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The following result is due to Grothendieck and Artin and can be found in SGA 1. Let $X$ be a normal scheme of finite type over $\mathbb{C}$. Let $\mathfrak{X}'$ be a normal complex analytic space, together with a finite morphism $\mathfrak{f}: \mathfrak{X}' \to X_h$. (We define a finite morphism of analytic spaces to be a proper morphism with finite fibers.) Then there is a unique normal scheme $X'$ and a finite morphism $g: X' \to X$ such that $X_h' \cong \mathfrak{X}'$ and $g_h = \mathfrak{f}$.

I was wondering if anybody could supply me their intuition behind Grothendieck and Artin's proof of the Riemann existence theorem in this generality?

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    $\begingroup$ The result you state (without precise reference) is false: you omitted a crucial etaleness hypothesis! Indeed, Remark 5.5 in Exp. XII gives a counterexample without that condition: let $X = \mathbf{A}^1 - \{0\}$ with coordinate $z$, and $\mathfrak{X}' \rightarrow X_h$ be the cover defined by $T^2 = \sin(1/z)$. The non-etale locus for this finite morphism is the infinite set of points $z = 1/(\pi n)$ for $n \in \mathbf{Z}$, so this cannot be the analytification of a finite map $Y \rightarrow X$, as the formation of the non-etale locus in the base is compatible with analytification. $\endgroup$
    – nfdc23
    Commented Nov 13, 2016 at 23:23
  • $\begingroup$ Typo: I meant to write "$z = 1/(\pi n)$ for $n \in \mathbf{Z} - \{0\}$" above. $\endgroup$
    – nfdc23
    Commented Nov 14, 2016 at 1:34

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