Lemma 100.46.1. Let $\mathcal{X}$ be an algebraic stack. The following are equivalent

1. there is a surjective smooth morphism $U \to \mathcal{X}$ where $U$ is a scheme such that every quasi-compact open of $U$ has finitely many irreducible components,

2. for every scheme $U$ and every smooth morphism $U \to \mathcal{X}$ every quasi-compact open of $U$ has finitely many irreducible components,

3. for every algebraic space $Y$ and smooth morphism $Y \to \mathcal{X}$ the space $Y$ satisfies the equivalent conditions of Morphisms of Spaces, Lemma 66.49.1, and

4. for every quasi-compact algebraic stack $\mathcal{Y}$ smooth over $\mathcal{X}$ the space $|\mathcal{Y}|$ has finitely many irreducible components.

Proof. The equivalence of (1), (2), and (3) follow from Descent, Lemma 35.16.3, Properties of Stacks, Lemma 99.7.1, and Morphisms of Spaces, Lemma 66.49.1. It is also clear from these references that condition (4) implies condition (1). Conversely, assume the equivalent conditions (1), (2), and (3) hold and let $\mathcal{Y} \to \mathcal{X}$ be a smooth morphism of algebraic stacks with $\mathcal{Y}$ quasi-compact. Then we can choose an affine scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$ by Properties of Stacks, Lemma 99.6.2. Since $V$ has finitely many irreducible components by (2) and since $|V| \to |\mathcal{Y}|$ is surjective and continuous, we conclude that $|\mathcal{Y}|$ has finitely many irreducible components by Topology, Lemma 5.8.5. $\square$

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