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I encountered this issue recently, but do not know of any general results to deal with it, so I would appreciate any pointers.

Let $\mathbb T=\{z\in\mathbb C\mid |z|=1\}$, and let $f:\mathbb T\to\mathbb C$ be continuous and injective, so its image $\mathbb T'$ is a Jordan loop. Under what (general) conditions can we ensure that there is a homeomorphism between the unit disc and the interior of $\mathbb T'$ whose extension to the boundary is $f$?

Moreover, if there are reasonable conditions that ensure this, and $f$ is $C^\infty$, can we further require some nice regularity (perhaps even $C^\infty$) of the extension as well?

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  • $\begingroup$ Doesn't Schoenflies give what you want? Or Google Schoenflies extension theorem. $\endgroup$
    – Bill Johnson
    Commented Sep 22, 2012 at 16:06
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    $\begingroup$ en.wikipedia.org/wiki/Schoenflies_problem $\endgroup$
    – Bill Johnson
    Commented Sep 22, 2012 at 16:18
  • $\begingroup$ I should have mentioned Schoenflies. I am asking for a sort of converse. Schoenflies theorem ensures that a homeomorphism of the interiors can be extended to the boundary and that, in general, given a Jordan loop $\mathbb T'$, there is an $f$ with range $\mathbb T'$ that can be extended. Here I am starting with a given $f$ and want to extend it to the interior. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 22, 2012 at 16:19
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    $\begingroup$ Since you asked for pointers: There is an extension version of Schoenflies stating that every homeomorphism $f\colon \mathbb{T} \to \mathbb{T}'$ can be extended to a homeomorphism of $\mathbb{C}$. This is mentioned (without proof) as a consequence of an extension theorem of Carathéodory in Remmert, Classical Topics in Complex Function Theory, page 187: books.google.com/books?id=BHc2b0iCoy8C&pg=PA187 and the book refers to Pommerenke, Boundary Behavior of Conformal Maps, Springer 1991 for details. $\endgroup$
    – Theo Buehler
    Commented Sep 23, 2012 at 13:07

1 Answer 1

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What you're asking is equivalent to asking whether any homeomorphism $g : S^1 \rightarrow S^1$ can be extended to a homeomorphism of the disc. This is easy -- write the disc in polar coordinates $(t,\theta)$ with $\theta \in S^1$, and define an extension $G(t,\theta) = (t,g(\theta))$.

The question about whether this can be done smoothly if $g$ is smooth is more subtle. Observe that the above also works for $S^k$ with $k > 1$. The smooth version fails in higher dimensions and is responsible for the existence of exotic spheres. However, for $k=1,2$ there is no problem. For $k=1$, this is a theorem of Smale; see

Smale, Stephen Diffeomorphisms of the 2-sphere. Proc. Amer. Math. Soc. 10 1959 621–626.

For $k=2$, it is a much deeper theorem of Hatcher; see

Hatcher, Allen E. A proof of the Smale conjecture, Diff(S3)≃O(4). Ann. of Math. (2) 117 (1983), no. 3, 553–607.

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  • $\begingroup$ I don't think this is what the question is asking. I think the question refers to the interior of the curve, which is a subset of the same complex plane where the curve is embedded, and which is guaranteed to exist by the Jordan curve theorem. The question asks for a homeomorphism to this interior, not just to any other space whose boundary is the curve. $\endgroup$
    – Zsbán Ambrus
    Commented Sep 22, 2012 at 19:03
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    $\begingroup$ @Zsbán, it is equivalent. $\endgroup$
    – Anton Petrunin
    Commented Sep 22, 2012 at 19:08
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    $\begingroup$ Hi Andy. Thanks! This does it nicely. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 22, 2012 at 19:25

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