About sufficient condition for smoothness

Question

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...

Answer 1

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 $k>1$...