I would like to prove that if a function $f(z)$ is holomorphic on $\overline{D(P,r)}$ and one to one on $\partial D(P,r)$ then $f$ is one to one on $D(P,r)$. I noticed that for $w \in f(D(P,r))$ with $w \notin f(\partial D(P,r))$
$\frac{1}{2\pi i}\int_{D(P,r)}\frac{f'(s)}{f(s) - w}ds = \frac{1}{2\pi i}\int_{\gamma}\frac{1}{s-w}ds$ where $\gamma(\theta) = f(re^{\theta i} + P)$
Then it could be argued that since $f(re^{\theta i} + P)$ is one to one, it can only go around $w$ once.
The problem is actually proveing this(unless there is a better route to proveing it).
Also if we knew beforehand that $f(D(P,r))$ was disjoint from $f(\partial D(P,r))$ we could use a continuity argument where the independent variable was $w$