An analytic function is onto All sets are subsets of $\mathbb{C}$. Suppose $f: U \to D$ is analytic where $U$ is bounded and open, and $D$ is the open unit disk.
Now suppose we can continuously extend $f$ to $\bar{f}: \bar{U} \to \bar D$, such that $\bar{f}(\partial U) \subseteq \partial D$.  To show that $f$ is onto, I was thinking maybe I could show that $f(U)$ is  a dense subset of $\bar{D}$ , and since $\bar{f}(U) = f(U)$ is open by the open mapping theorem, it must be $D$.  But to do this I would need to know that $f(\partial U) = \partial D$.  Is this true?
Some advice or other approaches would be greatly appreciated. Thank you.

Hint: If $f$ is not onto then $D$ containts a point $w$ which is on the boundary of $f(U)$. Take a sequence in $U$ with $f(z_k)\to w$, and use compactness.
