## in the analytic category, finite morphisms are open maps?

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Work locally, suppose X and Y are open subsets of C^n where C is the complex number field. Suppose f: X--->Y is a map given by n polynomials. If f is quasi-finite (i.e. each fiber is a finite set) and surjective, then is f an open map?

Another question is about finite morphism in the analytic category. Let X and Y be complex manifolds and f:X--->Y an analytic (i.e. holomorphic) map. I guess there is a notion of f being finite, and a definition is that the induced maps between the local rings of germs of analytic functions are finite ring homomorphisms. Are there equivalent and more transparent characterizations of f being finite?

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For finiteness, your proposed criterion only implies the "right" notion locally on source and target (think of an open embedding). The very self-contained book "Coherent Analytic Sheaves" answers these (in the affirmative) and many other related questions. (A more transparent definition of finiteness in the connected manifold setting is being either proper with finite fibers or being "classified" by a coherent sheaf of $O_Y$-algebras that is locally free of finite rank; equivalences among possible definitions of analytic finiteness are much deeper than in the algebraic theory.) – BCnrd Apr 17 2010 at 17:47

## 2 Answers

Open Mapping Theorem [Grauert-Remmert: Coherent analytic sheaves, p.107] Let $X,Y$ be pure $d$-dimensional complex spaces and assume that $Y$ is locally irreducible. Then any holomorphic map $f:X\to Y$ with discrete fibers is open.

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cf. same book p.69: Criterion of Openness. – Sándor Kovács Oct 11 2010 at 1:03

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A finite morphism $f:X-->S$ between complex manifolds is open. Because, in the general case, if the target is pure dimensionnal, the base locally irreducible and the fibers have the same dimension then the map is open (open mapping Remmert theorem in Fischer book). If the base is not locally irreducible but we can endowed the fibers with appropriate multiplicities so that becamme analytic family of cycles then the map is necessarily open. In the finite case, it means that we have a trace map compatible with cycle structure $f_{*}O_X---->O_S$ or a holomorphic map $S--> Sym^{k}(X)$ where k is the generic degre of the branched covering defined by f.

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The (strong) normalization is finite but no open. The weak normalization is open..... – kaddar Jun 11 2010 at 20:07
Read " source" instead "target" below! – kaddar Jun 11 2010 at 20:28

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In the first statement you may want to add dim(X)=dim(S), as closed immersions are finite but rarely open. – Qing Liu Jun 14 2010 at 21:20
Excuse me. Yes, of course, $dim(X)=dim (S)$. – kaddar Jun 18 2010 at 9:22
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