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

\[ \xymatrix{ Z \times _ Y X \ar[r] \ar[d] & X \ar[d] \\ Z \ar[r] & Y } \]

be a cartesian diagram of algebraic spaces over $S$. Then the map of sets of points

\[ |Z \times _ Y X| \longrightarrow |Z| \times _{|Y|} |X| \]

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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## Comments (2)

Comment #2039 by Keenan Kidwell on

Comment #2077 by Johan on

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