Retraction

A retraction of a topological space $X$ onto a subspace $A$ is a continuous map $r:X\to A$ whose restriction to $A$ is the identity.

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 $F(x)=\sqrt{1-\Vert x{\Vert }^2}{e}_0+Sx$. Then $\Vert F(x)\Vert =1$ and $F(x)\ne x$: equality would force all coordinates of $x$ to coincide, hence vanish, contradicting $F(0)=e_0$.

For $d=x-F(x)$, setThe nonvanishing denominator makes $r$ a continuous map. Expanding its squared norm gives $\Vert r(x)\Vert =1$. On $\Vert x\Vert =1$, the fraction is one, so $r(x)=x$. Thus the finite-dimensional no-retraction theorem does not extend to this closed unit ball.

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.

A once-punctured genus-$h$ closed orientable surface retracts to a wedge of $2h$ circles. A retraction from ${\Sigma }_g$ would inject its degree-one cohomology as an isotropic subspace of a symplectic vector space for the Poincare duality pairing, since the punctured surface has zero degree-two cohomology. The maximal isotropic dimension is $g$, giving $2h\le g$. Equality is realized by folding the double of the punctured surface onto one copy. Additional handles in the other copy can be pinched away while fixing the joining boundary, proving sufficiency for every embedding of such a one-boundary subsurface when $g\ge 2h$.