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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