Proof. It is probably better for the reader to find their own explanation of this than to read the proof here. In fact the lemma is true for the same reason as why the presheaf of continuous function is a sheaf, see Example 6.7.3 (and this analogy can be made precise using the “espace étalé”).

Anyway, let $U = \bigcup U_ i$ be an open covering. Suppose that $s_ i = (s_{i, u})_{u \in U_ i} \in \mathcal{F}^{\# }(U_ i)$ such that $s_ i$ and $s_ j$ agree over $U_ i \cap U_ j$. Because $\Pi (\mathcal{F})$ is a sheaf, we find an element $s = (s_ u)_{u\in U}$ in $\prod _{u\in U} \mathcal{F}_ u$ restricting to $s_ i$ on $U_ i$. We have to check property $(*)$. Pick $u \in U$. Then $u \in U_ i$ for some $i$. Hence by $(*)$ for $s_ i$, there exists a $V$ open, $u \in V \subset U_ i$ and a $\sigma \in \mathcal{F}(V)$ such that $s_{i, v} = (V, \sigma )$ in $\mathcal{F}_ v$ for all $v \in V$. Since $s_{i, v} = s_ v$ we get $(*)$ for $s$. $\square$

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