# Finite Blaschke product and proper maps on the unit disc

I have to present tomorrow on an introductory section in several complex variables about proper maps, and they gloss over a fact that seems important to me, but I do not know how to prove it.

Suppose $$f: \Bbb D \to \Bbb D$$ is an analytic proper map. Prove that $$f$$ is a finite Blaschke product.

That is, prove that $$f(z) = e^{i \theta} \prod_{j=1}^{k} {{z-a_j} \over {1- \overline a_jz}}$$

where, $$\theta$$ is real, and $$a_j \in \Bbb D$$.

They mention that to show this, you should consider the fiber of the origin. I'm not sure what to do with this information, though.

Since $f\colon \mathbb{D} \to \mathbb{D}$ is proper, for every $r < 1$, the preimage $f^{-1}(\overline{D_r(0)})$ of the closed disk with radius $r$ is compact, thus contained in $D_\rho(0)$ for some $\rho < 1$. Hence we have

$$\lim_{\lvert z\rvert \to 1} \lvert f(z)\rvert = 1.\tag{1}$$

In particular, $f$ is not constant. So the fibre above the origin, the zero set of $f$ is compact and discrete, hence finite.

Now let $g$ be the product of the Blaschke factors for the zeros of $f$ (according to multiplicity) and consider the function

$$h\colon z \mapsto \frac{f(z)}{g(z)}.$$

Then $h$ is a zero-free holomorphic function on $\mathbb{D}$, and since finite Blaschke products have constant modulus $1$ on the unit circle, we have

$$\lim_{\lvert z\rvert\to 1} \lvert h(z)\rvert = 1.\tag{2}$$

The minimum modulus principle implies that $h$ is a constant of modulus $1$, i.e. $h(z) = e^{i\theta}$ for some $\theta\in\mathbb{R}$.

• Ok...I knew that fibers were compact, but I have never heard of Blaschke factors, so this function $g$ is new to me. – Johnny Apple Sep 16 '14 at 19:34
• Very nice proof! – Bombyx mori Sep 16 '14 at 19:35
• The $\frac{z-a_j}{1-\overline{a_j}z}$ are often called Blaschke factors, @JohnnyApple, because they are factors of Blaschke products. – Daniel Fischer Sep 16 '14 at 19:37
• So, where have we used compactness of the fibers here? – Johnny Apple Sep 16 '14 at 19:37
• @Sushil Thus $$\rho(r) := \sup \{ \lvert z\rvert : \lvert f(z)\rvert \leqslant r\} < 1\,,$$ and that means $\lvert z\rvert > \rho \implies \lvert f(z)\rvert > r$. – Daniel Fischer Mar 23 '18 at 15:35