A holomorphic function $f$, injective on $\partial D$, must be injective in $\bar{D}$?

Prove: If $f$ is holomorphic on a neighborhood of the closed unit disc $\bar{D}$, and if $f$ is one-to-one on $\partial D$, then $f$ is one-to-one on $\bar{D}$.

(Greene and Krantz's Function Theory of One Complex Variable (3rd), Ch. 5, Problem 17.)

Can anyone provide a clue as to how to attack this problem ?

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Some hints:

The function $f$ restricted to $\partial D$ is an injective continuous map of $\partial D\sim S^1$ into ${\mathbb C}$. By the Jordan curve theorem the curve $\gamma:=f(\partial D)$ separates ${\mathbb C}\setminus\gamma$ into two connected domains $\Omega_{\rm int}$ and $\Omega_{\rm ext}$, called the interior and the exterior of $\gamma$. The points $a\in \Omega_{\rm int}$ are characterized by the fact that $\gamma$ has winding number $\pm1$ around them.

Create a connection with the argument principle.

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Let $b\in D$ and $f(b)=a\in \Omega_{int}$. Now the integral $$\frac{1}{2\pi i} \int_{\partial D} \frac{f'(z)}{f(z)-a}\, dz$$ equals the number of zeros (counting multiplicities) of $f(z)-a$. On the other hand, as @Christian implies it equals the winding number of $\gamma$ around $a$, which equals $\pm 1$. Hence $f(z)=a$ has a unique solution, for each $a\in f(D)=\Omega_{int}$, and so $f$ is injective in $D$. Still missing: (i) it should be injective in $\bar{D}$. –  Teddy Oct 10 '12 at 12:51
Since $\bar{D}\subseteq U$, there exists an $\epsilon$ such that $\bar{D}(0,1+\epsilon)\subseteq U$, and the argument in my previous comment may be duplicated to give that $f$ is injective in $\bar{D}$. –  Teddy Oct 10 '12 at 13:07

Here's an idea, but I'm not sure, I didn't make any mistakes. It's also incomplete.

Try to argue that $f$ doesn't branch within some region $U \supseteq\overline{D}$. (I don't know how to do this.)

Assume a holomorphic map $f$ on $U$ which doesn't branch. This makes $f$ a topological covering map on $U$. Assume there are $z_0$ and $z_1$ in $\overline{D}$ such that $w := f(z_0) = f(z_1)$ and $z_0 \neq z_1$. They can be joined by an arc $\gamma\colon [0,1] \to \overline{D}$. Also, use the homotopy lifting property of coverings and the uniqueness of liftings to lift another arc $\delta\colon [0,1] \to f(\overline{D})$ joining $w$ and and a boundary point of $f(D)$ to two different arcs $\hat\delta_i\colon [0,1] \to \overline{D}$ beginning at $z_i$ for $i=0,1$. Notice that those arcs have different end points $\hat\delta_0(1) \neq \hat\delta_1(1)$ by the uniqueness of lifting. They also lie on $\partial D$, because $f$ is open and can't map elements of the open disk $D$ to the boundary of its image $f(D)$. But since both arcs lift $\delta$, those end points are both mapped to the same point, viz the end point of the arc $\delta$, so you can conclude that $f$ isn't one-to-one on $\partial D$.

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Oh, I think you don't actually need $f$ to not branch on some region $U \supseteq \overline{D}$, but only to not branch on some region $U$ which contains $z_0$, $z_1$ and a line segment of $\partial D$ which should be possible. You also don't need arcs, but it's sufficient to just use paths if I'm not mistaken. –  k.stm Oct 10 '12 at 10:19