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Suppose I have a morphism of sheaves of abelian groups $\phi: \mathcal{F} \longrightarrow \mathcal{G}$ with induced morphisms of stalks $\phi_{p}: \mathcal{F}_{p} \longrightarrow \mathcal{G}_{p}$. We then have an exact sequence of presheaves $$ 0 \longrightarrow \ker\phi \longrightarrow \mathcal{F} \longrightarrow \mathcal{F}/\ker \phi \longrightarrow 0. $$ We further have that taking the stalks preserves exactness, so that we have an exact sequence of abelian groups $$ 0 \longrightarrow (\ker\phi )_{p} \longrightarrow \mathcal{F}_{p} \longrightarrow (\mathcal{F}/\ker \phi )_{p} \longrightarrow 0. $$ How do we use this information to show that $\ker( \phi_{p}) \simeq (\ker \phi)_{p}$?

My attempt was to first make an elementary diagram chase to find that $(\mathcal{F} / \ker \phi)_{p} \simeq \mathcal{F}_{p} / \ker (\phi_{p})$. We know further that as groups, $\mathcal{F}_{p} / \ker (\phi_{p}) \simeq \mathcal{G}_{p}$. We then have an exact sequence $$ 0 \longrightarrow (\ker\phi )_{p} \longrightarrow \mathcal{F}_{p} \longrightarrow \mathcal{G}_{p} \longrightarrow 0. $$ However, is this enough to show that $\ker(\phi_{p}) \simeq (\ker \phi)_{p}$? I feel like I need the surjective map in that last one to be exactly $\phi_{p}$, which it isn't. It is $\phi_{p}$ composed with some other isomorphisms.

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In general, an exact sequence $$0 \to A \to B \xrightarrow{f} C$$ is enough to show that $A \cong \operatorname{ker} f$. You do not need surjectivity of $f$. That is simply the universal property of the kernel.

In particular you have already proven what you wanted to prove. Exactness of talking stalks is all you need.

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