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Let $\mathscr{F}$ be a presheaf and $\mathscr{F}^+$ its sheafification, with the universal morphism $\theta:\mathscr{F}\rightarrow\mathscr{F}^+$. Question is: is $\theta$ always an inclusion? I'm pretty sure it isn't, but in many cases it seems that it is. For example, if $\phi:\mathscr{F}\rightarrow\mathscr{G}$ is a morphism of sheaves, then $\operatorname{ker}\phi$, $\operatorname{im}\phi$, $\operatorname{cok}\phi$ all seem to have this property. If it isn't always the case, then is there any characterization of such presheaves? A "subpresheaf" of a sheaf certainly has this property (e.g., $\operatorname{im}\phi$), but what about $\operatorname{cok}\phi$?

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Just to make sure, you are asking if the kernel of $\theta$ is always trivial? –  M Turgeon Jun 3 '12 at 15:06
@M Turgeon: Yes. –  ashpool Jun 3 '12 at 15:07
Dual question here. –  Zhen Lin Jun 3 '12 at 16:56

3 Answers 3

up vote 9 down vote accepted

No, $\mathcal F \to \mathcal F^+$ is not injective in general.

Consider the sheaf $\mathcal C$ of continuous functions on $\mathbb R$, its subpresheaf $\mathcal C_b\subset \mathcal C$ of bounded continuous functions and the quotient presheaf $\mathcal F$, characterized by $\mathcal F(U)= \mathcal C (U)/\mathcal C_b(U)$.
For every open subset $U\subset \mathbb R$, we have $\mathcal F(U)\neq 0$ but the associated sheaf is $\mathcal F^+=0$, so that
the morphism $\mathcal F \to \mathcal F^+=0$ is definitely not injective.

Injectivity of $\mathcal F \to \mathcal F^+$ is equivalent to requesting that whenever you have compatible gluing data $s_i\in \mathcal F(U_i)$ on an open covering $(U_i)$ of an open $U$ of your space, they can glue to at most one $s\in \mathcal F(U)$ : one half of the conditions for a presheaf to be a sheaf must be satisfied (the other half is to require that $s$ always exist)

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Cokernels do not have this property. The example to look at is the exponential map from the sheaf of holomorphic functions on $\mathbf C \setminus \{0\}$ to the sheaf of non vanishing holomorphic functions on that same domain. This is not surjective on global sections, since for example the identity map does not have a logarithm, but it is surjective as a morphism of sheaves, so the sheaf cokernel is trivial.

What distinguishes the kernel and image sheaves, I think, is that they satisfy the identity axiom [I think this is Hartshorne's (5)] because they sit inside of sheaves.

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Ah, sorry, I was thinking of $\mathscr{F}/\mathscr{G}$, where $\mathscr{F}\subset\mathscr{G}$ is a subsheaf. –  ashpool Jun 3 '12 at 15:44

The answer to the question is NO since the global information about a presheaf can not be determined by its local information. In fact, this is the essential point of the definition of a sheaf. In above, Georges Elencwajg has given a nice counter-example which is quite natural. We can also contruct a counter-example directly as follows. Take a topological space $X$ which is not empty, define a presheaf $\mathcal{F}$ of abelian groups in the following way: for any proper open subset $U$ define $\mathcal{F}(U)$ to be zero while define $\mathcal{F}(X)$ to be any nontrival abelian group, say $\mathbb{Z}$. Then the sheafication $\mathcal{F}^+$ is zero. Consider the global sections we will find it is not an inclusion.

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