Spring 2009 - Problem 11

Solution.

Suppose otherwise, and that there exist $z_1, z_2 \in \D$ such that $f\p{z_1} = f\p{z_2} = w$. Picking $R \in \p{0, 1}$ large enough, we have $z_1, z_2 \in B\p{0, R}$. Thus, if we set $\gamma = \partial B\p{0, R}$ oriented counter-clockwise with domain $\br{0, 2\pi}$, the argument principle tells us that

But this implies that $f \circ \gamma$ is not simple. Indeed, by the Jordan curve theorem, a simple closed curve has winding number in $\set{-1, 0, 1}$, depending on the orientation. Thus, there exist $0 \leq t_1 < t_2 < 2\pi$ such that $f\p{\gamma\p{t_1}} = f\p{\gamma\p{t_2}}$. Since $\gamma$ is in the annulus where $f$ is injective, we have $\gamma\p{t_1} = \gamma\p{t_2}$. This can only happen if $\gamma\p{t_1} = \gamma\p{t_2} = \gamma\p{0}$, which is impossible since $0 < t_2 < 2\pi$. Thus, $f$ must have been injective in $\D$ to begin with, as required.