscr{O}_{X,x}}(\mathscr{F}_x,\mathscr{G}_x)$?

Question

Fix a topological space $X$, and fix abelian sheaves $\mathscr{F}$ and $\mathscr{G}$ on $X$, and recall the definition of sheaf Hom, $\operatorname{\mathscr{H}\!\mathit{om}}_{\mathscr{O}_X}(\mathscr{F},\mathscr{G})$. Here, I am using $\operatorname{Hom}_{\mathscr{O}_{X,x}}(\mathscr{F}_x,\mathscr{G}_x)$ simply to refer to the Hom in the category of $\mathscr{O}_{X,x}$-modules. Note that we have a canonical map $$\operatorname{\mathscr{H}\!\mathit{om}}_{\mathscr{O}_X}(\mathscr{F},\mathscr{G})\to\operatorname{Hom}_{\mathscr{O}_{X,x}}(\mathscr{F}_x,\mathscr{G}_x)$$ sending $\langle U,\varphi\rangle\in \operatorname{\mathscr{H}\!\mathit{om}}_{\mathscr{O}_X}(\mathscr{F},\mathscr{G})$ to $\varphi_x\in\operatorname{Hom}_{\mathscr{O}_{X,x}}(\mathscr{F}_x,\mathscr{G}_x)$. I suspect that this map is not an isomorphism, but I can't prove it.

Answer 1

This is not an isomorphism in general. Here is an example in the category of sheaves (so with $\mathcal{O}_X=\mathbb{Z}$).

Let $\mathcal{F}=i_*\mathbb{Z}$ where $i:\{0\}\rightarrow\mathbb{R}^n$ is the inclusion of the origin. Let $\mathcal{G}=\mathbb{Z}$. I claim that $\mathcal{H}om(i_*\mathbb{Z},\mathbb{Z})=0$. Indeed, obviously $\operatorname{Hom}(i_*\mathbb{Z}_{|U},\mathbb{Z}_{|U})=0$ for all $U\not\ni 0$, and in the case where $0\in U$, we can assume that $U$ is an open ball, and we can assume that this ball is $\mathbb{R}^n$ (by homeomorphism). So we are reduced to prove that $\operatorname{Hom}(i_*\mathbb{Z},\mathbb{Z})=0$. Let $j:\mathbb{R}^n\setminus\{0\}\rightarrow\mathbb{R}^n$ be the inclusion of the complement. By the short exact sequence $$0\rightarrow j_!\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow i_*\mathbb{Z}\rightarrow 0$$ we see that $\operatorname{Hom}(i_*\mathbb{Z},\mathbb{Z})$ is the kernel of $\operatorname{Hom}(\mathbb{Z},\mathbb{Z})\rightarrow\operatorname{Hom}(j_!\mathbb{Z},\mathbb{Z})$. But this map is injective (even an isomorphism if $n\geq 2$), because $\operatorname{Hom}(j_!\mathbb{Z},\mathbb{Z})=\operatorname{Hom}(\mathbb{Z},j^{-1}\mathbb{Z})=\mathbb{Z}^{\pi_0(\mathbb{R}^n\setminus\{0\})}$.

This prove that $\mathcal{H}om(i_*\mathbb{Z},\mathbb{Z})=0$ and so are its stalks. However, $\operatorname{Hom}((i_*\mathbb{Z})_x,\mathbb{Z}_x)=\operatorname{Hom}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$.

Answer 2

For schemes, this is just a special case of Does localisation commute with Hom for finitely-generated modules? (recall how modules correspond to quasi-coherent sheaves). In general the answer is no, but yes for coherent sheaves (more generally, finitely presented sheaves).