No-retraction theorem

Remarks

Suppose such a retraction $r$ exists, and let $i \colon S^{n-1} \to D^n$ be the inclusion, so that $r \circ i = \id_{S^{n-1}}$. Apply reduced homology in degree $n - 1$: the identity of $\tilde{H}_{n-1}(S^{n-1}; \Z) \cong \Z$ factors through $\tilde{H}_{n-1}(D^n; \Z) = 0$, which is impossible. For $n = 1$ the argument is simply that $D^1 = [-1, 1]$ is connected and $S^0 = \set{-1, 1}$ is not. For smooth retractions there is also a proof by Sard's theorem: the preimage of a regular value would be a compact $1$-manifold with only one boundary point.

Example of use: the theorem gives the Brouwer fixed-point theorem. If $f \colon D^n \to D^n$ had no fixed point, the ray from $f(x)$ through $x$ would meet $S^{n-1}$ in a point $r(x)$, and $r$ would be a retraction of $D^n$ onto $S^{n-1}$.

The ball cannot be replaced by any space: the cylinder $S^{n-1} \times [0, 1]$ retracts onto its end $S^{n-1} \times \set{0}$ by $(x, t) \mapsto (x, 0)$.

Going deeper

For a student

The no-retraction theorem is equivalent to the Brouwer fixed-point theorem, and each implies the other with a short argument. In homotopy terms, it says that the identity map of $S^{n-1}$ is not null-homotopic, and so it is a statement about the degree of the identity. Similar retraction arguments, with homology or cohomology, show that the boundary of a compact manifold is never a retract of the manifold.