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
 * $$ F\colon D^n \to D^n \text{ with } F(rx) = rf(x) \text{ for all } r \in [0,1] \text{ and } x \in S^{n-1}$$ defines a homeomorphism of the ball.

Exotic spheres
The failure of smooth radial extension and the success of PL radial extension yield exotic spheres via twisted spheres.