# Non-proper surjective holomorphic map from the unit disk to itself

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 $$\lim\limits_{\lvert z \rvert \rightarrow 1} \lvert f(z) \rvert = 1$$?

• Corrected. Thank you. – v_lentin May 27 at 12:02
• 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. – DLeMeur May 27 at 12:15
• 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. – v_lentin May 27 at 12:36

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 $$f(z) \to \frac 18 \notin \partial \Bbb D$$ .