If $f$ is injective on $B_r(z_0)$ and holomorphic on $\overline{B_r(z_0)} $, then $f(B_r(z_0))\cap f(\partial B_r(z_0) )=\emptyset$

Question

I need to show that if $f$ is holomorphic on an open set $U$, $\overline{B_r(z_0)} \subseteq U$ and $f|_{B_r(z_0)}$ is injective, then $$ f^{-1}(w)= \frac{1}{2\pi i} \int_{\partial B_r(z_0) } \frac{zf'(z)}{f(z)-w}dz$$ for $w \in f(B_r(z_0))$. But I don't know how can I show that $w \notin f(\partial B_r(z_0))$. Can you help me, please?

Answer 1

That follows from the open mapping theorem. Suppose that there is a $z\in B_r(z_0)$ and a $w\in\partial B_r(z_0)$ such that $f(z)=f(w)$. Take $r'<r$ so that $z\in B_{r'}(z_0)$. You know that $f\bigl(B_{r'}(z_0)\bigr)$ is an open set and $f(w)=f(z)\in f\bigl(B_{r'}(z_0)\bigr)$. By continuity, there is then a $s>0$ such that $f\bigl(B_s(w)\bigr)\subset f\bigl(B_{r'}(z_0)\bigr)$. But so, if you take $s$ small enough so that $B_{r'}(z_0)\cap B_s(w)=\emptyset$, this goes against the injectivity of $f|_{B_r(z_0)}$.