must a surjective sheaf map be surjective on some open set?

Question

A surjective (i.e. surjective on stalks) map of sheaves $F\to G$ on a space $X$ need not be surjective on global sections, but is it true that for every $p\in X$, there exists a neighborhood $U$ of $p$ on which $F(U)\to G(U)$ is surjective?

Answer 1

No, this need not be the case. Consider for example a metric space $(X,d)$, and take ${\mathscr F} := {\mathscr C}_X$ the sheaf of continuous functions on $X$ and ${\mathscr G} := \widetilde{\text{C}(X;{\mathbb R})}$ is the constant sheaf on the globally defined continuous functions on $X$. Then the canonical restriction morphism ${\mathscr G}\to{\mathscr F}$ is surjective, since any continuous function on some open set extends to a globally defined function after suitable restriction. However, for any open $U\neq\emptyset$ with nonempty boundary, the function $x\mapsto d(x,X\setminus U)$ cannot be extended to $X$.

Answer 2

Let $F$ be pretty much anything, and let $x$ be a point, and let $G$ be the skyscraper sheaf at $x$ associated to $F_x$. Then $G \rightarrow F_x$ is surjective, because any element of $F_x = \lim_{x \in U} F(U)$ is in the image of $F(U)$ for some open $U$. However, for a typical $F$, the map $F(U) \rightarrow F_x$ is never surjective. For an explicit example, take $F$ to be the sheaf of holomorphic functions on $\mathbb{C}$, and take $x = 0$. Then, for $U$ containing $0$, $G = F_0(U)$ contains $$\frac{1}{z - \epsilon}$$ for any $\epsilon > 0$, but these will not be sections of $F(U)$ for $\epsilon \in U$.