3
$\begingroup$

Given a polynomial f with degree d, we can define a dynamical system $(\mathbb{C}, f)$. If we have a proper continuum $N \subset \mathbb{C}$, it is known that the set $f^{-1}(N)$ has at most $d$ components, with each component being mapped onto $N$ by $f$. This fact can be found in [pp95, Lemma 5.7.2, Beardon, Alan F.'s “Iteration of Rational Functions” in the book “Complex Analytic Dynamical Systems” (Graduate Texts in Mathematics, 132, Springer-Verlag, New York, 1991)].

The question now arises: if we further assume that the continuum $N$ is full (meaning both $N$ and its complement are connected) and does not contain the image of critical points (referred to as critical values and the cricital points mean the points at which the derivative of $f$ is zero), can we then conclude that there are exactly $d$ components of $f^{-1}(N)$, and that $f$ is injective when restricted to each component?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Yes, you can. First construct a Jordan domain $D$ which contains your continuum and does not contain any critical values. (To do this map the simply connected domain $\overline{\mathbf{C}}\setminus N$ conformally onto the unit disk by a function $\phi$, then $\phi^{-1}(\{ z:|z|=r\})$ will be a Jordan curve if $r$ is close enough to $1$, and the interior region $D$ of this curve will contain $N$ and contain no critical values). Then the restriction $$f: f^{-1}(D)\to D$$ is an (unramified) covering map of degree $d$, therefore it consists on $d$ simply connected components. Each component contains exactly one component of $f^{-1}(N)$.

Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ TeX note: \setminus spaces better than \backslash for complements, e.g., $\mathbb C \backslash f^{-1}(D)$ \mathbb C \backslash f^{-1}(D) vs. $\mathbb C \setminus f^{-1}(D)$ \mathbb C \setminus f^{-1}(D). I edited accordingly. $\endgroup$
    – LSpice
    Commented Jan 19 at 14:27
  • 1
    $\begingroup$ @LSpice: Thanks, I will try to remember this. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 20 at 2:09
  • $\begingroup$ thanks very much for your answer, here the restriction should be $f: f^{-1}(D)\rightarrow D$, right? So the fact is that for any (unramified) covering map $f: X \subset\mathbb{C}\rightarrow X$ of degree $d$, the set $f^{-1}(N)$ has eaxctly $d$ components for any full continuum $N \subset X$. $\endgroup$
    – Yee Neil
    Commented Jan 20 at 9:21
  • $\begingroup$ @Yee Neil: Thanks. You are right. I corrected the formula. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 20 at 15:00
  • $\begingroup$ Thank you very much. I have remembered an important proposition $\endgroup$
    – Yee Neil
    Commented Jan 20 at 23:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .