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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 $\operatorname{cok}\phi$?

ashpool
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    Just to make sure, you are asking if the kernel of $\theta$ is always trivial? – M Turgeon Jun 03 '12 at 15:06
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    @M Turgeon: Yes. – ashpool Jun 03 '12 at 15:07
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    Dual question here. – Zhen Lin Jun 03 '12 at 16:56
  • It might be relevant to note that I've seen presheaves with this property be called "separated". It's equivalent to half of the definition of sheaf: that if ${ U_i : i \in I }$ is a cover of $U$, $f,g \in \mathscr{F}(U)$, and $f \mid_{U_i} = g \mid_{U_i}$ for each $i \in I$, then $f = g$. – Daniel Schepler Feb 05 '18 at 19:15
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    Another example : consider the presheaf $F : U \mapsto O_X(1)(U) \otimes_{O_X(U)} O_X(1)(U)$ where $(X, O_X) = \Bbb P^1_k$. Then $F^+ \cong O_X(2)$ but $F \to F^+$ is not injective, since $F(X) \cong k^4$ but $F^+(X) \cong k^3$. The key notion is separated presheaf ; see here. – Watson May 20 '18 at 12:07

4 Answers4

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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)

Georges Elencwajg
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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.

Joy-Joy
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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.

Dylan Moreland
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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 03 '12 at 15:44
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This is not a full proof, but just a quick intuition I find interesting, as to why it isn’t in general true and when it is true.

Suppose the canonical map $\theta \colon \mathcal F \to \mathcal F ^+$ is injective. This means that we can see $\mathcal F$ as a subpresheaf of $\mathcal F ^+$: $$\mathcal F \subseteq \mathcal F ^+.$$

Now, we know that $\mathcal F ^+$ is a sheaf, therefore it satisfies the sheaf axioms. As you recall, there are two sheaf axioms: one is locality/uniqueness, and the other one is glueing/existence.

If you think about it, logically speaking, the glueing axiom contains existential quantifiers, so we have no certainty it will hold on smaller objects. On the other hand, the locality axiom only contains universal quantifiers, so we’re sure it will also hold for a “smaller” object.

In other words, when the canonical map is an embedding, the presheaf $\mathcal F$ is already “half” of a sheaf, namely it satisfies the locality axiom, but not necessarily the glueing axiom.

It’s not hard to prove that the converse is also true: namely, that the fact that a presheaf satisfies the locality condition ensures the canonical map $\theta$ is injective. So the locality axiom perfectly characterizes the injectivity of $\theta$.

The reason why, as you said in the question, in many cases it appears to be true, is the following. Consider a presheaf that is not a sheaf: then it is so, because the presheaf fails to satisfy either one of the locality condition or the glueing condition. But the majority of the presheaves we encounter, which aren’t sheaves, only fail in the glueing condition.

Something more is true, and it is rather remarkable. The same way that locality characterizes the injectivity of $\theta$, the glueing condition almost perfectly characterizes the surjectivity of $\theta$. However, there is a little asymmetry here: one can prove that $\theta$ is surjective if and only if $\mathcal F$ already satisfies both glueing and locality. I guess it would be prettier if the surjectivity of $\theta$ were equivalent just to glueing, without locality, but I’ve never seen a proof of such a fact, without including locality, nor have I ever been able to provide one.

Melanzio
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