I am looking for a reference to something like that: if $f\colon X\to Y$ is a finite (i.e., proper with finite fibers) morphism of reduced and irreducible normal (or at least smooth) complex spaces such that $f$ is 1-1 over $U\subset Y$, where $U$ is open and dense, then $f$ is an isomorphism.

Could somebody help me?

Thanks in advance, Serge

  • $\begingroup$ "Irreducible normal"="connected normal", and $U$ non-empty open suffices. Tensor the injection $O_Y\rightarrow f_{\ast}O_X$ between coherent sheaves of algebras (with reduced stalks) by the sheaf $M_Y$ of "meromorphic functions", so $M_Y \hookrightarrow M_Y\otimes_{O_Y}f_{\ast}(O_X)\hookrightarrow f_{\ast}(M_X)$ with reduced stalks. Stalks of $M_Y$ are fields, so stalks of the middle term are finite products of fields, so the 2nd map is an isomorphism. Make local $M_Y$-bases from stalks to show $f_{\ast}(M_X)$ is locally free over $M_Y$. The rank $r$ is constant by connectedness...[cont'd] $\endgroup$
    – user29720
    Commented Jun 20, 2013 at 14:30
  • $\begingroup$ Added the tag ag.algebraic-geometry, since this question is surely of interest for those working in that field. $\endgroup$ Commented Jun 20, 2013 at 14:47
  • $\begingroup$ and it suffices to show $r=1$ (as then $f_{\ast}(O_X)$ has reduced stalks that contain $O_Y$ with the same total ring of fractions, so $f_{\ast}(O_X)=O_Y$ by normality of $Y$, so $f$ is an isomorphism by the link of finiteness and coherent sheaves of algebras). The reduced stalks of $f_{\ast}(M_X)$ are products of finite separable extensions of stalk fields of $M_Y$. Since char. 0, write it in "primitive element" form and denominator-chase to get a dense open over which $f_{\ast}(O_X)$ is $O_Y$-finite etale of rank $r$. Restrict over $U$ to get $r=1$. $\endgroup$
    – user29720
    Commented Jun 20, 2013 at 16:43

1 Answer 1


One reference is Proposition 14.7 in Remmert's paper Local Theory of complex analytic spaces, Several complex variable VII, Encyclopaedia of Math. Sci. vol 74. For the reader's convenience I will restate the result here.

Recall that a finite, surjective, holomorphic map $\eta \colon X \to Y$ between complex spaces is called a one sheeted (analytic) covering if there exists a (critical) thin set $A \subset Y$ such that $\eta^{-1} A$ is thin in $X$ and $\eta \colon X \setminus \eta^{-1}(A) \to Y \setminus A$ is biholomorphic. Then Remmert's statement essentially says that normalizations "dominate" all one-sheeted coverings:

Proposition. Let $\eta \colon X \to Y$, $\xi \colon Z \to Y$ be one-sheeted coverings. If $X$ is normal, there exists a unique holomorphic map $g \colon X \to Z$ such that $\eta= \xi \circ g$. If moreover $Z$ is normal, the map $g$ is biholomorphic.

In your setting, take $\eta=f$ and $g=1_Y$. Then if both $X$ and $Y$ are normal Remmert's Proposition tells us that $f$ is a biholomorphism.

  • $\begingroup$ To apply this, it seems one has to verify that $Y-U$ is a "thin set". That can certainly be done by using "generic flatness" results in analytic geometry. The version of the argument in my comment was meant to avoid that (depending on where one is getting $U$ from). But perhaps Serge knows thinness is the situation of interest. $\endgroup$
    – user29720
    Commented Jun 20, 2013 at 16:48
  • $\begingroup$ Right. But in many natural contexts (for instance, when $Y-U$ is the zero locus of a holomorphic function) thinnes condition is automatically satisfied. $\endgroup$ Commented Jun 20, 2013 at 17:46

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