Exercise 111.32.10. Let $X$, $A_ x$, $\mathcal{G}$ be as in Exercise 111.32.9. Let $\mathcal{B}$ be a basis for the topology of $X$, see Topology, Definition 5.5.1. For $U \in \mathcal{B}$ let $A_ U$ be a subgroup $A_ U \subset \mathcal{G}(U) = \prod _{x \in U} A_ x$. Assume that for $U \subset V$ with $U, V \in \mathcal{B}$ the restriction maps $A_ V$ into $A_ U$. For $U \subset X$ open set
\[ \mathcal{F}(U) = \left\{ (s_ x)_{x \in U} \middle | \begin{matrix} \text{ for every }x\text{ in }U\text{ there exists } V \in \mathcal{B}
\\ x \in V \subset U\text{ such that } (s_ y)_{y \in V} \in A_ V
\end{matrix} \right\} \]
Show that $\mathcal{F}$ defines a sheaf of abelian groups on $X$. Show, by an example, that it is usually not the case that $\mathcal{F}(U) = A_ U$ for $U \in \mathcal{B}$.
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