= Solution
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, $r\circ\iota=\operatorname{id}_Y$, where $\iota:Y\hookrightarrow X$ is the inclusion.
The <no-retraction theorem> for the finite-dimensional <closed unit ball> says
$$
\boxed{\text{For }n\geq1,\text{ no retraction }\mathbb B^n\longrightarrow\partial\mathbb B^n\text{ exists}.}
$$
Indeed, 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>.
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