Lemma 68.15.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is integral and induces a bijection $|X| \to |Y|$. Then $X$ is a scheme if and only if $Y$ is a scheme.

**Proof.**
An integral morphism is representable by definition, hence if $Y$ is a scheme, so is $X$. Conversely, assume that $X$ is a scheme. Let $U \subset X$ be an affine open. An integral morphism is closed and $|f|$ is bijective, hence $|f|(|U|) \subset |Y|$ is open as the complement of $|f|(|X| \setminus |U|)$. Let $V \subset Y$ be the open subspace with $|V| = |f|(|U|)$, see Properties of Spaces, Lemma 64.4.8. Then $U \to V$ is integral and surjective, hence $V$ is an affine scheme by Proposition 68.15.2. This concludes the proof.
$\square$

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