 Solution

The Brouwer fixed-point theorem says that every continuous map from a closed Euclidean ball to itself has a fixed point. For the closed disc $D\subset {\mathbb{R}}^2$ its equivalent no-retraction theorem says that there is no continuous $r:D\to \partial D$ satisfying $r(z)=z$ on $\partial D$.

Assume the fixed-point theorem and suppose such a retraction exists. Then $F(x)=-r(x)$ maps $D$ continuously into itself. A fixed point must lie on the boundary, where $r(x)=x$, giving $x=-x$, impossible on the unit circle.

Conversely suppose $F:D\to D$ is continuous and has no fixed point. From $F(x)$ draw the ray through $x$ and let $r(x)$ be its exit point on the circle. To check continuity rather than merely relying on the picture, put $v=x-F(x)\ne 0$. The exit point is $F(x)+t(x)v$, whereThis is continuous, lies on the circle and has $t(x)\ge 1$. On the boundary the exit point is $x$, so $r$ is a retraction, contradicting the assumed theorem. Hence the two disc formulations are equivalent; the same argument works for a ball in any dimension.