Lemma 6.11.1. Let \mathcal{F} be a sheaf of sets on the topological space X. For every open U \subset X the map
is injective.
Lemma 6.11.1. Let \mathcal{F} be a sheaf of sets on the topological space X. For every open U \subset X the map
is injective.
Proof. Suppose that s, s' \in \mathcal{F}(U) map to the same element in every stalk \mathcal{F}_ x for all x \in U. This means that for every x \in U, there exists an open V^ x \subset U, x \in V^ x such that s|_{V^ x} = s'|_{V^ x}. But then U = \bigcup _{x \in U} V^ x is an open covering. Thus by the uniqueness in the sheaf condition we see that s = s'. \square
Comments (0)
There are also: