= 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 $v=x-F(x)\ne0$. The exit point is $F(x)+t(x)v$, where
$$
t(x)=\frac{-F(x)\cdot v+\sqrt{(F(x)\cdot v)^2+(1-|F(x)|^2)|v|^2}}{|v|^2}.
$$
This is continuous, lies on the circle and has $t(x)\geq1$. On the boundary the exit point is $x$, so $r$ is a <retraction>, contradicting the assumed theorem. Hence \b[the two disc formulations are equivalent]; the same argument works for a ball in any dimension.
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