Lemma 37.11.5. 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 smooth, so is $f|_ U : U \to V$.
Proof. Consider a solid diagram
as in Definition 37.11.1. If $f$ is formally smooth, then there exists an $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Since the underlying sets of $T$ and $T'$ are the same we see that $a'$ is a morphism into $U$ (see Schemes, Section 26.3). And it clearly is a $V$-morphism as well. Hence the dotted arrow above as desired. $\square$
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