The Stacks project

Lemma 66.6.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. The following are equivalent

  1. $f$ is universally open (in the sense of Section 66.3), and

  2. for every morphism of algebraic spaces $Z \to Y$ the morphism of topological spaces $|Z \times _ Y X| \to |Z|$ is open.

Proof. Assume (1), and let $Z \to Y$ be as in (2). Choose a scheme $V$ and a surjective ├ętale morphism $V \to Y$. By assumption the morphism of schemes $V \times _ Y X \to V$ is universally open. By Properties of Spaces, Section 65.4 in the commutative diagram

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

the horizontal arrows are open and surjective, and moreover

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

is surjective. Hence as the left vertical arrow is open it follows that the right vertical arrow is open. This proves (2). The implication (2) $\Rightarrow $ (1) is immediate from the definitions. $\square$


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