Definition 66.6.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

1. We say $f$ is open if the map of topological spaces $|f| : |X| \to |Y|$ is open.

2. We say $f$ is universally open if for every morphism of algebraic spaces $Z \to Y$ the morphism of topological spaces

$|Z \times _ Y X| \to |Z|$

is open, i.e., the base change $Z \times _ Y X \to Z$ is open.

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