Lemma 101.36.9. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

$f$ is unramified, and

$f$ is locally of finite type and its diagonal is étale.

Lemma 101.36.9. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

$f$ is unramified, and

$f$ is locally of finite type and its diagonal is étale.

**Proof.**
Assume $f$ is unramified. Then $f$ is DM hence we can choose algebraic spaces $U$, $V$, a smooth surjective morphism $V \to \mathcal{Y}$ and a surjective étale morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ (Lemma 101.21.7). Since $f$ is unramified the induced morphism $U \to V$ is unramified. Thus $U \to V$ is locally of finite type (Morphisms of Spaces, Lemma 67.38.6) and we conclude that $f$ is locally of finite type. The diagonal $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is a morphism of algebraic stacks over $\mathcal{Y}$. The base change of $\Delta $ by the surjective smooth morphism $V \to \mathcal{Y}$ is the diagonal of the base change of $f$, i.e., of $\mathcal{X}_ V = \mathcal{X} \times _\mathcal {Y} V \to V$. In other words, the diagram

\[ \xymatrix{ \mathcal{X}_ V \ar[r] \ar[d] & \mathcal{X}_ V \times _ V \mathcal{X}_ V \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

is cartesian. Since the right vertical arrow is surjective and smooth it suffices to show that the top horizontal arrow is étale by Properties of Stacks, Lemma 100.3.4. Consider the commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r] & U \times _ V U \ar[d] \\ \mathcal{X}_ V \ar[r] & \mathcal{X}_ V \times _ V \mathcal{X}_ V } \]

All arrows are representable by algebraic spaces, the vertical arrows are étale, the left one is surjective, and the top horizontal arrow is an open immersion by Morphisms of Spaces, Lemma 67.38.9. This implies what we want: first we see that $U \to \mathcal{X}_ V \times _ V \mathcal{X}_ V$ is étale as a composition of étale morphisms, and then we can use Properties of Stacks, Lemma 100.3.5 to see that $\mathcal{X}_ V \to \mathcal{X}_ V \times _ V \mathcal{X}_ V$ is étale because being étale (for morphisms of algebraic spaces) is local on the source in the étale topology (Descent on Spaces, Lemma 74.19.1).

Assume $f$ is locally of finite type and that its diagonal is étale. Then $f$ is DM by definition (as étale morphisms of algebraic spaces are unramified). As above this means we can choose algebraic spaces $U$, $V$, a smooth surjective morphism $V \to \mathcal{Y}$ and a surjective étale morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ (Lemma 101.21.7). To finish the proof we have to show that $U \to V$ is unramified. We already know that $U \to V$ is locally of finite type. Arguing as above we find a commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r] & U \times _ V U \ar[d] \\ \mathcal{X}_ V \ar[r] & \mathcal{X}_ V \times _ V \mathcal{X}_ V } \]

where all arrows are representable by algebraic spaces, the vertical arrows are étale, and the lower horizontal one is étale as a base change of $\Delta $. It follows that $U \to U \times _ V U$ is étale for example by Lemma 101.35.6^{1}. Thus $U \to U \times _ V U$ is an étale monomorphism hence an open immersion (Morphisms of Spaces, Lemma 67.51.2). Then $U \to V$ is unramified by Morphisms of Spaces, Lemma 67.38.9.
$\square$

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