Let $f: B(0,1) \rightarrow \mathbb{C}$ be holomorphic and suppose $\exists\ r \in (0,1)$ such that $f$ is injective in $A = \{z \in \mathbb{C} : r < |z| < 1\}$. Prove that $f$ is injective.
I tried using Rouché theorem, or the identity theorem, but I don't know what to do. Any hints? :)
This is quite non-trivial. Suppose $z_1,z_2\in B(0,1)$ are such that $f(z_1)=f(z_2)=z_0$. Take a circle $\gamma\in A$ such that $z_1,z_2\in Int(\gamma)$. Then observe that $f(\gamma)$ is a Jordan curve (this is where you use the injectivity of $f$ in $A$) and thus by Jordan Curve Theorem, $\text{Ind}(f(\gamma),z_0)\leq1$, ignoring orientation. However by the argument principle, $$\text{Ind}(f(\gamma),z_0)=\dfrac{1}{2\pi i} \int_\gamma\dfrac{f'}{f-z_0}\geq2$$
This gives the required contradiction.