Understanding surjective morphism of sheaves

Question

Given a surjective morphism of sheaves $\varphi:\mathcal{A}\to \mathcal{B}$ on topological space $X$, it may not be surjective on sections,i.e., on an open set $U$, $\varphi(U):\mathcal{A}(U)\to \mathcal{B}(U)$ may not be surjective.

I have trouble understanding the definiton of surjective morphism of sheaves. In Hartshorne's, he says that if image sheaf $im\varphi$, which is the sheaf associated to presheaf image $U\to \cup _{p\in U}im(\varphi(p))$ is equal to $\mathcal{B}$, then $\varphi$ is surjective. But this means that $\cup _{p\in U}im(\varphi(p))=\mathcal{B}(U)$. On the other hand, if a presheaf $\mathcal{A}$ is a sheaf, then the sheaf associated to presheaf $\tilde{\mathcal{A}}=\mathcal{A}$, i.e, $\mathcal{A}(U)=\cup _{p\in U}\mathcal{A}_p$. Also, $\varphi$ is surjective iff it's surjective at stalks, i.e, $\mathcal{A}_p\to im(\varphi(p))=\mathcal{B}_p$ is surjective. Combining the arguments above, it seems that we have $\mathcal{A}(U)=\mathcal{B}(U)$, which can't be right. So I hope someone can point out my mistakes. Thanks!

Answer 1

In Hartshorne the definition of presheaf image is $U\mapsto im\varphi(U)$. $\varphi:\mathcal A\rightarrow \mathcal B$ is surjective means that $\mathcal B$ is associated to the presheaf image.

Answer 2

The Hartshorne definition of surjectivity of a morphism of sheaves $\phi: \mathcal{F} \to \mathcal{G}$ is that $\mathcal{G}$ is a sheaf associated to the image presheaf of $\phi$. But one important thing to note is that the univeral arrow that goes along with it is the inclusion morphism, not just any morphism. This occurs precisely because the image presheaf is inside a sheaf.

If the universal property holds with the inclusion morphism, then it is easy to see using the definition of associated sheaf that every section $s \in \mathcal{G}(U)$ is locally a section of the image presheaf, i.e. for every $z \in U$, there is a neighbourhood $V \subset U$ of $z$ such that $s|_V \in \operatorname{im}(f_V)$.

This allows us to see that, in the case where $\phi$ is injective as well, it will be an isomorphism, i.e. the presheaf image will actually equal $\mathcal{G}$.