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Suppose $\gamma:[0,1]\to\mathbb{R}^2$ is a closed path, and let $D$ denote some bounded component of $\gamma^c$ (writing "$\gamma$" for the trace of $\gamma$ as well). Must there be a parameterization for $\partial D$?

Related: Is there a characterization for what $\gamma^{-1}(\partial D)$ can look like in $[0,1]$?

EDIT: To say that $\partial D$ is parameterizable is to say that there is a continuous surjective map from $[0,1]$ to $\partial D$.

Trevor J Richards
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  • I'm sorry, what is meant by $\gamma^{c}$? And bounded component? I'm probably not able to help, just trying to understand the question. – Shoutre Aug 15 '16 at 18:09
  • If you want the boundary to be non-parametrizable, we just have to choose one such that for some $p \in \partial D$, there exists no neighborhood which is diffeomorphic to $\mathbb{R}$. Why not just take any lemniscate in the complement which doesn't intersect the curve? Just fill in the region is bounds and now the lemniscate is the boundary and it is not parametrizable with this definition given above. – Faraad Armwood Aug 15 '16 at 18:10
  • @Shoutre, ${}^c$ typically denotes complement. –  Aug 15 '16 at 18:29
  • @FaraadArmwood That won't work. Let the path $\gamma $ start at $(1,0),$ go to $(0,0)$ then back to $(1,0)$ then once around the unit circle. There is no neighborhood of $(1,0)$ in $\partial D$ homeomorphic to $\mathbb R,$ but obviously $\partial D$ can be parameterized. – zhw. Aug 15 '16 at 18:35
  • @zhw. The traveling back to $(1,0)$ does not give a diffeomorphism for any point on that segment, so this doesn't work. – Faraad Armwood Aug 15 '16 at 19:41
  • @zhw. Also, I was providing an example of a bounded component in the complement, who's boundary is not parametrizable which is clear by looking at the lemniscate. There is an obvious point for which the lemniscate has no neighborhood which is diffeomorphic to $\mathbb{R}$. Do you see a problem with this example? – Faraad Armwood Aug 15 '16 at 19:43
  • @FaraadArmwood Yes, I see a problem. To say $\partial D$ is parameterizable, to me, means there is a map $f:[0,1] \to \partial D$ whose range is $ \partial D.$ – zhw. Aug 15 '16 at 19:46
  • @zhw. I was reading a few hours ago about how something being parametrizable is a very unclear matter. The definition I used to come up with his example was the local parametrization definition. In the sense that $W \subset M$ is parametrizable if for each $x \in W, \exists$ a diffeomorphism $\phi_U: U \to \mathbb{R}^n$ where $M$ is a $n$-manifold and $U$ is an open set in $M$. – Faraad Armwood Aug 15 '16 at 19:49
  • I should have stated $f$ is continuous in my last comment. – zhw. Aug 15 '16 at 19:58
  • @Faraad: Maybe I'm missing something, but you don't get to pick another curve or region. You need to work with one of the components of the complement of the given $\gamma$. So the curve in the OP's question is a subset of $\gamma$. – Ted Shifrin Aug 16 '16 at 00:41

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First, one needs to specify the meaning of "there exists a parameterization of $X$". In the context of your question, the only meaningful interpretation of this is that there exists a continuous surjective mapping $[0,1]\to X$. With this interpretation the answer to your question is positive. The key to the proof is

Theorem. (Hahn and Mazurkiewicz, see e.g. theorem 31.5 in Willard's book "General Topology") A compact metrizable space $X$ is a continuous image of the interval $[0,1]$ if and only if $X$ is connected and locally connected.

In your situation, $\gamma([0,1])$ is connected and locally connected, from which one concludes that the boundary $\partial A$ of each component $A$ of ${\mathbb R}^2 - \gamma([0,1])$ satisfies the same property. Therefore, $\partial A$ is parameterizable.

Edit. Here is an additional argument used to conclude local connectivity of $\partial A$. The argument is taken from my answer here. Set $K:=\gamma([0,1])$. It is a connected, locally connected subset of $S^2= R^2 \cup \{\infty\}$. Therefore, every component of $S^2 - K$ is simply connected. The subset $A$ is one of these components. Its complement, $A^c=S^2 - A$, is the union of $K$ with some open subsets of $S^2$. Open subsets of $S^2$ are, of course, locally connected. Since $K$ is also locally connected, so is $A^c$. Now, the Caratheodory-Torhorst extension theorem quoted in my answer implies that $\partial A$ is also locally connected.

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Moishe Kohan
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  • Why must $\partial A$ be locally connected? I do not see that it follows directly from the fact that $\gamma([0,1])$ is locally connected. – Trevor J Richards Aug 18 '16 at 03:16
  • @TrevorRichards: See my answer here: http://math.stackexchange.com/questions/1781090/converse-to-the-jordan-curve-theorem/1782617#1782617, I will add more details though. – Moishe Kohan Aug 18 '16 at 05:36
  • Thank you very much for extending you answer. In your Edit you seem to by using the property that the disjoint union of two locally connected sets is locally connected. Is that accurate? – Trevor J Richards Sep 18 '16 at 01:55
  • @TrevorRichards: No, this more general statement is false (you can build an example using the topologist's sine curve). I will add more details when I have a bit more time. – Moishe Kohan Sep 20 '16 at 15:13