Lemma 68.22.9. Let $S$ be a scheme. Let $f : X \to Y$ be an integral birational morphism of decent algebraic spaces over $S$ which have finitely many irreducible components. Then there exists a factorization $Y^\nu \to X \to Y$ and $Y^\nu \to X$ is the normalization of $X$.

**Proof.**
Consider the map $X^\nu \to Y^\nu $ of Lemma 68.22.7. This map is integral by Morphisms of Spaces, Lemma 67.45.12. Hence it is an isomorphism by Lemma 68.22.8.
$\square$

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