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

$\mathcal{F}(U) \longrightarrow \prod \nolimits _{x \in U} \mathcal{F}_ x$

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 $s = s'$. $\square$

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