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Assume you have a sheaf morphism $f:\mathcal{F} \rightarrow \mathcal{G}$ and consider the sheafification of the presheaf $\textrm{im}(f)$. I want to prove that it's isomorphic to a sub sheaf of $\mathcal{G}$. I basically do this by showing that whenever there is an injective morphism of presheaves $\mathcal{F} \rightarrow \mathcal{G}$, the morphism induced on the sheafifications is still injective, hence I can conclude. It's basically Exercise 2.1.4 in Hartshorne. The inclusion $i :\textrm{im}(f)(U) \rightarrow \mathcal{G}(U)$ is injective for all $U$ so the argument applies to my situation. The thing is that the proof is pretty long in this way and I was wondering if there is a more direct way of proving it exploiting the specifics of this context rather than prove it for every injective morphism of presheaves. Thanks in advance.

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  • $\begingroup$ Did you show part (a) in exercise 2.1.4 ? Because then you have a morphism of presheaves $im(f) \to G$ which is injective on each open set $U$, so by (a) you get an injective morphism of sheaves $im(f)^+ \to G^+ \cong G$. $\endgroup$
    – Watson
    Oct 27, 2018 at 9:12
  • $\begingroup$ Yes, but this is the path that I feel makes it too long. Proving part a relies on other exercises as well. Also, it requires proving that the stalks of the presheaf and the stalks of the sheafification are isomorphic. I was wondering if there's a way to exploit the fact that I'm talking about the image sheaf and an inclusion here, rather than a random presheaf and injective presheaf morphism $\endgroup$
    – Dalamar
    Oct 27, 2018 at 9:14

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If you showed part (a) in exercise 2.1.4, then you are almost done for (b) ! And I don't think there is any easier way that this. It might look too long for you now, but in a few months, I believe that you will actually find this proof rather short !

Indeed, you have a morphism of presheaves $$\mathrm{im}(f) \to \mathscr G,$$ which is injective on each open set $U$, so by (a) you get an injective morphism of sheaves $$\mathrm{im}(f)^+ \to \scr G^+.$$

But $\scr G$ is a sheaf by assumption, so that $\scr G^+ \cong G$. Therefore, the image sheaf $\mathrm{im}(f)^+$ is isomorphic to a subsheaf of $\scr G$

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  • $\begingroup$ Well then maybe there's a quicker way to prove part a)? Do you think it's worth it for me to ask another question about it? $\endgroup$
    – Dalamar
    Oct 27, 2018 at 9:25
  • $\begingroup$ @Dalamar : it is completely normal to find these kind of things a bit difficult at the beginning, while learning sheaf theory. But I really believe that to solve part a), you need to use that sheafification preserves the stalks. This is actually one of the main properties of sheafification ! :-) $\endgroup$
    – Watson
    Oct 27, 2018 at 9:30
  • $\begingroup$ I already proved that :). However after that there's a small problem for me. How can I say that a morphism on the stalks induces a morphism on the whole sheaf and take the final step? Is it just obvious? $\endgroup$
    – Dalamar
    Oct 27, 2018 at 9:51
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    $\begingroup$ If $F \to G$ is a morphism of presheaves, then composing with the natural map $G \to G^+$, you get a morphism $F \to G^+$ from a presheaf to a sheaf. By the universal property of sheafification, this morphism factors through a morphism of sheaves $F^+ \to G^+$. Is that answering your comment? $\endgroup$
    – Watson
    Oct 27, 2018 at 9:56
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    $\begingroup$ @Dalamar : see here ; if all the induced maps on the stalks are injective, then the morphism of sheaves is injective. $\endgroup$
    – Watson
    Oct 27, 2018 at 10:10

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