An analytic function on the disk that sends the boundary into the boundary sends the interior onto the interior Consider $f$ analytic on $B(0; 1) = \{z \in \mathbb{C}\ |\ |z| < 1\}$ and continuous in $D(0; 1) = \{z \in \mathbb{C}\ |\ |z| \leq 1\}$ such that $f(S^1) \subset S^1$. I must show that $f(B(0; 1)) = B(0; 1)$.
I've shown that $f$ must have the form $f(z) = \omega \prod_{j = 1}^{n} \frac{z - a_j}{1 - \overline{a_j}z}$ where $a_1, \ldots, a_n$ are the zeros of $f$ inside $B(0; 1)$ and $\omega \in S^1$. In particular, we can see that $f$ admits analytic extension to an open neighborhood of $D(0; 1)$. Furthermore, I know that each term in the product for $f$ is an automorphism of $D(0; 1)$. However, I still can't seem to conclude that $f(B(0; 1)) = B(0; 1)$. How might I proceed?

Let $\mathbb {D}$ be the open unit disc. Assume $f$ is nonconstant with the stated properties. Suppose $f(\bar {\mathbb {D}}) \ne \bar {\mathbb {D}}.$ Then there exists $a \in \mathbb {D}$ not in $f(\bar {\mathbb {D}}).$ It follows that $\varphi_a \circ f$ is never $0$ (where you can guess what I mean by $\varphi_a$). We've thus arrived at a nonconstant holomorphic function on $\mathbb {D},$ continuous on $\bar {\mathbb {D}},$ of modulus $1$ on the boundary, that is never $0,$ contradiction. That $f$ maps $\mathbb {D}$ onto $\mathbb {D}$ follows easily.
