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Let $S^2 = \{x\in \mathbf R^3\colon |x|=1\}$ be the unit sphere, $S^2_+ = S^2 \cap \{x_3 \ge 0\}$ be the upper hemisphere and $S^1 = S^2 \cap \{x_3 = 0\}$ be the unit circle. Let $u\colon S^2_+\to \mathbf R^3$ be a continuous function. Then it is easy to extend $u$ to the whole sphere $S^2$ preserving continuity: just patch with a cone constructed on the curve $u(S^1)$ i.e. identify the lower hemisphere with a unit disk and extend $u$ by $1$-homogenuity on this disk.

My problem: if $u\colon S^2_+\to \mathbf R^3$ is continuous and injective, is it possible to extend it to a continuous and injective map defined on the whole sphere $S^2$?

I think the answer should be yes. Notice that since $u$ is injective, continuous, and defined on a compact set it is an open map, and hence an homeomorphism with its image. I imagine that, from a topological point of view, there is a unique way to embed a 2-disk into $\mathbf R^3$ (cannot make a knot)... so my intuition says that there should even exist an extension of $u$ as an homeomorphism of the whole $\mathbf R^3$ into itself.

Moreover, if the above is possible, I would like to have an explicit construction because I actually have a Lipschitz map and I would like to have a Lipschitz extension and also some control on the gradient of the extension, (as I have with the 1-homogeneous extension in the non injective case).

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  • $\begingroup$ Unless I am confused, $u$ is closed but not open. $\endgroup$
    – Gregory Arone
    Commented Aug 3, 2016 at 20:29
  • $\begingroup$ When I say it is "open" and hence "homeomorphism" i mean that we restrict the codomain to be equal to the image of the map, so it becomes bijective. A closed bijective map is also open. I added some words to make it clearer. $\endgroup$
    – Emanuele Paolini
    Commented Aug 4, 2016 at 9:21

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I think that if you only assume that $u$ is injective and continuous, then the answer to your main question is no, because of things like the Alexander horned sphere. The "Alexander horned disk" is an example of an embedded disk in ${\mathbb R}^3$ whose complement is not simply connected. You can find a picture in Bredon's Topology and Geometry, page 232. In this picture, the boundary of the disk defines the trivial element in the fundamental group of the complement, so it does not give a counterexample to your question. But I think a slight modification of this construction does. Namely, imagine one of the horns growing down rather than up, then coming around the boundary of the disk to meet the other horn. Here is what I have in mind.

Modified horned disk

It seems pretty clear to me that the boundary of the embedded disk defines a non-trivial element of the fundamental group of the complement of the open disk. In particular it does not bound an embedded disk in the complement.

I am guessing that if $u$ a (bi)-Lipshitz map the answer is yes, because of the generalized Schoenflies theorem. For example see this paper http://arxiv.org/pdf/1008.3544.pdf But I have not thought about it enough to be sure.

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  • $\begingroup$ I think that this surface is the limit of disks but is not itself a disk. In fact the little horns, in the limit, do have points in common (we agree that this set is compact) and we can find loops which are not contractible. The example was new to me and I looked the original article here: math.ku.dk/~moller/f03/algtop/notes/alexander.pdf If I understand it correctly the claim is about the interior of the surface (which is homeomorphic to an open 3-ball) not its boundary. $\endgroup$
    – Emanuele Paolini
    Commented Aug 4, 2016 at 9:38
  • $\begingroup$ No, the original article is about an embedding of the two-dimensional sphere, not of an open 3-ball. So if you remove a small open disk from this sphere, you will get an exotic embedding of a disk in ℝ3, whose complement is not simply connected. My picture is a slight modification of this embedding. It is a topological embedding of a disk, which can not be drawn satisfactorily (google image will bring up better attempts than mine). You can read about this in many introductory texts on algebraic topology. For example, have a look at Bredon's Topology and Geometry, page 232, and compare. $\endgroup$
    – Gregory Arone
    Commented Aug 4, 2016 at 10:16
  • $\begingroup$ The example of Alexander seems wrong to me. However Bredon's book gives also another example namely the "wild arc of Fox and Artin" which I can understand and which can be easily extended to a disk (just take a strip which becomes thinner and thinenr at the ends). So I agree that the example exists. I'm also very interested to what happens to lipschitz disks, so your link is a very good starting point. $\endgroup$
    – Emanuele Paolini
    Commented Aug 4, 2016 at 11:41

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