Lemma 67.22.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

1. $X$ and $Y$ are decent and have finitely many irreducible components,

2. $f$ is integral and birational,

3. $Y$ is normal, and

4. $X$ is reduced.

Then $f$ is an isomorphism.

Proof. Let $V \to Y$ be an étale morphism with $V$ affine. It suffices to show that $U = X \times _ Y V \to V$ is an isomorphism. By Lemma 67.22.6 and its proof we see that $U$ and $V$ are decent and have finitely many irreducible components and that $U \to V$ is birational. By Properties, Lemma 28.7.5 $V$ is a finite disjoint union of integral schemes. Thus we may assume $V$ is integral. As $f$ is birational, we see that $U$ is irreducible and reduced, i.e., integral (note that $U$ is a scheme as $f$ is integral, hence representable). Thus we may assume that $X$ and $Y$ are integral schemes and the result follows from the case of schemes, see Morphisms, Lemma 29.54.8. $\square$

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