TITLE: Bi-Lipschitz extension
QUESTION [8 upvotes]: Given a bi-Lipschitz homeomorphism 
$\Phi:\mathbb{B}^n(0,1)\to\mathbb{R}^n$, (that is a bi-Lipschitz map onto the image), can one find a bi-Lipschitz homeomorphism $\Psi:\mathbb{R}^n\to\mathbb{R}^n$ such that
$$
\Psi|_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\frac{1}{2})}\ \ ?
$$

REPLY [2 votes]: Let me describe a baby case of Sulivan's construction which is beautiful.
Assume I have a homeomorphism $h$ from pentagon to itself that sends each side to itself.
Let us assume that the pentagon has right angles in the conformal disc model of Lobachevsky plane. Extend the map $h$ to the disc applying reflections in the sides of pentagon, so the obtained map $\tilde h$ commutes with and any reflection in a side of pentagon.
Note that $\tilde h$ is bi-Lipschitz and it is identity on the boundary of the disc. So we can extend $\tilde h$ by identity map outside of disc.
It is problematic to do the same for general $h$ and higher dimensions, but all this seem to be solved by Sullivan.