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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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    This seems to be problem 6 of this: http://www.dpmms.cam.ac.uk/study/II/AlgebraicTopology/2010-2011/ATqns1.pdf2011-01-29
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    It is, that's why I labeled it homework - I don't want a solution, just advice.2011-01-29
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    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?2011-01-30

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