Lemma 66.22.5. Let $S$ be a scheme. Let $f : X \to Y$ be an étale morphism of algebraic spaces over $S$. Let $x \in X$. Then (1) $\dim _ x(X) = \dim _{f(x)}(Y)$ and (2) the dimension of the local ring of $X$ at $x$ equals the dimension of the local ring of $Y$ at $f(x)$. If $f$ is surjective, then (3) $\dim (X) = \dim (Y)$.

**Proof.**
Choose a scheme $U$ and a point $u \in U$ and an étale morphism $U \to X$ which maps $u$ to $x$. Then the composition $U \to Y$ is also étale and maps $u$ to $f(x)$. Thus the statements (1) and (2) follow as the relevant integers are defined in terms of the behaviour of the scheme $U$ at $u$. See Definition 66.9.1 for (1). Part (3) is an immediate consequence of (1), see Definition 66.9.2.
$\square$

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