Let $\mathbb{D}$ denote the open unit disk of the complex plane. Does there exist a non-proper surjective holomorphic map $f \colon \mathbb{D} \rightarrow \mathbb{D}$? In other words, does every surjective holomorphic map $f \colon \mathbb{D} \rightarrow \mathbb{D}$ satisfy $\lim\limits_{\lvert z \rvert \rightarrow 1} \lvert f(z) \rvert = 1$?

  • $\begingroup$ Corrected. Thank you. $\endgroup$ – v_lentin May 27 at 12:02
  • $\begingroup$ Using the fact that $f$ is proper iff $f$ is a (ramified) covering, the question is equivalent to find a surjective holomorphic map which is not a ramified covering... Don't know if it helps. $\endgroup$ – DLeMeur May 27 at 12:15
  • $\begingroup$ The proper holomorphic maps $\mathbb{D} \rightarrow \mathbb{D}$ are exactly the finite Blaschke products. Therefore, my question is equivalent to asking whether every surjective holomorphic map $\mathbb{D} \rightarrow \mathbb{D}$ is a finite Blaschke product. $\endgroup$ – v_lentin May 27 at 12:36

A counter-example:

Let $g$ be a conformal map from the unit disk $\Bbb D$ to the semi-disk $$ G = \{ z : |z| < 1, \operatorname{Im}(z) > 0 \} $$ which exists according to the Riemann mapping theorem, and has a continuous one-to-one extension to the closed disk (Carathéodory's theorem). It is not difficult to find an explicit expression for this mapping, but that is not needed here.

Now let $f(z) = g(z)^3$. $f$ is a surjective mapping from the unit disk onto itself, but for $z \to g^{-1}(\frac 12) \in \partial \Bbb D$ we have $f(z) \to \frac 18 \notin \partial \Bbb D$ .


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