Lemma 37.8.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open subschemes such that $f(U) \subset V$. If $f$ is formally étale, so is $f|_ U : U \to V$.
Proof. Consider a solid diagram
as in Definition 37.8.1. If $f$ is formally ramified, then there exists exactly one $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Since $|T'| = |T|$ we conclude that $a'(T') \subset U$ which gives our unique morphism from $T'$ into $U$. $\square$
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