Lemma 6.18.1. Let $X$ be a topological space. Let $\mathcal{F}$ be a presheaf of sets on $X$. Let $U \subset X$ be open. There is a canonical fibre product diagram

\[ \xymatrix{ \mathcal{F}^\# (U) \ar[d] \ar[r] & \Pi (\mathcal{F})(U) \ar[d] \\ \prod _{x \in U} \mathcal{F}_ x \ar[r] & \prod _{x \in U} \Pi (\mathcal{F})_ x } \]

where the maps are the following:

The left vertical map has components $\mathcal{F}^\# (U) \to \mathcal{F}^\# _ x = \mathcal{F}_ x$ where the equality is Lemma 6.17.2.

The top horizontal map comes from the map of presheaves $\mathcal{F} \to \Pi (\mathcal{F})$ described in Section 6.17.

The right vertical map has obvious component maps $\Pi (\mathcal{F})(U) \to \Pi (\mathcal{F})_ x$.

The bottom horizontal map has components $\mathcal{F}_ x \to \Pi (\mathcal{F})_ x$ which come from the map of presheaves $\mathcal{F} \to \Pi (\mathcal{F})$ described in Section 6.17.

**Proof.**
It is clear that the diagram commutes. We have to show it is a fibre product diagram. The bottom horizontal arrow is injective since all the maps $\mathcal{F}_ x \to \Pi (\mathcal{F})_ x$ are injective (see beginning proof of Lemma 6.17.2). A section $s \in \Pi (\mathcal{F})(U)$ is in $\mathcal{F}^\# $ if and only if $(*)$ holds. But $(*)$ says that around every point the section $s$ comes from a section of $\mathcal{F}$. By definition of the stalk functors, this is equivalent to saying that the value of $s$ in every stalk $\Pi (\mathcal{F})_ x$ comes from an element of the stalk $\mathcal{F}_ x$. Hence the lemma.
$\square$

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