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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!

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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.

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