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I am helping some students study for an exam. We have a two part problem. The first part is to prove that any smooth map $f: S^k \to \mathbb{R}^n$ extends to a smooth map $F: D^{k + 1} \to \mathbb{R}^n$. We tried to define $$ F(x) := |x|f\left( \frac{x}{|x|} \right) $$ and $F(0) := 0$. This is certainly continuous. But, we couldn't show that it is smooth. Is this the correct map?

The second part of the problem says, if $M \subseteq \mathbb{R}^n$ is a closed manifold of dimension $m$ such that $k + 1 < n - m$, then any smooth map $S^k \to \mathbb{R}^n - M$ extends to a smooth map $D^{k + 1} \to \mathbb{R}^n - M$. I thought maybe I could use the dimension condition to get that the image is contractible, and use some version of the first part. But, that's all I've got.

I should mention that the class is just a graduate introduction course. So, they've got the basics of manifolds, Sard's Theorem, etc. No Morse Theory though.

J126
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  • I don't know if your $F$ is smooth, but if it's not then I'll bet that $|x|^2f(x/|x|)$ is. – JonathanZ Jul 04 '19 at 20:31
  • Why don't you use that $\mathbb{R}^n$ is contractible? Any linear homotopy from such $f$ to a point in its image will be smooth and will provide such an extension. That is for the first part. – Laz Jul 04 '19 at 20:49
  • The second part implies that $\pi_k(\mathbb{R}^n-M)=0$ for all those $k$. So maybe that is something you could start with. It seems trickier, though. – Laz Jul 04 '19 at 21:02
  • I'm confused. For the second part, does the extended map have range $\Bbb R^n$ or $\Bbb R^n-M$? It seems to me that, as written, the problem requires no hypotheses other than to apply the first part directly. – Ted Shifrin Jul 06 '19 at 00:07
  • @Laz Their class did not cover higher homotopy groups. So, I think that's not a fair approach. Though it does seem promising. – J126 Jul 06 '19 at 22:39
  • @TedShifrin It should be $\mathbb{R}^n - M$. I have corrected it. – J126 Jul 06 '19 at 22:39
  • OK, Joe. I just said "homotopy" because that is it's name, but you don't have to mention that word if you want. – Laz Jul 06 '19 at 23:04
  • Hmm, if we had $k+1<m$ we could do it by the transversality extension theorem (since the map is transverse to $M$ by default on the boundary), extending as a map to $\Bbb R^n$ and then wiggling to make it transverse to $M$. The $n-m$ suggests that we should be thinking about deforming the given smooth map into a fiber of the normal bundle of $M$. – Ted Shifrin Jul 06 '19 at 23:05

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Given an application $f : \mathbb{S}^{k} \longrightarrow \mathbb{R}^{n}$, define its radial extension by $F : \mathbb{R}^{k+1} \longrightarrow \mathbb{R}^{n}$, with : $F(x) = |x|.f \left(\dfrac{x}{|x|} \right)$ if $x \neq 0$ and $F(0) = 0$.

Then, $F$ is differentiable at the point $0 \in \mathbb{R}^{k+1}$ if, and only if, $f$ is (the restriction to $\mathbb{S}^{k}$) of a linear transformation.

Indeed, by definition of $F$, we have: $\dfrac{F \left(tx \right)}{t}$, if $ t > 0$ and $\dfrac{F \left(tx \right)}{t} = - F(-x)$, if $t < 0$.

As $F(0) = 0$, suppose $F$ differentiable at point $x = 0$, there exists $\displaystyle \lim_{t \to 0} \dfrac{F \left(tx \right)}{t}$. Therefore $F(x) = - F(-x)$. $$F(x) = \displaystyle \lim_{t \to 0} \dfrac{F \left(tx \right)}{t} = F'(0).x$$

therefore, F coincides with the linear transformation $F'(0) : \mathbb{R}^{k+1} \longrightarrow \mathbb{R}^{n}$. The reciprocal is true.

Allan Ramos
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