= Retraction of a Hilbert ball onto its sphere
The <closed unit ball> $B$ in the infinite-dimensional real <Hilbert space> $\ell^2$ admits a <retraction> onto its <unit sphere>. Write $S$ for the forward <unilateral shift operator> and set $F(x)=\sqrt{1-\|x\|^2}\,e_0+Sx$. Then $\|F(x)\|=1$ and $F(x)\ne x$: equality would force all coordinates of $x$ to coincide, hence vanish, contradicting $F(0)=e_0$.
For $d=x-F(x)$, set
$$
r(x)=F(x)+\frac{2(1-\langle F(x),x\rangle)}{\|d\|^2}d.
$$
The nonvanishing denominator makes $r$ a <continuous map>. Expanding its squared <norm> gives $\|r(x)\|=1$. On $\|x\|=1$, the fraction is one, so $r(x)=x$. Thus the finite-dimensional <no-retraction theorem> does not extend to this <closed unit ball>.
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