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 ?

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 $\partial D$.
