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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 $S^$ 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 $f\left(x\right) = x \text x \in S^$, then an isotopy connecting ''f'' to the identity is given by : $J\left(x,t\right) = \begin tf\left(x/t\right), & \text 0 \leq \|x\| < t, \\ x, & \text t \leq \|x\| \leq 1. \end$ 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 $\left(x,t\right)=\left(0,0\right)$. 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 $S^$, then $g^f$ is the identity on $S^$, so we have an isotopy $J$ from the identity to $g^f$. The map $gJ$ is then an isotopy from $g$ to $f$.

Some authors use the term ''Alexander trick'' for the statement that every homeomorphism of $S^$ 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 $f\colon S^ \to S^$ be a homeomorphism, then : 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 [[exotic sphere#Twisted spheres|twisted spheres.