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

Question

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.

Answer 1

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 $\scr G$