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Let $\phi,\psi:\mathscr{F}\rightarrow\mathscr{G}$ be morphisms of sheaves on $X$ with values in a category $\mathbf{C}$. Let's assume that $\mathbf{C}$ is nice enough to have products, equalizers, etc. Is it true that if $\phi_x=\psi_x:\mathscr{F}_x\rightarrow\mathscr{G}_x\ \forall x\in X$, then $\phi=\psi$ ? The proof of this is easy in a concrete category, but it doesn't seem so easy to do that without using elements of sets and instead using things like equalizers, in terms of which sheaf is defined (Wikipedia).

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    You also need to assume $\mathbf{C}$ has, say, filtered colimits in order to make sense of stalks at all. I suspect if you could make filtered colimits in $\mathbf{C}$ behave sufficiently badly, then you could construct counterexamples. But why on earth would you want to do that? The reason why things are easy in the case when $\mathbf{C}$ is a category of "algebras" is because then it can be embedded nicely in the topos of sheaves of sets, and then properties like "having enough points" are just inherited from the topos.2012-10-28

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