Radial retraction of a punctured ball ($r(x)=a+\tau (x)(x-a)$)

Removing any interior point $a$ from the Euclidean closed unit ball allows a retraction onto its unit sphere. For $v=x-a\ne 0$, the positive root of $|a+tv{|}^2=1$ isThen $r(x)=a+\tau (x)v$ is a continuous map on the punctured closed unit ball, has norm one, and fixes its boundary because $\tau (x)=1$ there. The quadratic equation has one positive and one negative root since $|a|<1$, so the chosen root is unambiguous.