No differentiable retraction on unit ball implies no continuous retraction.

Question

I have seen it mentioned in multiple proofs of the no retraction theorem (e.g. here) that the fact that there exists no differentiable retraction $r$ from the closed unit ball $B^n$ to the unit sphere $S^{n-1}$ implies the non-existence of a continuous (i.e. "normal") retraction $q$ between these two spaces.

My best guess so far is that this can be done with a proof by contradiction and some variation of the Stone-Weierstrass theorem if that's what's meant by "a suitable approximation argument" in the paper linked above.

For this we'd assume that $\not\exists r \land \exists q$. By Stone-Weierstrass (if I understand that theorem correctly) there'd then be a sequence of differentiable functions converging uniformly towards $q$, none of which can be a retraction as per the assumption. However, I don't see how that would yield a contradiction so I'm really not sure if my idea is any good.

Answer 1

Here is an elementary but tricky argument I'm pilfering from class notes of Robert Bruner that I unearthed with a Google search.

Let $B$ be the closed unit ball and let $B_2$ be the closed ball of radius $2$. Given a continuous retraction $r\colon B\to\partial B$, we can define a continuous function $g\colon B_2\to B_2-\{0\}$ by $$g(x) = \begin{cases} r(x), & |x|\le 1 \\ x, & 1\le |x|\le 2\end{cases}.$$ Now set $R(x)= 2g(x)/\|g(x)\|$. $R$ is a continuous retraction $B_2\to\partial B_2$ that is obviously smooth in a neighborhood of $\partial B_2$. Thus, by Whitney extension, there is a smooth retraction $\tilde R\colon B_2\to \partial B_2$. Oops.