Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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.

share|improve this answer
    
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:

$f$ 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.

share|improve this answer
    
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).

share|improve this answer
    
: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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.