Prove that $S^{n-1} \simeq \Bbb R^n - \{ 0 \}$.

Question

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 $z$.

Answer 1

$f:S^{n-1}\to \Bbb R^{n}\setminus \{0\}$ , by $f(x)=x$

and $g:\Bbb R^n\setminus \{0\}\to S^{n-1}$ , by $f(x)=\dfrac{x}{\|x\|}$.

$g\circ f =Id_{S^{n-1}}$ and $(f\circ g)(x)=\dfrac{x}{\|x\|}$ which is homotopic to $Id_{R^n\setminus \{0\}}$, by the homotopy $H(x,t)=tx+(1-t)\frac{x}{\|x\|}$.

Note that $\forall t\in [0,1]$, $H(x,t)\neq 0$: In fact if $tx+(1-t)x/\|x\|=0$ then $tx=-(1-t)x/\|x\|$, hence (applying $\|.\|$) $\|x\|=1$, and $(1-t)=t$ that is $t=\frac12$, but $H(x,\frac12)=x\neq O$.

Answer 2

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.