Example 6.7.5. Let $X$ be a topological space. Let $(A_ x)_{x \in X}$ be a family of sets $A_ x$ indexed by points $x \in X$. We are going to construct a sheaf of sets $\Pi$ from this data. For $U \subset X$ open set

$\Pi (U) = \prod \nolimits _{x \in U} A_ x.$

For $V \subset U \subset X$ open define a restriction mapping by the following rule: An element $s = (a_ x)_{x\in U} \in \Pi (U)$ restricts to $s|_ V = (a_ x)_{x \in V}$. It is obvious that this defines a presheaf of sets. We claim this is a sheaf. Namely, let $U = \bigcup U_ i$ be an open covering. Suppose that $s_ i \in \Pi (U_ i)$ are such that $s_ i$ and $s_ j$ agree over $U_ i \cap U_ j$. Write $s_ i = (a_{i, x})_{x\in U_ i}$. The compatibility condition implies that $a_{i, x} = a_{j, x}$ in the set $A_ x$ whenever $x \in U_ i \cap U_ j$. Hence there exists a unique element $s = (a_ x)_{x\in U}$ in $\Pi (U) = \prod _{x\in U} A_ x$ with the property that $a_ x = a_{i, x}$ whenever $x \in U_ i$ for some $i$. Of course this element $s$ has the property that $s|_{U_ i} = s_ i$ for all $i$.

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