Lemma 37.6.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open such that $f(U) \subset V$. If $f$ is formally unramified, so is $f|_ U : U \to V$.

Proof. Consider a solid diagram

$\xymatrix{ U \ar[d]_{f|_ U} & T \ar[d]^ i \ar[l]^ a \\ V & T' \ar[l] \ar@{-->}[lu] }$

as in Definition 37.6.1. If $f$ is formally ramified, then there exists at most one $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Hence clearly there exists at most one such morphism into $U$. $\square$

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