Is every open and connected set in $\mathbb C$ the continuous image of the open unit disk?

Question

Let $$ \mathbb D=\{z\in\mathbb C\ |\ |z|<1\} $$ be the open unit disk in $\mathbb C$. It is well known that an open (nonempty) set $U\subseteq\mathbb C$ is simply connected if and only if it is homeomorphic to the unit disk. One can show that this is equivalent to the fact that there is a continuous injective function $f:\mathbb D\to\mathbb C$ such that $f(\mathbb D)=U$.

Therefore, my question is: If $U$ is a (nonempty) open and connected subset of $\mathbb C$, is there always a continuous (but not necessarily injective) function $f:\mathbb D\to\mathbb C$ such that $f(\mathbb D)=U$? It sufficices to assume $U\subseteq\mathbb D$.

If not, how do images of such functions look like?

I know that any compact connected and locally connected subset of $\mathbb C$ is the image of a continuous function defined on $[0,1]$ and hence also the image of a continuous function on $\overline{\mathbb D}$. However, the closure of an open and connected set does not have to be locally connected.

I believe that the answer to my first question is negative (although I hope that it is not ;) ).

Any help is highly appreciated. Thank you very much in advance!

Answer 1

The Hahn-Mazurkiewicz theorem, as pointed out by Mathlover in the comments, is enough to show my educated guess actually works. In particular, it implies that closed balls in $\Bbb{C}$ are the continuous image of a compact interval. We will need a lemma:

Lemma $\quad$ Suppose $\{C_n\}_{n=0}^\infty$ is a countable collection of subspaces of a topological space $X$ (e.g. $\Bbb{C}$) which are continuous images of a compact interval and $$C := \bigcup_{n=0}^\infty C_n$$ is path-connected. Then $C$ is the continuous image of $[0, \infty)$.

Proof. Consider the intervals $I_n := [2n, 2n + 1]$ and $J_n := [2n + 1, 2n + 2]$ for $n \in \Bbb{N} \cup \{0\}$. Let $\mathfrak{i}_n : I_n \to C_n$ be surjective and continuous, and let $\mathfrak{j}_n : J_n \to C$ be a continuous function such that $\mathfrak{j}_n(2n + 1) = \mathfrak{i}_n(2n + 1)$ and $\mathfrak{j}_n(2n + 2) = \mathfrak{i}_{n+1}(2n + 2)$. Then, the map $$\phi : [0, \infty) = \bigcup_{n=0}^\infty (I_n \cup J_n) \to C$$ defined by $$\phi(x) = \begin{cases} \mathfrak{i}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n \le x < 2n + 1 \\ \mathfrak{j}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n + 1 \le x < 2n + 2. \end{cases}$$ This function is made up of countably infinitely many continuous pieces, joined at common points, making the function continuous. It's clearly surjective (even just restricting to $\bigcup I_n$), hence $C$ is the continuous image of $[0, \infty)$ under $\phi$. $\square$

Note that every ball in $\Bbb{C}$ is a countable union of closed balls, and recall that open subsets of $\Bbb{C}$ are a countable union of open balls. Also, in a locally path-connected space, open and connected implies path-connected. Thus, $U$ is the path-connected countable union of closed balls, so by the lemma, it is a continuous image of $[0, \infty)$.

So, to wrap up the proof, as in my comment, simply project the unit disk of $\Bbb{C}$ onto $(-1, 1)$, which is homeomorphic to $\Bbb{R}$. You can simply keep the map constant for points in $(-\infty, 0]$, and then use $[0, \infty)$ to map continuously onto $U$.

Answer 2

One can prove a stronger claim: given a nonempty open connected $U\subseteq \mathbb{C}$, there's a holomorphic $\varphi:\mathbb{D}\to \mathbb{C}$ such that $\varphi(\mathbb{D})=U$.

First, let $U\neq \mathbb{C},\neq \mathbb{C}-\{z_0\}$. Thanks to the Riemann uniformization theorem, every such open connected subset of $\mathbb{C}$ is hyperbolic, i.e. has $\mathbb{D}$ as its universal cover. Thus there exists a surjective holomorphic map $\mathbb{D}\to U$. Note that there's a surjective holomorphic map $(\exp(z)+z_0)$ from $\mathbb{C}$ to $\mathbb{C}-\{z_0\}$, and thus it suffices to prove the result for $\mathbb{C}$. Now, let $f(z)=z^3-1$. It is easy to see that $f:\mathbb{C}-\{1,\exp(2\pi i/3)\}\twoheadrightarrow\mathbb{C}$. Precomposing with the covering map $\pi:\mathbb{D}\to \mathbb{C}-\{1,\exp(2\pi i/3)\}$, we get the claim.

One can also explore "how much" this function fails to be injective. Indeed, the uniformization theorem implies that, for $U\neq\mathbb{C},U\neq \mathbb{C}-\{z_0\}$, we have $U\simeq \mathbb{D}/\Gamma$, where $\Gamma$ is a discrete subgroup of $Aut(\mathbb{D})$ (complex automorphisms) isomorphic to $\pi_1(U)$. Thus two elements $z_1,z_2$ have the same image iff $\exists \gamma\in \Gamma:\gamma(z_1)=z_2$ (note that with $U$ simply connected we get injectivity).