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Nov 23, 2010 at 19:40 vote accept Kevin M Pilgrim
Nov 23, 2010 at 19:27 history edited Sergei Ivanov CC BY-SA 2.5
fixed tupos
Nov 23, 2010 at 19:22 comment added BCnrd Dear Sergei: Shouldn't $\delta$ and $\varepsilon$ also be swapped at the end of the first sentence of (5)? (That is, for $S$ of diameter at least $\varepsilon$, $f(S)$ has diameter at least $\delta$. We seek a uniform positive lower bound on the diameter of $f(S)$, given such a lower bound on the diameter of $S$.)
Nov 23, 2010 at 18:55 comment added Sergei Ivanov @BCnrd: thanks, renamed $\delta_0$ to $\varepsilon_0$. For finiteness of pre-images, I had in mind the following argument: the first coordinates of roots of $f$ are roots of some polynomial $f_1:\mathbb C\to\mathbb C$ (that can be written explicitly via resultants or whatever). If this polynomial is zero, then the set of roots of $f$ is non-compact and hence non-proper. Otherwise the number of possible values of first coordinates of roots of $f$ is bounded by the degree of $f_1$ which is bounded in terms of the degree of $f$. Applying this argument to other coordinates yields the result.
Nov 23, 2010 at 18:46 history edited Sergei Ivanov CC BY-SA 2.5
clarified notation
Nov 23, 2010 at 17:48 comment added BCnrd Dear Buschi: Your example $f$ is not proper. Sergei is correct about (1) because a proper affine morphism has finite fibers (and is even a finite morphism) and a map between separated finite type $\mathbf{C}$-schemes is proper if and only if the map on $\mathbf{C}$-points is topologically proper (for the analytic topology). There is a small typo in the argument, however: near the end of the first sentence of (5) the roles of $\varepsilon$ and $\delta$ should be swapped (and later in (5) the role of $\delta_0$ should be called $\varepsilon_0$, though logically it is OK to write $\delta_0$).
Nov 23, 2010 at 14:14 comment added Buschi Sergio You write: (1) The pre-image of every point if finite... but let $g: C^n\to C$ polynomial, and $f:=(g,..,g): C^n\to C^n$ then $f^{-1}(z,...,z)=g^{-1}(z)$ is a (n-1)-variety.
Nov 23, 2010 at 8:06 history answered Sergei Ivanov CC BY-SA 2.5