# Image sheaf isomorphism : another proof? (Exercise 2.1.4 in Hartshorne)

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.

• 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 $im(f)^+ \to G^+ \cong G$. – Watson Oct 27 '18 at 9:12
• 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 – Dalamar Oct 27 '18 at 9:14

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 $$\scr G$$
• 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? – Watson Oct 27 '18 at 9:56