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Suppose that $Y$ is an algebraic space over $\mathbb{C}$ which is "nice" (say, separated and of finite type). Let $R\subset Y\times Y$ be a closed algebraic subspace which forms a categorical equivalence relation. Now, suppose that the quotient of $Y$ by $R$ makes sense in the analytic category. That is, there is a complex analytic space $Z^{an}$ and a map $f:Y^{an}\rightarrow Z^{an}$ such that

  1. $f$ is the categorical quotient of $Y^{an}$ by $R^{an}$,
  2. The complex points of $Z^{an}$ parametrize the $R^{an}$-equivalence classes of $Y^{an}$, and
  3. The sheaf $O_{Z^{an}}$ coincides with the subsheaf of elements of $f_*O_{Y^{an}}$ that have the same pullback along the two projections $\pi_i:R^{an}\to Y^{an}$. (This can be ignored if we assumed $Y,R$ are reduced)

Does it follows that there is a categorical quotient $Z$ of $Y$ by $R$ in the category of algebraic spaces, whose analytification is $Z^{an}$?

I'd also greatly appreciate any references to "GAGA" results of this type. Thanks!

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    $\begingroup$ Is your categorical quotient $f$ a uniform categorical quotient? If so, then $f$ is flat over a dense open subset of $Z^{\text{an}}$. This implies that $\pi_1:R\to Y$ is flat over a dense open subset of $Y$. Then you can play the "usual game": choose $i:Y\hookrightarrow \overline{Y}$, a Nagata compactification; form the closure $\overline{R}$ of $R$ in $Y\times \overline{Y}$; this is flat over a dense open $U$ of $Y$; use this to embed a dense open subset of $Z^{\text{an}}$ as a locally closed subspace of the Hilbert algebraic space of $\overline{Y}$. $\endgroup$
    – Jason Starr
    Commented Jul 11, 2018 at 20:36
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    $\begingroup$ @JasonStarr Why can't one take the quotient in the category of algebraic stacks? Call this $\mathcal{Z}$. By the "niceness" assumptions this should be a finitely presented algebraic stack. now, by assumption, $\mathcal{Z}^{an}$ is a complex analytic space. Thus, the inertia groups are trivial of $\mathcal{Z}^{an}$ hence $\mathcal{Z}$ are trivial. Thus, $\mathcal{Z}$ is an algebraic space. Where does this line of reasoning go wrong? $\endgroup$
    – Ariyan Javanpeykar
    Commented Jul 11, 2018 at 21:50
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    $\begingroup$ @AriyanJavanpeykar. That might work. However, the OP did not state that $R$ is flat over (all of) $Y$. So it is not immediately clear how to form an fppf stack out of $Y$ and $R$. My comment is that, if there is a "good" quotient in the category of analytic spaces, then that good quotient is bimeromorphic to a subspace of the Hilbert algebraic space of any Nagata compactification. It should be possible to use this to prove that the quotient is an algebraic space. $\endgroup$
    – Jason Starr
    Commented Jul 11, 2018 at 21:58
  • $\begingroup$ @JasonStarr, I'm not familiar with the notion of uniform categorical quotients. But I don't know how to proceed once one gets a dense open subset of Z^{an} is, say, even an affine scheme. In the case I'm thinking about, thats actually easy to see. $\endgroup$
    – jacob
    Commented Jul 12, 2018 at 1:07
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    $\begingroup$ @jacob. Since you clearly have additional hypotheses you are willing to impose, I suggest you tell us what they are. Anyway, the "usual game" goes as follows: for the dense open subscheme U of Y, form the morphism $g:U\to \text{Hilb}_{\overline{Y}}$. Now define $\widetilde{Y}$ to be the closure of the graph of $U$ in the product $\overline{Y}\times \text{Hilb}_{\overline{Y}}$. The first projection, $p_1:\widetilde{Y}\to \overline{Y}$ is proper and birational. The image of the second projection, $p_2:\widetilde{Y}\to \widetilde{Z}$, is a proper algebraic space bimeromorphic to $Z$. $\endgroup$
    – Jason Starr
    Commented Jul 12, 2018 at 6:28

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