Lemma 6.17.2. Let X be a topological space. Let \mathcal{F} be a presheaf of sets on X. Let x \in X. Then \mathcal{F}_ x = \mathcal{F}^\# _ x.
Proof. The map \mathcal{F}_ x \to \mathcal{F}^\# _ x is injective, since already the map \mathcal{F}_ x \to \Pi (\mathcal{F})_ x is injective. Namely, there is a canonical map \Pi (\mathcal{F})_ x \to \mathcal{F}_ x which is a left inverse to the map \mathcal{F}_ x \to \Pi (\mathcal{F})_ x, see Example 6.11.5. To show that it is surjective, suppose that \overline{s} \in \mathcal{F}^\# _ x. We can find an open neighbourhood U of x such that \overline{s} is the equivalence class of (U, s) with s \in \mathcal{F}^\# (U). By definition, this means there exists an open neighbourhood V \subset U of x and a section \sigma \in \mathcal{F}(V) such that s|_ V is the image of \sigma in \Pi (\mathcal{F})(V). Clearly the class of (V, \sigma ) defines an element of \mathcal{F}_ x mapping to \overline{s}. \square
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