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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