Lemma 99.3.5. In Situation 99.3.1 let \{ X_ i \to X\} _{i \in I} be an fppf covering and for each i, j \in I let \{ X_{ijk} \to X_ i \times _ X X_ j\} be an fppf covering. Denote \mathcal{F}_ i, resp. \mathcal{F}_{ijk} the pullback of \mathcal{F} to X_ i, resp. X_{ijk}. Similarly define \mathcal{G}_ i and \mathcal{G}_{ijk}. For every scheme T over B the diagram
\xymatrix{ \mathit{Hom}(\mathcal{F}, \mathcal{G})(T) \ar[r] & \prod \nolimits _ i \mathit{Hom}(\mathcal{F}_ i, \mathcal{G}_ i)(T) \ar@<1ex>[r]^-{\text{pr}_0^*} \ar@<-1ex>[r]_-{\text{pr}_1^*} & \prod \nolimits _{i, j, k} \mathit{Hom}(\mathcal{F}_{ijk}, \mathcal{G}_{ijk})(T) }
presents the first arrow as the equalizer of the other two.
Proof.
Let u_ i : \mathcal{F}_{i, T} \to \mathcal{G}_{i, T} be an element in the equalizer of \text{pr}_0^* and \text{pr}_1^*. Since the base change of an fppf covering is an fppf covering (Topologies on Spaces, Lemma 73.7.3) we see that \{ X_{i, T} \to X_ T\} _{i \in I} and \{ X_{ijk, T} \to X_{i, T} \times _{X_ T} X_{j, T}\} are fppf coverings. Applying Descent on Spaces, Proposition 74.4.1 we first conclude that u_ i and u_ j restrict to the same morphism over X_{i, T} \times _{X_ T} X_{j, T}, whereupon a second application shows that there is a unique morphism u : \mathcal{F}_ T \to \mathcal{G}_ T restricting to u_ i for each i. This finishes the proof.
\square
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