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It is very well known that if $\alpha : \mathscr{F} \to \mathscr{G}$ is a morphism of sheaves, then it induces homomorphisms on the stalks. I have been wondering for a while if given a collection of homomorphisms between the stalks of two sheaves and some suitable patching condition, can we construct a homomorphism of sheaves?

I have been thinking about this problem a little bit lately, but I have not been able to come up with a "canonical" solution. In other words I have found ways of producing sheaf morphisms from stalk morphisms but they have always felt very unnatural.

Let me phrase the question a little bit more formally. Let $X$ be a topological space and $\mathscr{F}, \mathscr{G}$ sheaves on $X$. Let $\alpha_p : \mathscr{F}_p \to \mathscr{G}_p$ be a homomorphism for each $p \in X$. Is there a suitable condition on the $\alpha_p$s such that there exists a sheaf morphism $\alpha : \mathscr{F} \to \mathscr{G}$ whose induced maps on the stalks are exactly the $\alpha_p$s?

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up vote 7 down vote accepted

There is a simple condition. Stalk maps $\alpha_p : F_p \to G_p$ yield a map of sheaves $\alpha : F \to G$ if and only if for all open $U \subseteq X$ and sections $s \in F(U)$ there is an open covering $U = \cup_i U_i$ and sections $t_i \in G(U_i)$ such that $\alpha_p(s_p) = (t_i)_p$ for all $p \in U_i$ and all $i$. This can also be stated as continuity of the $\alpha_p$ (see the answer of David Roberts).

This cirterion is sometimes useful, for example in the construction of fibered products of locally ringed spaces (German:

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This answer is in the context of sheaves of sets. If you meant sheaves of groups, please say so.

If you look at the 'espace etale' (wikipedia) of a sheaf $\mathrm{Open}(X)^{op} \to Set$ (which is defined by taking the union of stalks and topologising appropriately - fibres of the canonical map to $X$ are then discrete spaces), then continuous maps of these over $X$ correspond to sheaf maps. The maps of stalks give a map of the underlying sets. An obvious sufficient condition that the collection of maps of stalks gives a map of sheaves is that this map is continuous.

(If you actually have a sheaf of groups, then you get a group object over $X$. All you need to check is the continuity - the homomorphisms of stalks give a homomorphism of the underlying set of the group object)

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Are you saying that there is a bijection between sheaf maps $\alpha : \mathscr{F} \to \mathscr{G}$ and continuious maps $f : {\rm Spe}(\mathscr{F}) \to {\rm Spe}(\mathscr{G})$? – Daniel Barter Jan 12 '11 at 1:23
Continuous maps $f$ over $X$ - there is an equivalence of categories between sheaves on a space and the category of spaces etale over that space. This is in MacLane-Moerdijk, for one. – David Roberts Jan 12 '11 at 1:38

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