Lemma 100.35.6. Let $\mathcal{X}, \mathcal{Y}$ be algebraic stacks étale over an algebraic stack $\mathcal{Z}$. Any morphism $\mathcal{X} \to \mathcal{Y}$ over $\mathcal{Z}$ is étale.

Proof. The morphism $\mathcal{X} \to \mathcal{Y}$ is DM by Lemma 100.4.12. Let $W \to \mathcal{Z}$ be a surjective smooth morphism whose source is an algebraic space. Let $V \to \mathcal{Y} \times _\mathcal {Z} W$ be a surjective étale morphism whose source is an algebraic space (Lemma 100.21.7). Let $U \to \mathcal{X} \times _\mathcal {Y} V$ be a surjective étale morphism whose source is an algebraic space (Lemma 100.21.7). Then

$U \longrightarrow \mathcal{X} \times _\mathcal {Z} W$

is surjective étale as the composition of $U \to \mathcal{X} \times _\mathcal {Y} V$ and the base change of $V \to \mathcal{Y} \times _\mathcal {Z} W$ by $\mathcal{X} \times _\mathcal {Z} W \to \mathcal{Y} \times _\mathcal {Z} W$. Hence it suffices to show that $U \to W$ is étale. Since $U \to W$ and $V \to W$ are étale because $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$ are étale, this follows from the version of the lemma for algebraic spaces, namely Morphisms of Spaces, Lemma 66.39.11. $\square$

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