image of a morphism of sheaves

Question

i want to be sure that the following assertion is true :

let be

$\phi : \mathcal{F}\rightarrow \mathcal{G}$

an injective morphism of sheaves then the presheaf

$U \rightarrow im(\phi(U)) $

not need to sheafify

i will be grateful if someone could confirm me the truthy of the assertion. Thanks in advance !

Answer 1

That's true. Let's denote your presheaf $(U \mapsto \text{im}(\phi(U)))$ by $\text{im}(\phi)$ (this is slightly abusive notation, since $\text{im}(\phi)$ typically refers to the sheafification of the image presheaf, but it shouldn't cause confusion in the context of this example). There is a natural map of presheaves $\mathcal{F} \rightarrow \text{im}(\phi)$ induced by $\phi$ (on any open $U \subset X$, the map $\mathcal{F}(U) \rightarrow \text{im}(\phi)(U) = \text{im}(\phi(U))$ is given by $\phi(U)$). It is by construction surjective on every open, and by assumption injective. Therefore, $\phi$ is an isomorphism of presheaves. Since $\mathcal{F}$ is a sheaf, the same is automatically true of the isomorphic presheaf $\text{im}(\phi)$.