continuous injective extension of a map defined on a hemisphere 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).
 A: 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.

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.
