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My question is the following:

Let $b:X \rightarrow Y$ be a proper holomorphic mapping between two Stein manifolds $X$ and $Y$. For obvious reasons, $b^{-1}(y)$ are finite subsets of $X$. Is the set $\lbrace \# b^{-1}(y)\, \colon\, y \in Y\rbrace $ bounded?

Here is the background: I'm trying to understand the equisingular stratification lemma in the paper "Eliashberg and Gromov: Embeeding of Stein manifold of dimension $n$ into the affine space of dimension $\frac{3}{2}n+1$, Annals of mathematics 1992".The anwser to the above question is yes according to their lemma.And if the anwser is yes,I can prove the existence of equisingular stratification.

A possible way to solve the question is: Denote by $A_i$ the subset of $Y$ consists of points $y \in Y$ with $\#b^{-1}(y) \geq i$, then they are analytic subsets of $Y$. If the number of irreducible components of $A_i$ are finite,then the anwser is yes. The problem is that $Y$ is not a compact manifold and $A_i$ may have infinite many components.

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  • $\begingroup$ You must be assuming $Y$ is connected. Are you assuming $X$ is connected or $b$ is surjective? It is false without at least one of those, so please state all of your hypotheses. $\endgroup$
    – user30180
    Commented May 31, 2013 at 14:48
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    $\begingroup$ If $X$ is connected and $b$ is surjective then $X$ and $Y$ have the same dimension at all points, so $b$ is necessarily flat (in the sense of maps of local rings), so the coherent sheaf $b_{\ast}(O_X)$ is a vector bundle on $Y$. The rank of this vector bundle must be constant since $Y$ is connected, and it provides a uniform upper bound on the size of fibers. Are you interested in cases with either $X$ disconnected or $b$ not surjective? $\endgroup$
    – user30180
    Commented May 31, 2013 at 15:09
  • $\begingroup$ I'm sorry for the ambiguous.The map $b$ is in general not surjective.$X$ and $Y$ are both smooth and connected.Thank you for the comments. $\endgroup$
    – Jun Li
    Commented Jun 2, 2013 at 6:05

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