Lemma 65.5.11. Let S be a scheme contained in \mathit{Sch}_{fppf}. Let F be a presheaf of sets on (\mathit{Sch}/S)_{fppf}. Let \mathcal{P} be a property as in Definition 65.5.1. If for every U, V \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf}) and a \in F(U), b \in F(V) we have
h_ U \times _{a, F, b} h_ V is representable, say by the scheme W, and
the morphism W \to U \times _ S V corresponding to h_ U \times _{a, F, b} h_ V \to h_ U \times h_ V has property \mathcal{P},
then \Delta : F \to F \times F is representable and has property \mathcal{P}.
Proof.
Observe that \Delta is representable by Lemma 65.5.10. We can formulate condition (2) as saying that the transformation h_ U \times _{a, F, b} h_ V \to h_{U \times _ S V} has property \mathcal{P}, see Lemma 65.5.3. Consider T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf}) and (a, b) \in (F \times F)(T). Observe that we have the commutative diagram
\xymatrix{ F \times _{\Delta , F \times F, (a, b)} h_ T \ar[d] \ar[r] & h_ T \ar[d]^{\Delta _{T/S}} \\ h_ T \times _{a, F, b} h_ T \ar[r] \ar[d] & h_{T \times _ S T} \ar[d]^{(a, b)} \\ F \ar[r]^\Delta & F \times F }
both of whose squares are cartesian. In this way we see that the morphism F \times _{F \times F} h_ T \to h_ T is the base change of a morphism having property \mathcal{P} by \Delta _{T/S}. Since \mathcal{P} is preserved under base change this finishes the proof.
\square
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