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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    $\begingroup$ Just to make sure, you are asking if the kernel of $\theta$ is always trivial? $\endgroup$
    – M Turgeon
    Jun 3, 2012 at 15:06
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    $\begingroup$ @M Turgeon: Yes. $\endgroup$
    – ashpool
    Jun 3, 2012 at 15:07
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    $\begingroup$ Dual question here. $\endgroup$
    – Zhen Lin
    Jun 3, 2012 at 16:56
  • $\begingroup$ 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 $f = g$. $\endgroup$ Feb 5, 2018 at 19:15
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    $\begingroup$ 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. $\endgroup$
    – Watson
    May 20, 2018 at 12:07

3 Answers 3


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)


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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    $\begingroup$ Ah, yes: yours is a great, simple and amusing example. Bravo: +1 $\endgroup$ Oct 6, 2015 at 7:50
  • $\begingroup$ @Georges Elencwajg: Is it enough to show that $\mathcal{F}^{+}(X)=0 $ ? so that $\mathcal{F}(X) \to \mathcal{F}^{+}(X)$ is not injective group homomorphism. $\endgroup$
    – user371231
    Oct 2, 2020 at 9:44

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.

  • $\begingroup$ Ah, sorry, I was thinking of $\mathscr{F}/\mathscr{G}$, where $\mathscr{F}\subset\mathscr{G}$ is a subsheaf. $\endgroup$
    – ashpool
    Jun 3, 2012 at 15:44

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