# Homotopy equivalence of Unit Balls [duplicate]

Possible Duplicate:
$S^n \backslash S^m$ homotopy equivalent to $S^{n-m-1}$

I'm trying to show that, if we embed $S^m$ in $S^n$ as the subspace $\{ (x_1,x_2, \ldots, x_{m+1}, 0, 0, \ldots, 0) \}$, $S^n-S^m$ is homotopy equivalent to $S^{n-m-1}$.

In order to show the homotopy equivalence, I tried to use the inclusion map $S^{n-m-1}$ into $S^n - S^m$, including in the last $n-m$ coordinates, and I'm trying to use functions along the lines of

$$f(x_1,x_2, \ldots,x_{n+1}) = (x_{m+1}, x_{m+2}, \ldots, x_n, \sqrt{x_1^2+...x_m^2+x_{n+1}^2}) .$$

Then $f \circ g$ is easily shown to be the identity map except with issues of positivity in the last coordinate, so I tried

$$f(x_1,x_2, \ldots, x_{n+1}) = (x_{m+1}, x_{m+2}, \ldots, x_n, a(x_{n+1}) \sqrt{x_1^2+...x_m^2+x_{n+1}^2}) ,$$

where $a$ is either $\pm 1$ or $0$ depending on the sign of the final coordinate, but then there's an issue in the other direction, i.e. that the final coordinate can only be $0$ when $x_{n+1}=0$.

Is it possible to amend my approach somehow so that $g \circ f$ is also homotopic to the identity map? Thank you for any thoughts or advice.

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## marked as duplicate by Zhen Lin, t.b., Srivatsan, Guess who it is., Henning MakholmNov 15 '11 at 0:47

Here is a hint: there is one case that can be visualized, namely that $S^2 \setminus S^1 \cong S^0$. Can you write down a homotopy between them in coordinates, and then generalize your construction to arbitrary dimensions? –  Dan Petersen Jan 30 '11 at 0:00
Hint: If $(x_{1}, \ldots, x_{m+1},y_{m+2},\ldots,y_{n+1}) \in S^{n} \smallsetminus S^{m}$ then $\frac{(\lambda x, y)}{\|(\lambda x,y)\|} \in S^{n}$ for all $\lambda$. Use this to retract $S^{n} \smallsetminus S^{m}$ to $S^{n-m-1}$ (embedded as $(0,y)$).