# 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.

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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.

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Got it!Thank you very much! –  user14242 Aug 9 '11 at 15:14

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:


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I would like to draw that colimit condition, can somebody please tell me how to write commutative diagrams on stackexange? I personaly use xy package, but it doesn't seem to work here. –  rafaelm Aug 9 '11 at 19:12
See this meta thread. If you want to use Xy-pic, you might have to create an image file and upload it somewhere else. –  Dylan Moreland Aug 9 '11 at 19:46

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).

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:very nice answer!A little more question:what is your definition for epimorphism of sheaves?Not the one you listed as an equivalence? –  user14242 Aug 9 '11 at 14:51