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Given a presheaf $F'$ of rings, say, we have a morphism $F' \rightarrow F$ where $F$ is the sheafification of $F'$. I see no reason this morphism should be injective or surjective. Is that right?

user524014
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Here is an example where the map $F' \to F$ is not injective : take a ring $R$, $X = \{0,1\}$ with discrete topology (for example) and $F'(U) = 0$ if $U \neq X$ and $F'(X) = R$ with zero restrictions morphisms. Then, $F$ is the zero sheaf and in particular the canonical map $F' \to F$ is not injective.

For surjectivity, take $F'$ the presheaf of bounded functions on $\Bbb R$. Then, $F$ is the sheaf of locally bounded functions, in particular $\text{id}_{\Bbb R} \in F(X)$ but is not in $F'(X)$. So the map $F' \to F$ is not surjective.

Nicolas Hemelsoet
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If you notice on page 81 of the PDF I linked in my comment, there isn't necessarily an injectivity or surjectivity of sheaves in the universal mapping property of a sheafification. However, do note that the sheafification of a sheaf returns the sheaf itself. The definition of sheafification in some sense (if I remember right), makes the smallest sheaf which somehow contains the original presheaf. (I hope that doesn't make things worse.)

Tanner Strunk
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    actually thats super helpful. the book im reading didnt state the universal property for some reason. – user524014 Jan 23 '18 at 05:44
  • By smallest sheaf containing the original presheaf, I am referring to the universal property of a sheafification. Any morphism from the original presheaf $\mathscr{F}$ to a presheaf (or sheaf) $\mathscr{G}$ will give a morphism from $\mathscr{F}^{sh}$ to $\mathscr{G}^{sh}$. (Again, $\mathscr{G}^{sh} \cong \mathscr{G}$ if $\mathscr{G}$ is already a sheaf, so in that case the morphism would be a morphism from sheaf $\mathscr{F}^{sh}$ to sheaf $\mathscr{G}$.) – Tanner Strunk Jan 23 '18 at 05:45
  • Oh super! Yeah um...always go to Vakil ha...for category theoretic stuff anyway (e.g. sheafification). What book have you been reading? There is a fantastic post on here (actually by one of my peers now) about algebraic geometry resources that I will try to find and link. – Tanner Strunk Jan 23 '18 at 05:47
  • I'm reading Eisenbud and Harris, Geometry of Schemes – user524014 Jan 23 '18 at 05:48
  • Ah! Also a good book, and Harris was Vakil's PhD advisor if I remember right. Still, I liked all the exercises in Vakil and felt that he provided a firm technical foundation even if I was lacking a bit in geometric intuition at the time (which took a year to develop and is...well still developing hopefully haha). Here is a great list of other resources for diving into AG: https://math.stackexchange.com/questions/255063/why-study-algebraic-geometry/257528#257528 – Tanner Strunk Jan 23 '18 at 05:50
  • Here is another great road map for this purpose (again by the same person): https://math.stackexchange.com/questions/285201/path-to-basics-in-algebraic-geometry-from-hs-algebra-and-calculus/285355#285355 – Tanner Strunk Jan 23 '18 at 05:51
  • cool. thanks a bunch – user524014 Jan 23 '18 at 05:51
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    It's not exactly the "smallest sheaf" since sections might die in the process. For example, consider the following presheaf $F$ on $X$ : take a group $G$ and the sections are $F(U) = G$ if $U = X$ and $F(U) = 0$ else with zero restrictions morphisms. Then the associated sheaf is the zero sheaf. – Nicolas Hemelsoet Jan 23 '18 at 10:16
  • @NicolasHemelsoet Ok, so that answers half of my question. Is there an example which is not surjective? – user524014 Jan 23 '18 at 12:58
  • @user524014 : I made an answer with the two examples. – Nicolas Hemelsoet Jan 23 '18 at 13:02