= Solution
The planar <Brouwer fixed-point theorem> says that every continuous map from the closed disk $D$ to itself has a <fixed point>. If a <retraction> $r:D\to\partial D$ existed, $x\mapsto-r(x)$ would be a continuous self-map of $D$. At a <fixed point> $x=-r(x)$, $x$ lies on the boundary, where $r(x)=x$; this would give $x=-x$, impossible on the unit circle.
Conversely, suppose a continuous self-map $f:D\to D$ has no <fixed point>. For each $x$, take the ray starting at $f(x)$ through $x$, and let $r(x)$ be its exit point on $\partial D$. The denominator $|x-f(x)|$ never vanishes, and solving the quadratic intersection with the circle shows that the exit point depends continuously on $x$. More explicitly, with $v=x-f(x)$,
$$
r(x)=f(x)+\lambda(x)v,\qquad \lambda(x)=\frac{-f(x)\cdot v+\sqrt{(f(x)\cdot v)^2+(1-|f(x)|^2)|v|^2}}{|v|^2}.
$$
The ray exits beyond or at $x$, so $\lambda\ge1$, and when $x\in\partial D$ its exit point is $x$. Hence $r$ is a <retraction>. \b[Nonexistence of a boundary retraction and the planar fixed-point theorem are equivalent.]
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