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.
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.