Lemma 100.23.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

1. $f$ is quasi-DM,

2. for any morphism $V \to \mathcal{Y}$ with $V$ an algebraic space there exists a surjective, flat, locally finitely presented, locally quasi-finite morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ where $U$ is an algebraic space, and

3. there exist algebraic spaces $U$, $V$ and a morphism $V \to \mathcal{Y}$ which is surjective, flat, and locally of finite presentation, and a morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ which is surjective, flat, locally of finite presentation, and locally quasi-finite.

Proof. The implication (2) $\Rightarrow$ (3) is immediate.

Assume (1) and let $V \to \mathcal{Y}$ be as in (2). Then $\mathcal{X} \times _\mathcal {Y} V \to V$ is quasi-DM, see Lemma 100.4.4. By Lemma 100.4.3 the algebraic space $V$ is DM, hence quasi-DM. Thus $\mathcal{X} \times _\mathcal {Y} V$ is quasi-DM by Lemma 100.4.11. Hence we may apply Theorem 100.21.3 to get the morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ as in (2).

Assume (3). Let $V \to \mathcal{Y}$ and $U \to \mathcal{X} \times _\mathcal {Y} V$ be as in (3). To prove that $f$ is quasi-DM it suffices to show that $\mathcal{X} \times _\mathcal {Y} V \to V$ is quasi-DM, see Lemma 100.4.5. By Lemma 100.4.14 we see that $\mathcal{X} \times _\mathcal {Y} V$ is quasi-DM. Hence $\mathcal{X} \times _\mathcal {Y} V \to V$ is quasi-DM by Lemma 100.4.13 and (1) holds. This finishes the proof of the lemma. $\square$

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