TITLE: Does sheafification preserve surjectivity?
QUESTION [5 upvotes]: Let $\mathcal{F}$ and $\mathcal{G}$ be presheaves over a topological space X and $f$ be a surjective morphism of presheaves, meaning for every open subset $U\subset X$ the homomorphism $\mathcal{F}(U)\to\mathcal{G}(U)$ is surjective. Then we know there is corresponding sheaf morphism of the sheafifications $f^{+}: \mathcal{F}^+\to \mathcal{G}^+$ and $f^+$ is surjective morphism of sheaves, i.e. surjective of every stalk.
My question is whether $f^{+}$ is surjective in the sense of presheaves, i.e. whether for every open subset $U\subset X$, whether $\mathcal{F}^+(U)\to\mathcal{G}^+(U)$ is still surjective.
I cannot prove it and my guess is not but I couldn't figure out a counterexample. The question partly come from the feather that injectivity (as presheaves) is preserved after sheafification so how about surjectivity.
Any help is appreciated. Thanks!
PS: First time posting a question so welcome to pointing out any mistake or improperness in format!

REPLY [5 votes]: No it is not. To see this, let 
$$ 0\rightarrow \mathcal{F}\rightarrow\mathcal{G}\rightarrow\mathcal{H}\rightarrow 0$$
be your favorite exact sequence of sheaf. For example the exponential sequence $0\rightarrow\mathbb{Z}\rightarrow\mathcal{C}\rightarrow\mathcal{C}^\times\rightarrow 0$ where $\mathcal{C}$ is the sheaf of continuous section on a space $X$ and $\mathcal{C}^\times$ the sheaf of nowhere vanishing sections.
Now let $\mathcal{K}$ be the presheaf $U\mapsto\operatorname{coker}(\mathcal{F}(U)\rightarrow\mathcal{G}(U))$ so $\mathcal{G}\rightarrow\mathcal{K}$ is a surjective morphism of presheaves, and its sheafification is $\mathcal{G}\rightarrow\mathcal{H}$. But $\mathcal{G}\rightarrow\mathcal{H}$ is not necessarily a surjective morphism of presheaves.
Indeed with the exponential sequence on $X=\mathbb{C}^*$, the section $(z\mapsto z)\in\mathcal{C}^\times(\mathbb{C})$ is not in the image of $\exp:\mathcal{C}(\mathbb{C}^*)\rightarrow\mathcal{C}^\times(\mathbb{C})$. (In fact the cokernel of $\mathcal{C}(\mathbb{C}^*)\rightarrow\mathcal{C}^\times(\mathbb{C})$ is $H^1(\mathbb{C}^*,\mathbb{Z})\simeq\mathbb{Z}$ with generator the above section)