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Dear Brian.

The dimension 1 case is very special. We assume $X$ no compact (Stein if we want), normal and of dimension >1...

In fact, i want to prove the following:

Let $f:X\rightarrow S$ be an open and proper surjective map of reduced complex spaces with $n$-dimensional fibers. Assume moreover that $f$ is geometrically flat which means that the fibers can be endowed with cycle structure s.t $X$ can be seen as a graph of analytic family of cycles. I can prove that $X$ normal (resp. weakly normal) implies $S$ normal (resp. weakly normal). Analogon of this can be found in EGA4, chap.6 for FLAT morphism.

I want to prove now that we have a similar result with smoothness.

For this, i reduce the problem to a finite, open surjective map $f:X\rightarrow S$ with $S$ normal. Then GEOMETRICALLY FLATNESS is giving by a holomorphic map $F:S\rightarrow {\rm Sym}^{k}(X)$ where ${\rm Sym}^{k}(X):=X^{k}/\sigma_{k}$ and then equivalent to say that $f$ is $k$-branched covering. With this, the fundamental question is:

If $f:X\rightarrow S$ is $k$-branched covering on normal complex space $S$, it is true that $X$ smooth implies $S$ smooth ?

Rk: I have many problem with the Tex font and to add a comments...

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I believe that this is a continuation of mathoverflow.net/questions/40099/… in which case any answers or further comments should go there. If you (kaddar) are having problems with posting or commenting then that should be reported on meta or directly to the moderators. If you wish to clarify your original question, you should edit that rather than reposting. –  Andrew Stacey Sep 27 '10 at 9:11
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1 Answer

The answer is no. We can take the projection $f:({\Bbb C}^{p})^{k}\rightarrow {\rm Sym}^{k}({\Bbb C}^{p})$ Then $f$ is a branched covering of degre $k!$ on normal space which is never smooth for $k>1$...

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