That is prove that the $n-1$ sphere is homotopic to the Euclidean space without the origin.
Two topological spaces $X, Y$ are said to be homotopic if there are maps $f: X \to Y$ and $g: Y \to X $ such that $f \circ g \simeq Id_Y$ and $g \circ f \simeq Id_X.$
I thought that we could define (when $n = 3$, for example)
$f:= f(x,y,z) = (\frac{x}{\sqrt{x^2 + y^2 + z^2 – 1}},\frac{y}{\sqrt{x^2 + y^2 + z^2 – 1}},z)$ this ensure that we remove the origin. So $f: \Bbb S^2 \to \Bbb R^3 – \{ 0 \} $
On the other hand,
$g: g(x,y,\pm\sqrt{ (x^2 + y^2)-1}) = (x,y,\pm\sqrt{ 1-(x^2 + y^2)}) $ where $g:\Bbb R^3 – \{ 0 \} \to S^2$.
I have trouble making the last component in $f$ maps back to $z$.
Best Answer
The idea is to show that $S^{n}$ is a deformation retract of $\mathbb R^{n+1}\setminus\{0\}$. It is easy to see that $H(x,t)=tx+(1-t)\frac{x}{\|x\|}$ works.