Lemma 105.6.2. For any morphism (105.6.1.2) the map $f' : V' \to U'$ is étale.

**Proof.**
Namely $f : V \to U$ is étale as a morphism in $W_{spaces, {\acute{e}tale}}$ and we can apply Lemma 105.5.2 because $U' \to \mathcal{X}'$ and $V' \to \mathcal{X}'$ are smooth and $U = \mathcal{X} \times _{\mathcal{X}'} U'$ and $V = \mathcal{X} \times _{\mathcal{X}'} V'$.
$\square$

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