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Consider $S^m$ embedded in $S^n$ $ (m < n ) $ as the subspace $ \left\{ (x_1, x_2, ..., x_{m+1}, 0,...0) | \sum x_i^2 = 1 \right\} $. Show that $ S^n \backslash S^m $ is homotopy equivalent to $ S^{n-m-1} $

My thoughts:

I've first considered the case $ n = 2 $, $m = 1 $ (as this is the only case I can actually draw). So, we have the unit sphere (variables $x, y , z$ ) minus the unit circle (variables $x,y$), and want to show it's homotopy equivalent to the two point space (I think).

So think of $S^0$ as $ \left\{ (0,0,-1) , (0,0,1) \right\} $. Then we have the inclusion $ i : S^0 \to S^2 \backslash S^1$ as a potential homotopy equivalence in one direction.

For the other, let $ f : S^2 \backslash S^1 \to S^0 $ be defined by $ f ( (x,y,z) ) = (0,0, \frac{z}{||z||}) $. This is continuous thanks to the removal of $ S^1 $.

Now, $ fi = {id}_{S^0} $. It remains to be shown that $ if \simeq {id}_{S^2 \backslash S^1} $.

The linear map $ H : S^2 \backslash S^1 \times [0,1] \to S^2 \backslash S^1 $ defined by $ H(x,t) = tx + (1-t)f(x) $ doesn't work, as $ S^2 \backslash S^1 $ isn't convex. I'm not sure what else could work as a homotopy, or if what I've done makes any sense.

Any help would be appreciated. Thanks

EDIT: I now realise I could have started with $ n = 1$ , $ m = 0$ ...

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    Consider $(x_1,\ldots,x_n)\mapsto((t-1)x_1,\ldots,(t-1)x_{m+1},h(t)x_{m+2},h(t)x_{n+1})$ with $h$ chosen for each point such that $h(0)=1$ and $h$ increases just quickly enough to compensate for the die-off of the first $n+1$ coordinates. Prove that the combination of the right $h$ choices for each point actually fit together continuously.2011-10-13

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