Lemma 66.4.3. Let $S$ be a scheme. Let
be a cartesian diagram of algebraic spaces over $S$. Then the map of sets of points
is surjective.
Lemma 66.4.3. Let $S$ be a scheme. Let
be a cartesian diagram of algebraic spaces over $S$. Then the map of sets of points
is surjective.
Proof. Namely, suppose given fields $K$, $L$ and morphisms $\mathop{\mathrm{Spec}}(K) \to X$, $\mathop{\mathrm{Spec}}(L) \to Z$, then the assumption that they agree as elements of $|Y|$ means that there is a common extension $M/K$ and $M/L$ such that $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(K) \to X \to Y$ and $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(L) \to Z \to Y$ agree. And this is exactly the condition which says you get a morphism $\mathop{\mathrm{Spec}}(M) \to Z \times _ Y X$. $\square$
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Comment #2039 by Keenan Kidwell on
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