Does sheafification preserve surjectivity? 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!
 A: 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)
A: I think it is true by the following argument:
EDIT: My argument actually fails, as pointed out by Roland (I marked the spot). 
Take a section of $G^+(U)$, i.e. $s:U\to \prod_{x\in U} G_x$ such that the coherence condition is fullfilled. This means around each point $P\in U$, we can find an open neighbourhood $U_P$ around P, s.t. $g_p\in G(U_P)$ and $s(P)=[U_P,g_P]\in G_P$. Now for each of these $g_P$ we can a preimage $h_P \in F(U_P)$  with $f(h_P)=g_P$ (as we assume that $f$ is surjective on all open sets). 
Here the argument fails! There is no gurantee for such a choice to exist
Note that we can choose the preimage s.t.  $h_P|_{U_P\cap U_Q}=h_Q|_{U_P\cap U_Q}$. Now define $t:U\to \prod_{x\in U} F_x$ as $t(P)=[U_P,h_P]\in F_P$. Then $t$ is a well-defined element in $F^+(U)$ and $f^+(t)=s$ per construction.
