Let $U\subset \mathbb{C}$ be a domain (i.e. open and connected). Is it always the case that there exists a holomorphic map $f:U\to \mathbb{C}$ such that $f(U)=\mathbb{C}$?
I managed to prove it for the upper half plane ($(z-i)^2$) (and so for every simply connected set by the Riemann mapping theorem)
and for $\mathbb{C}-\{z_0,\dots,z_n\}$ (just take $f(z)=\prod_{i=0}^{n+1} (z-z_i)$, where $z_{n+1}$ is a point in $\mathbb{C}-\{z_0,\dots,z_n\}$) but I do not seem to be able to prove it in general.
Best Answer
If $U=\mathbb C,$ then $f(z)=z$ does the job. If $U\ne \mathbb C,$ then $\partial U$ is nonempty. In this case take any $a\in U.$ Then $d(a,\partial U)$ is a finite positive number $r,$ and there is $b\in \partial U$ such that $|a-b|=r.$ It follows that $D(a,r)\subset U.$
For $z\in \mathbb C\setminus \{b\}\supset U,$ set $f(z) = \dfrac{1}{z-b}.$ Then $f(D(a,r))$ is an open half plane. As you pointed out, this leads to a holomorphic function from $U$ to $\mathbb C$ that is surjective.