Are surjective sheaf morphisms locally surjective? Let $\mathcal F$ and $\mathcal G$ be sheaves of groups of some topological space $X$ and let $\varphi : \mathcal F \to \mathcal G$ be a surjective morphism. That means that for each $x \in X$ the induced group morphism $\varphi_x : \mathcal F_x \to \mathcal G_x$ is surjective.
Does this imply that for each $x\in X$ there exists an open $U \subseteq X$ with $x \in U$ such that $\varphi_U : \mathcal F(U) \to \mathcal G(U)$ is surjective?
I don't think so but would be glad to see an easy counter example.
 A: Consider $X=\mathrm{Spec}(\Bbb Z)$, pick a prime number $p$ and consider the closed point $x=(p) \in X$. Then we have the skyscraper sheaf $\mathcal G=i_{x,*}\Bbb Z_{(p)}$ which sends an open subset $U \subset X$ to $0$ if $x \notin U$ and to $\Bbb Z_{(p)}$ if $x \in U$. All restrictions are either the identity on $\Bbb Z_{(p)}$ or zero. Now consider the structure sheaf $\mathcal F$ of the scheme $X$. For any open subset $U \subset X$, we have a morphism of groups $\mathcal F(U) \to \mathcal G(U)$. If $x \notin U$, this is the zero map.
If $x \in U$, then we can assume $U=D(f)$ with $f \notin (p)$ (as $\Bbb Z$ is a PID), this implies that we have a canonical morphism $\mathcal{F}(U)=\Bbb Z_f \to \Bbb Z_{(p)}$ by the universal property of localization. All these morphisms are compatible and form a morphism of sheaves of groups (even commutative rings). As one can easily check, all the stalks of $\mathcal G$ are zero except at $x$, but the induced map at $x$ is an isomorphism, so the morphism is surjective. But for no open subset $U \subset X$ containing $x$, the map $\mathcal{F}(U) \to \mathcal{G}(U)$ is surjective, because we any non-empty open subset is of the form $U=D(f)$ for $f \neq 0$ (using the fact that $\Bbb Z$ is a PID) and we have that $\mathcal{F}(U) = \Bbb Z_f$ . Thus it suffices to show that $\Bbb Z_{f} \to \Bbb Z_{(p)}$ is not surjective for any $f \in \Bbb Z \setminus \{0\}$ such that $(p) \in D(f)$. But this is clear: $\Bbb Z_{f}$ is a finitely generated $\Bbb Z$-algebra and $\Bbb Z_{(p)}$ isn't. Or more directly: let $q$ be any prime distinct from $p$, not dividing $f$, then $\frac{1}{q}$ is in $\Bbb Z_{(p)}$, but not in $\Bbb Z_{f}$. (Note that this argument, amusingly, uses the infinitude of the prime numbers.)
