Alexander's trick

Alexander's trick, also known as the Alexander trick, is a basic result in geometric topology, named after J. W. Alexander.

Statement

Two homeomorphisms of the n-dimensional ball $D^{n}$ which agree on the boundary sphere $S^{n-1}$ are isotopic.

More generally, two homeomorphisms of $D^{n}$ that are isotopic on the boundary are isotopic.

Proof

Base case: every homeomorphism which fixes the boundary is isotopic to the identity relative to the boundary.

If $f\colon D^{n}\to D^{n}$ satisfies $f(x)=x{\text{ for all }}x\in S^{n-1}$ , then an isotopy connecting f to the identity is given by

J(x,t)={\begin{cases}tf(x/t),&{\text{if }}0\leq \|x\|<t,\\x,&{\text{if }}t\leq \|x\|\leq 1.\end{cases}}

Visually, the homeomorphism is 'straightened out' from the boundary, 'squeezing' $f$ down to the origin. William Thurston calls this "combing all the tangles to one point". In the original 2-page paper, J. W. Alexander explains that for each $t>0$ the transformation $J_{t}$ replicates $f$ at a different scale, on the disk of radius $t$ , thus as $t\rightarrow 0$ it is reasonable to expect that $J_{t}$ merges to the identity.

The subtlety is that at $t=0$ , $f$ "disappears": the germ at the origin "jumps" from an infinitely stretched version of $f$ to the identity. Each of the steps in the homotopy could be smoothed (smooth the transition), but the homotopy (the overall map) has a singularity at $(x,t)=(0,0)$ . This underlines that the Alexander trick is a PL construction, but not smooth.

General case: isotopic on boundary implies isotopic

If $f,g\colon D^{n}\to D^{n}$ are two homeomorphisms that agree on $S^{n-1}$ , then $g^{-1}f$ is the identity on $S^{n-1}$ , so we have an isotopy $J$ from the identity to $g^{-1}f$ . The map $gJ$ is then an isotopy from $g$ to $f$ .

Radial extension

Some authors use the term Alexander trick for the statement that every homeomorphism of $S^{n-1}$ can be extended to a homeomorphism of the entire ball $D^{n}$ .

However, this is much easier to prove than the result discussed above: it is called radial extension (or coning) and is also true piecewise-linearly, but not smoothly.

Concretely, let $f\colon S^{n-1}\to S^{n-1}$ be a homeomorphism, then