a counterexample for the morphism of sheaves Suppose $\textbf{F}$ and $\textbf{G}$ are two presheaves over a topological space $X$,and $\mu:\textbf{F}\longrightarrow \textbf{G}$ is a morphism of presheaves which is surjective.We have a naturally induced morphism of the associated sheaves $\mu^+:\textbf{F}^+\longrightarrow \textbf{G} ^+$.I feel that $\mu^+$ may not be surjective.Is it right?If so, can you give me an example for this?Thank you very much.
 A: Definition: $f$ is epimorphism if $h \circ f = k \circ f \Rightarrow h=k$, for any composable $h,k$ (definition works in any category).
In category of presheaves of sets or presheaves of abelian groups it can be seen that it is equivalent to surjectiviy on all sections. But not in category of sheaves.
In category of sheaves, being epimorphism is equivalent to surjectivity on all stalks.
If you have epimorphism of presheaves $f \colon F \to G$, then the induced morphism of the associated sheaves $f^+ \colon F^+ \to G^+$ is also epimorphism of sheaves.
It is because the sheafification functor is left adjoint (to the inclusion functor of sheaves to presheaves), so by abstract nonsense result it must preserve all colimits. And being epimorphism can be characterised as a certain colimit condition:
$f$ is epimorphism iff diagram $
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
\bullet & \ra{f} & \bullet & \\
\da{f} & & \da{1} \\
\bullet & \ra{1} & \bullet \\
\end{array}
$ is pushout. You can easily check this.
A: one can show that to have an epimorphism $\phi$ of presheaves $F$ and $G$ is equivalent to: $\phi(U)$ is surjective for all open U.
But then clearly also the map $\phi_x$ is for all x surjective as the direct limit is an exact functor (I suppose we are talking of abelian presheaves).
A: Well, you have the category of presheaves of abelian groups.
Whenever you have a category you can define "epimorphism" by a universal property (see any textbook or wiki). But one can show that this really is equivalent to surjectivity on the sections.
