**Proof.**
Assume (1). Then $f$ is DM and since being unramified is preserved by base change, we see that (2) holds.

Assume (2). Choose a scheme $V$ and a surjective étale morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a surjective étale morphism $U \to V \times _\mathcal {Y} \mathcal{X}$ (Lemma 100.21.7). Thus we see that (3) holds.

Assume $W \to \mathcal{Y}$ and $T \to W \times _\mathcal {Y} \mathcal{X}$ are as in (3). We first check $f$ is DM. Namely, it suffices to check $W \times _\mathcal {Y} \mathcal{X} \to W$ is DM, see Lemma 100.4.5. By Lemma 100.4.12 it suffices to check $W \times _\mathcal {Y} \mathcal{X}$ is DM. This follows from the existence of $T \to W \times _\mathcal {Y} \mathcal{X}$ by (the easy direction of) Theorem 100.21.6.

Assume $f$ is DM and $W \to \mathcal{Y}$ and $T \to W \times _\mathcal {Y} \mathcal{X}$ are as in (3). Let $V$ be an algebraic space, let $V \to \mathcal{Y}$ be surjective smooth, let $U$ be an algebraic space, and let $U \to V \times _\mathcal {Y} \mathcal{X}$ is surjective and étale (Lemma 100.21.7). We have to check that $U \to V$ is unramified. It suffices to prove $U \times _\mathcal {Y} W \to V \times _\mathcal {Y} W$ is unramified by Descent on Spaces, Lemma 73.11.27. We may replace $\mathcal{X}, \mathcal{Y}, W, T, U, V$ by $\mathcal{X} \times _\mathcal {Y} W, W, W, T, U \times _\mathcal {Y} W, V \times _\mathcal {Y} W$ (small detail omitted). Thus we may assume that $Y = \mathcal{Y}$ is an algebraic space, there exists an algebraic space $T$ and a surjective étale morphism $T \to \mathcal{X}$ such that $T \to Y$ is unramified, and $U$ and $V$ are as before. In this case we know that

\[ U \to V\text{ is unramified} \Leftrightarrow \mathcal{X} \to Y\text{ is unramified} \Leftrightarrow T \to Y\text{ is unramified} \]

by the equivalence of properties (1) and (2) of Lemma 100.34.1 and Definition 100.36.1. This finishes the proof.
$\square$

## Comments (2)

Comment #7770 by Anonymous on

Comment #8010 by Stacks Project on