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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.

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

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  • $\begingroup$ Yes, sure! In retrospective, it was easy. Many thanks. $\endgroup$ – Pelota Feb 20 at 20:24
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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 $U\twoheadrightarrow\text{something containing upper half-plane}\twoheadrightarrow\mathbb{C}$.

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    $\begingroup$ 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. $\endgroup$ – Caffeine Feb 20 at 18:42
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    $\begingroup$ And for the second point, how can the Mobius transformation send $U$ to a half-plane, if $U$ is not simply connected? $\endgroup$ – Caffeine Feb 20 at 18:49
  • $\begingroup$ @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 $\mathbb{C}$. $\endgroup$ – user10354138 Feb 21 at 0:52

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