Lemma 66.6.6. Let $S$ be a scheme. Let $p : X \to \mathop{\mathrm{Spec}}(k)$ be a morphism of algebraic spaces over $S$ where $k$ is a field. Then $p : X \to \mathop{\mathrm{Spec}}(k)$ is universally open.

Proof. Choose a scheme $U$ and a surjective étale morphism $U \to X$. The composition $U \to \mathop{\mathrm{Spec}}(k)$ is universally open (as a morphism of schemes) by Morphisms, Lemma 29.23.4. Let $Z \to \mathop{\mathrm{Spec}}(k)$ be a morphism of schemes. Then $U \times _{\mathop{\mathrm{Spec}}(k)} Z \to X \times _{\mathop{\mathrm{Spec}}(k)} Z$ is surjective, see Lemma 66.5.5. Hence the first of the maps

$|U \times _{\mathop{\mathrm{Spec}}(k)} Z| \to |X \times _{\mathop{\mathrm{Spec}}(k)} Z| \to |Z|$

is surjective. Since the composition is open by the above we conclude that the second map is open as well. Whence $p$ is universally open by Lemma 66.6.5. $\square$

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