Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2023 / 1 / b

A retraction from a topological space $X$ onto a subset $Y$ is a continuous map $r:X\to Y$ satisfying $r(y)=y$ for every $y\in Y$. Equivalently, $r\circ \iota ={\mathrm{id}}_Y$, where $\iota :Y\hookrightarrow X$ is the inclusion.

The no-retraction theorem for the finite-dimensional closed unit ball saysIndeed, if $r$ were such a retraction, $F(x)=-r(x)$ would be a continuous map from ${\mathbb{B}}^n$ into itself. The Brouwer fixed-point theorem would give $x=F(x)=-r(x)$. Since $|r(x)|=1$, $x$ lies on the boundary, where $r(x)=x$. Thus $x=-x$, contradicting $|x|=1$. Here the Brouwer fixed-point theorem asserts that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.