= Radial retraction of a punctured ball
{title2=$r(x)=a+\tau(x)(x-a)$}
Removing any interior point $a$ from the Euclidean <closed unit ball> allows a <retraction> onto its <unit sphere>. For $v=x-a\ne0$, the positive root of $|a+tv|^2=1$ is
$$
\tau(x)=\frac{-a\cdot v+\sqrt{(a\cdot v)^2+(1-|a|^2)|v|^2}}{|v|^2}.
$$
Then $r(x)=a+\tau(x)v$ 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.
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