Lemma 65.4.3. 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 $\mathop{\mathrm{Spec}}(M) \to Z \times _ Y X$. $\square$

Comment #2039 by Keenan Kidwell on

The scheme $S$ introduced in the first sentence is never mentioned again. Presumably one wants the algebraic spaces to be algebraic spaces over $S$?

There are also:

• 4 comment(s) on Section 65.4: Points of algebraic spaces

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).