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?

If $f(S^1) \subset S^1$, then the maximum principle implies that $f$ maps the interior of the disk to the interior of the disk. This implies that $f(S^1) = S^1$.
It's also true that, at points where $f^{\prime}(z) \neq 0$, $f(z)$ is an interior point of $f(B(0,1))$. So, the only possible interior points that get mapped to the boundary of $f(B(0,1))$ are those where $f^{\prime}(z) = 0$. Now, if this occurs at infinitely many points, then $f^{\prime}(z)$ is identically zero, meaning $f$ is constant. Otherwise, the zeroes form a discrete set.
Now, since the disk is compact, $f(D(0,1))$ is also compact. So, $B(0,1) - f(D(0,1))$ is open (note that it can't have any boundary points on $S^1$). A nonempty, bounded open set can't have a discrete boundary. Thus, $B(0,1) - f(D(0,1))$ is empty.
