Lemma 70.15.4. 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 66.4.8. Then U \to V is integral and surjective, hence V is an affine scheme by Proposition 70.15.2. This concludes the proof. \square
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