Proof.
By Lemma 38.11.2 the morphism f is locally of finite presentation. The image f(X) \subset S is open (Morphisms, Lemma 29.25.10) hence we may replace S by f(X). Thus we have to prove that f is an isomorphism. We may assume S is affine. We can reduce to the case that X is quasi-compact because it suffices to show that any quasi-compact open X' \subset X whose image is S maps isomorphically to S. Thus we may assume f is quasi-compact. All the fibers of f have dimension 0, see Lemma 38.11.3. Hence f is quasi-finite, see Morphisms, Lemma 29.29.5. Let s \in S. Choose an elementary étale neighbourhood g : (T, t) \to (S, s) such that X \times _ S T = V \amalg W with V \to T finite and W_ t = \emptyset , see More on Morphisms, Lemma 37.41.6. Denote \pi : V \amalg W \to T the given morphism. Since \pi is flat and locally of finite presentation, we see that \pi (V) is open in T (Morphisms, Lemma 29.25.10). After shrinking T we may assume that T = \pi (V). Since f is an isomorphism over U we see that \pi is an isomorphism over g^{-1}U. Since \pi (V) = T this implies that \pi ^{-1}g^{-1}U is contained in V. By Morphisms, Lemma 29.25.15 we see that \pi ^{-1}g^{-1}U \subset V \amalg W is scheme theoretically dense. Hence we deduce that W = \emptyset . Thus X \times _ S T = V is finite over T. This implies that f is finite (after replacing S by an open neighbourhood of s), for example by Descent, Lemma 35.23.23. Then f is finite locally free (Morphisms, Lemma 29.48.2) and after shrinking S to a smaller open neighbourhood of s we see that f is finite locally free of some degree d (Morphisms, Lemma 29.48.5). But d = 1 as is clear from the fact that the degree is 1 over the dense open U. Hence f is an isomorphism.
\square
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