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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. – Zhen Lin Oct 28 '12 at 19:04

First, in order to make sense of the sheaf condition, we must assume that the category $\mathcal{C}$ has enough limits; for simplicity we assume $\mathcal{C}$ has all limits. Next, to make sense of stalks, we must assume that $\mathcal{C}$ has enough filtered colimits; for simplicity we assume $\mathcal{C}$ has colimits for all small filtered diagrams.

Write $\textbf{Sh}(X; \mathcal{C})$ for the category of $\mathcal{C}$-valued sheaves on $X$. The key condition is the following:

  • A morphism $\phi : \mathscr{F} \to \mathscr{G}$ in $\textbf{Sh}(X; \mathcal{C})$ is an isomorphism if and only if the stalk $\phi_x : \mathscr{F}_x \to \mathscr{G}_x$ is an isomorphism for all points $x$ in $X$.

If $\mathcal{C} = \textbf{Set}$, this is just the fact that the topos $\textbf{Sh}(X)$ has enough points. This property is absolutely essential for doing anything useful with stalks.

Lemma. The functor $x^* : \textbf{Sh}(X; \mathcal{C}) \to \mathcal{C}$ that sends a sheaf $\mathscr{F}$ to its stalk $\mathscr{F}_x$ has a right adjoint.

Proof. The usual construction of the skyscraper sheaf goes through without problems.  ◼

Lemma. Let $\textbf{Psh}(X; \mathcal{C})$ be the category of $\mathcal{C}$-valued presheaves on $X$.

  • Limits of small diagrams in $\textbf{Psh}(X; \mathcal{C})$ exist and can be computed componentwise.
  • $\textbf{Sh}(X; \mathcal{C})$ is closed under small limits in $\textbf{Psh}(X; \mathcal{C})$.

Proof. The first claim is standard abstract nonsense, and the second claim is essentially a consequence of the fact that limits preserve limits. ◼

Proposition. Let $\phi, \psi : \mathscr{F} \to \mathscr{G}$ be a pair of parallel morphisms in $\textbf{Sh}(X; \mathcal{C})$. Suppose at least one of the following conditions holds:

  • Filtered colimits in $\mathcal{C}$ preserve equalisers.
  • $\textbf{Sh}(X; \mathcal{C})$ has coequalisers.

Then the following are equivalent:

  • $\phi = \psi$.
  • $\phi_x = \psi_x$ for all points $x$ in $X$.

Proof. If $\phi = \psi$ then obviously $\phi_x = \psi_x$ for all $x$. Conversely, suppose $\phi_x = \psi_x$ for all $x$. There are two cases:

  • Assume $\mathcal{C}$ has coequalisers. Let $\theta : \mathscr{G} \to \mathscr{H}$ be the coequaliser of $\phi$ and $\psi$; left adjoints always preserve coequalisers, so $\theta_x$ is the coequaliser of $\phi_x$ and $\psi_x$. Since $\phi_x = \psi_x$, $\theta_x$ is an isomorphism; but this is true for all $x$, so $\theta$ is also an isomorphism, by our assumption on $\textbf{Sh}(X; \mathcal{C})$. Thus $\phi = \psi$.

  • Assume instead that filtered colimits in $\mathcal{C}$ preserve equalisers. Equalisers exist in $\textbf{Sh}(X; \mathcal{C})$ and are computed componentwise, so this means $x^*$ preserves them. The rest of the argument is essentially the same. ◼

Of course, the real question is, when does $\textbf{Sh}(X; \mathcal{C})$ have enough points? This is actually fairly tricky and I don't see a good general argument. Here's one that works, but it is somewhat restrictive.

Proposition. Let $\mathcal{C}$ be a locally finitely-presentable (l.f.p.) category. Then:

  • Filtered colimits in $\mathcal{C}$ preserve finite limits.
  • $\textbf{Sh}(X; \mathcal{C})$ has enough points.

Proof. Both claims basically boil down to the representation theorem for l.f.p. categories: there exist a small category $\mathcal{A}$ and a fully faithful functor $N : \mathcal{C} \to [\mathcal{A}^\textrm{op}, \textbf{Set}]$ that preserves filtered colimits and has a left adjoint. There is then a fully faithful functor $\textbf{Sh}(X; \mathcal{C}) \to \textbf{Sh}(X; [\mathcal{A}^\textrm{op}, \textbf{Set}])$ obtained by applying $N$ componentwise, and it is not hard to check that $\textbf{Sh}(X; [\mathcal{A}^\textrm{op}, \textbf{Set}])$ and $[\mathcal{A}^\textrm{op}, \textbf{Sh}(X)]$ are equivalent as categories. Now, the fact that $N$ preserves filtered colimits means that the stalk functors fit into a commutative diagram $$\begin{array}{rcl} \textbf{Sh}(X; \mathcal{C}) & \rightarrow & \mathcal{C} \\ \downarrow & & \downarrow \\ [\mathcal{A}^\textrm{op}, \textbf{Sh}(X)] & \rightarrow & [\mathcal{A}^\textrm{op}, \textbf{Set}] \end{array}$$ and so ultimately it boils down to the fact that $\textbf{Sh}(X)$ has enough points.

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