# Surjective holomorphic map from a domain to $\mathbb{C}$

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.

• $z\mapsto z^3$ on open upper half plane misses $0$. Move down by a tiny bit and square instead. – user10354138 Feb 20 at 18:12
• @user10354138 Noted and edited. – Pelota Feb 20 at 19:15

## 2 Answers

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.

• Yes, sure! In retrospective, it was easy. Many thanks. – Pelota Feb 20 at 20:24

Either there exists an open disc $$\Delta\subseteq U$$ which gives $$\#(\partial\Delta\cap\partial U)\geq 2$$ or there isn't.

• If no such $$\Delta$$ exist, then $$U$$ misses at most one point and so $$U$$ contains a half-plane.
• If such $$\Delta$$ exists, pick two points in $$\partial\Delta\cap\partial U$$ and send by Mobius to $$0,\infty$$ so you have a $$U$$ contains a half-plane.

So you have $$U\twoheadrightarrow\text{something containing upper half-plane}\twoheadrightarrow\mathbb{C}$$.

• In the first case you do not necessarily have a half plane: for every $\vartheta$, $\{z:|\arg(z)-\vartheta|<\varepsilon\}$ satisfies $R=\infty$, if I understood what you mean. – Caffeine Feb 20 at 18:42
• And for the second point, how can the Mobius transformation send $U$ to a half-plane, if $U$ is not simply connected? – Caffeine Feb 20 at 18:49
• @Caffeine Sorry my bad. Simplified my argument the wrong way. Also, for the second point, by "have a half-plane" I mean $U$ contains a half-plane, which is all that matters to use $z\mapsto (z-z_0)^2$ to hit all of $\mathbb{C}$. – user10354138 Feb 21 at 0:52