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.
$\square$
Comments (0)