Lemma 99.3.2. In Situation 99.3.1 the functor $\mathit{Hom}(\mathcal{F}, \mathcal{G})$ satisfies the sheaf property for the fpqc topology.

Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $B$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and $\mathcal{F}_ i = u_{T_ i}$ and $\mathcal{G}_ i = \mathcal{G}_{T_ i}$. Note that $\{ X_ i \to X_ T\} _{i \in I}$ is an fpqc covering of $X_ T$, see Topologies on Spaces, Lemma 73.9.3. Thus a family of maps $u_ i : \mathcal{F}_ i \to \mathcal{G}_ i$ such that $u_ i$ and $u_ j$ restrict to the same map on $X_{T_ i \times _ T T_ j}$ comes from a unique map $u : \mathcal{F}_ T \to \mathcal{G}_ T$ by descent (Descent on Spaces, Proposition 74.4.1). $\square$

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