Lemma 68.9.1. Let $S$ be a scheme. Let $j : V \to Y$ be a quasi-compact open immersion of algebraic spaces over $S$. Let $\pi : Z \to V$ be an integral morphism. Then there exists an integral morphism $\nu : Y' \to Y$ such that $Z$ is $V$-isomorphic to the inverse image of $V$ in $Y'$.

**Proof.**
Since both $j$ and $\pi $ are quasi-compact and separated, so is $j \circ \pi $. Let $\nu : Y' \to Y$ be the normalization of $Y$ in $Z$, see Morphisms of Spaces, Section 67.48. Of course $\nu $ is integral, see Morphisms of Spaces, Lemma 67.48.5. The final statement follows formally from Morphisms of Spaces, Lemmas 67.48.4 and 67.48.10.
$\square$

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