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Suppose $f: \mathbb{C}^n \to \mathbb{C}^n$ is a proper polynomial mapping and $\gamma: [0,1] \to \mathbb{C}^n$ is a continuous path. Further, suppose $z_0 \in \mathbb{C}^n$ satisfies $f(z_0)=\gamma(0)$. Does there exist a (not necessarily unique) lift of $\gamma$ under $f$ based at $z_0$? Is there a published reference for this fact?

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  • $\begingroup$ If $n=1$, the answer is "yes", which makes me think that the answer is always "yes". Note: already for $n=1$, the question is quite non-trivial. But in that case, it is enough to look at the map $z\mapsto z^n$ (that's the local model). Given a continuous path $\gamma:[0,1]\to \mathbb C^n$, one considers the closed subset $\gamma^{-1}(0)$, and its complement, which is a countable disjoint union of intervals. The lifting problem can then be solved independently on each component of that complement. $\endgroup$ Nov 23, 2010 at 7:47

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The answer is yes, here is a proof.

Let $k=2n$. The polynomial, regarded as a map $f:\mathbb R^k\to\mathbb R^k$, has the following properties:

(1) The pre-image of every point if finite, moreover the cardinality of a pre-image is uniformly bounded by some constant $N$ (by Bezout's theorem, see comments).

(2) The set $\Sigma$ of singularities of $f$ (i.e. points where $\det df=0$) is a union of finitely many smooth submanifolds of (real) codimension at least 2.

This (along with smoothness and properness) implies the path-lifting property.

First, observe the following facts:

(3) $f(\Sigma)$ is closed (because $\Sigma$ is closed and $f$ is proper).

(4) $\mathbb R^k\setminus f(\Sigma)$ is path connected. Moreover any path in $\mathbb R^k$ is a uniform limit of paths avoiding $f(\Sigma)$. This follows from the fact that $f(\Sigma)$ has Hausdorff dimension at most $k-2$.

(5) For every compact set $K\subset\mathbb R^k$ and every $\varepsilon>0$, there exists $\delta>0$ such that for every connected set $S\subset K$ of diameter greater than $\varepsilon$, the diameter of $f(S)$ is greater than $\delta$. Indeed, if this is not the case, there would be a sequence $S_i$ of connected subsets of $K$ with diameters at least some $\varepsilon_0>0$ and diameters of images going to zero. By choosing a subsequence, we may assume that the images converge to some $y_0\in\mathbb R^k$. Since $S_i$ is connected and has diameter at least $\varepsilon_0$, it contains a finite subset $P_i$ of cardinality $N+1$ such that all points of $P_i$ are separated away from one another by distance at least $\varepsilon_0/2N$. A subsequence of $\{P_i\}$ converge to a set $P$ of cardinality $N+1$, and all points of $P$ are mapped by $f$ to $x_0$, contrary to (1).

Now, in order to lift a path $\gamma$, approximate it by paths $\gamma_i\subset\mathbb R^k\setminus f(\Sigma)$. The restriction of $f$ to the pre-image of $R^k\setminus f(\Sigma)$ is a covering map to $R^k\setminus f(\Sigma)$ because it is a proper local homeomorphism. So there are lifts $\tilde\gamma_i$ of $\gamma_i$. Now it suffices to find a converging subsequence of $\{\tilde\gamma_i\}$. To do this, it suffices to show that this sequence is equi-continuous. And this follows from (5): if you could find arbitrarily close pair of points on some $\gamma_i$ with lifts bounded away from each other, the lift of the segment of $\gamma_i$ between these points would have diameter bounded away from zero but the diameter of the image arbitrarily small, contrary to (5).

To make sure that the limit lift starts from $z_0$, choose $\gamma_i$'s so that their starting points have pre-images near $z_0$ and start $\tilde\gamma_i$'s from there.

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  • $\begingroup$ 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. $\endgroup$ Nov 23, 2010 at 14:14
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    $\begingroup$ 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 $\delta_0$). $\endgroup$
    – BCnrd
    Nov 23, 2010 at 17:48
  • $\begingroup$ @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. $\endgroup$ Nov 23, 2010 at 18:55
  • $\begingroup$ 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 $S$.) $\endgroup$
    – BCnrd
    Nov 23, 2010 at 19:22

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