**Proof.**
Assume (1). Then $f$ is quasi-DM by assumption. Let $V \to \mathcal{Y}$ and $U \to \mathcal{X} \times _\mathcal {Y} V$ be as in (2). By Lemma 101.23.5 the composition $U \to \mathcal{X} \times _\mathcal {Y} V \to V$ is locally quasi-finite. Thus (1) implies (2).

Assume (2). Let $V \to \mathcal{Y}$ be as in (3). By Lemma 101.23.6 we can find an algebraic space $U$ and a surjective, flat, locally finitely presented, locally quasi-finite morphism $U \to \mathcal{X} \times _\mathcal {Y} V$. By (2) the composition $U \to V$ is locally quasi-finite. Thus (2) implies (3).

It is immediate that (3) implies (4).

Assume (4). We will prove (1) holds, which finishes the proof. By Lemma 101.23.6 we see that $f$ is quasi-DM. To prove that $f$ is locally of finite type it suffices to prove that $g : \mathcal{X} \times _\mathcal {Y} V \to V$ is locally of finite type, see Lemma 101.17.6. Then it suffices to check that $g$ precomposed with $h : U \to \mathcal{X} \times _\mathcal {Y} V$ is locally of finite type, see Lemma 101.17.7. Since $g \circ h : U \to V$ was assumed to be locally quasi-finite this holds, hence $f$ is locally of finite type. Finally, let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ be a morphism. Then $V \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k)$ is a nonempty algebraic space which is locally of finite presentation over $k$. Hence we can find a finite extension $k'/k$ and a morphism $\mathop{\mathrm{Spec}}(k') \to V$ such that

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[r] \ar[d] & V \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & \mathcal{Y} } \]

commutes (details omitted). Then $\mathcal{X}_{k'} \to \mathcal{X}_ k$ is representable (by schemes), surjective, and finite locally free. In particular $|\mathcal{X}_{k'}| \to |\mathcal{X}_ k|$ is surjective and open. Thus it suffices to prove that $|\mathcal{X}_{k'}|$ is discrete. Since

\[ U \times _ V \mathop{\mathrm{Spec}}(k') = U \times _{\mathcal{X} \times _\mathcal {Y} V} \mathcal{X}_{k'} \]

we see that $U \times _ V \mathop{\mathrm{Spec}}(k') \to \mathcal{X}_{k'}$ is surjective, flat, and locally of finite presentation (as a base change of $U \to \mathcal{X} \times _\mathcal {Y} V$). Hence $|U \times _ V \mathop{\mathrm{Spec}}(k')| \to |\mathcal{X}_{k'}|$ is surjective and open. Thus it suffices to show that $|U \times _ V \mathop{\mathrm{Spec}}(k')|$ is discrete. This follows from the fact that $U \to V$ is locally quasi-finite (either by our definition above or from the original definition for morphisms of algebraic spaces, via Morphisms of Spaces, Lemma 67.27.5).
$\square$

## Comments (2)

Comment #788 by Matthew Emerton on

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