Recently, I read the relation between Shabat polynomials and trees. The book [0] says that if $p: \mathbb{C} \to \mathbb{C}$ is a degree-$n$ polynomial such that the segment $[-1,1]$ contains no critical value of $p$ then $p^{-1} ([-1,1])$ has $n$ connected components and each is homeomorphic to $[-1,1]$.

I can't argue why this should be true. Only things that I can say is that $p$ becomes a $n$-sheeted covering map from $p^{-1} ([-1,1])$ to $[-1,1]$. Any kind of help will be appreciated.

[0] Sergei K. Lando, Alexander K. Zvonkin, Graphs on Surfaces and their Applications. (Chapter 2: Dessins d’Enfants)

  • $\begingroup$ Can you describe some $n$-sheeted covering map of $[-1,1]$ not of that prescribed form? $\endgroup$ – Gerald Edgar Apr 7 '18 at 21:51
  • $\begingroup$ Although only tangentially related, it may be worth looking into Reeb's theorem for spheres to see how this works for the case in which we have only $2$ critical values for our polynomial and how we can use an automorphism on $/mathbb{C}$ to set them on $0$ and $1$ so we can use Shabat's theorem. $\endgroup$ – Casquibaldo Apr 7 '18 at 23:08

If $f$ has no critical values on $[-1,1]$ then the restriction of $$f:f^{-1}([-1,1])\to [-1,1]$$ is a covering map. This covering has degree $n=\deg f$, so there are $n$ inverse branches of $f$ on $[-1,1]$. The mages of $[-1,1]$ under these branches are $n$ distinct curves, components of $f^{-1}$. For elementary properties of covering maps see Ahlfors, Conformal invariants, or Massey, Algebraic Topology. An Introduction.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.