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

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!

• Not my field, but I'm pretty sure that the answer is positive to your first question. Here's my idea: first project $\Bbb{D}$ to the interval $(-1, 1)$, then expand to $\Bbb{R}$. From there, I'm pretty sure you can use a space-filling curve to fill any disk. Since $U$ is open and connected, it's path connected, and is made up of a countable union of open disks, which we can cover by a single space-filling curve (just join the endpoints for each of the countably many open disks by a path). The result will be heavily non-injective, but I think it works? Feb 14 at 11:46
• Feb 14 at 12:13

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 $$U$$.

• This is amazing, thank you! Feb 14 at 13:45

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).

• Beautiful answer, thank you very much! Feb 14 at 13:46