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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