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Let $X$ be a smooth irreducible variety over $\mathbb{C}$ and $Y$ be a smooth algebraic space proper over $\mathbb{C}$. Assume $f:X \to Y$ is a morphism such that bijective on $\mathbb{C}$-points.

Question: Is $f$ an isomorphism? Hence $Y$ is a variety?

When $Y$ is already a variety, this is true from Zariski's Main Theorem (see for example Bijection implies isomorphism for algebraic varieties).

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    $\begingroup$ $Y$ can be considered a compact complex manifold by Artin and then $f$ is a holomorphic bijection. Holomorphic bijections have holomorphic inverses. $\endgroup$
    – user178279
    Commented May 29, 2021 at 9:15
  • $\begingroup$ @virkkunen Thanks! Is there any reference about this fact? I mean consider $Y$ as a complex manifold? $\endgroup$
    – Kim
    Commented May 29, 2021 at 10:02
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    $\begingroup$ By miracle flatness (Matsumura, Comm. Ring Theory, Thm 23.1) the morphism $f$ is flat, hence open, hence a homeomorphism, so $Y$ is integral. $\endgroup$
    – Matthieu Romagny
    Commented May 29, 2021 at 10:51
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    $\begingroup$ The same proof from the answer that you link works for algebraic spaces as well as for schemes: Zariski's Main Theorem works for algebraic spaces. $\endgroup$
    – Jason Starr
    Commented May 29, 2021 at 12:34
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    $\begingroup$ @MatthieuRomagny the question is not whether $Y$ is integral (as opposed to just of finite type), but whether $Y$ is even a scheme. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 29, 2021 at 16:30

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Edit: Not an answer to the question. I confused the roles of $X$ and $Y$.

If $X\to Y$ is a quasi-finite separated morphism of algebraic spaces and $Y$ is a scheme, then $X$ is a scheme.

This is, as Jason Starr already commented, a consequence of Zariski's Main Theorem for algebraic spaces.

You are thus reduced to the case of varieties.

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    $\begingroup$ I believe the more relevant result has hypothesis that $X$ is a scheme, rather than that $Y$ is a scheme, cf. mathoverflow.net/questions/4573/… $\endgroup$
    – Jason Starr
    Commented May 30, 2021 at 23:11
  • $\begingroup$ @JasonStarr Whoops, you are right. I misread the roles of $X$ and $Y$. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 6, 2021 at 16:32

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