Lemma 65.11.4. Let S be a scheme contained in \mathit{Sch}_{fppf}. Let F, G be algebraic spaces over S. Let G \to F be a representable morphism. Let U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf}), and q : U \to F surjective and étale. Set V = G \times _ F U. Finally, let \mathcal{P} be a property of morphisms of schemes as in Definition 65.5.1. Then G \to F has property \mathcal{P} if and only if V \to U has property \mathcal{P}.
Proof. (This lemma follows from Lemmas 65.5.5 and 65.5.6, but we give a direct proof here also.) It is clear from the definitions that if G \to F has property \mathcal{P}, then V \to U has property \mathcal{P}. Conversely, assume V \to U has property \mathcal{P}. Let T \to F be a morphism from a scheme to F. Let T' = T \times _ F G which is a scheme since G \to F is representable. We have to show that T' \to T has property \mathcal{P}. Consider the commutative diagram of schemes
\xymatrix{ V \ar[d] & T \times _ F V \ar[d] \ar[l] \ar[r] & T \times _ F G \ar[d] \ar@{=}[r] & T' \\ U & T \times _ F U \ar[l] \ar[r] & T }
where both squares are fibre product squares. Hence we conclude the middle arrow has property \mathcal{P} as a base change of V \to U. Finally, \{ T \times _ F U \to T\} is a fppf covering as it is surjective étale, and hence we conclude that T' \to T has property \mathcal{P} as it is local on the base in the fppf topology. \square
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