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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. – Henning Makholm Oct 13 '11 at 18:59

I think you should tag your questions with "homework" if they are homework, as Zhen Lin pointed you out.

Anyway, here is my solution. It just needs a little bit of notation. Write the points of $\mathbb{R}^{n+1} = \mathbb{R}^{m+1} \times \mathbb{R}^{n-m}$ as $(x,y)$, with $x\in\mathbb{R}^{m+1}$ and $y \in \mathbb{R}^{n-m}$.

So, $S^m$ is embedded in $S^n \subset \mathbb{R}^{n+1}$ as those $(x,y) \in\mathbb{R}^{n+1} $ such that $y=0$ and $\| x \| = 1$. Likewise, embed $S^{n-m-1}$ in $S^n$ as those $(x,y)$ such that $x=0$ and $\|y\| =1$. So, $S^{n} \backslash S^m$ are those $(x,y)\in S^n$ such that $y\neq 0$. Briefly:

  1. $S^m = \left\{ (x,y) \in \mathbb{R}^{n+1} \ \vert \ y = 0 \ , \ \|x \|=1 \right\}$
  2. $S^{n-m-1} = \left\{ (x,y) \in \mathbb{R}^{n+1} \ \vert \ x = 0 \ ,\ \|y \|=1 \right\}$
  3. $S^n \backslash S^m = \left\{ (x,y) \in \mathbb{R}^{n+1}\ \vert \ y \neq 0 \ , \ \|x\|^2+\|y\|^2 =1 \right\} $

Then, our inclusion is just

$$ i : S^{n-m-1} \longrightarrow S^n\backslash S^m \ , \qquad i(0,y) = (0,y) \ . $$

In the other direction, now the map is pretty obvious: since points $(x,y) \in S^{n-m-1}$ are those with $x = 0$, given a point $(x,y) \in S^n\backslash S^m$, the simplest way to obtain a point in $S^{n-m-1}$ seems to put $x=0$ and then normalize the second coordinate:

$$ f: S^n \backslash S^m \longrightarrow S^{n-m-1} \ , \qquad f(x,y) = \left( 0,\frac{y}{\|y\|} \right) \ . $$

Notice that, since $y\neq 0$, $f$ is continuous and

$$ (f\circ i) (0,y) = f(0,y) = \left( 0,\frac{y}{\|y\|} \right) = (0,y) \ , $$

because $\|y\|=1$.

As for the other composition, we have

$$ (i\circ f)(x,y) = \left( 0,\frac{y}{\|y\|} \right) \ , $$

and we need now an homotopy between $\mathrm{id_{S^n\backslash S^m}}$ and $i\circ f$. For this, the old trick of the straight line homotopy normalized suffices. Define

$$ H : (S^n\backslash S^m)\times I \longrightarrow S^n\backslash S^m \ , \qquad H(x,y,t) = \frac{\left((1-t)x, (1-t)y + t\frac{y}{\|y\|} \right)}{\left\| \left((1-t)x, (1-t)y + t\frac{y}{\|y\|} \right) \right\|} \ . $$

And we have

$$ H(x,y,0) = \frac{( x,y )}{\|(x,y)\|} = (x,y) \ , $$

since $(x,y) \in S^n$ and

$$ H(x,y,1) = \frac{\left(0, \frac{y}{\|y\|} \right)}{\left\| \left(0, \frac{y}{\|y\|} \right) \right\|} = \left( 0, \frac{y}{\|y\|} \right) \ . $$

We need to prove two more things:

  • That $H(x,y,t) \in S^n\backslash S^m$ for all $(x,y,t) \in (S^n\backslash S^m)\times I$.
  • And that $H$ is continuous.

We can do it in one go, just showing that $(1-t)y + t\frac{y}{\|y\|} \neq 0 $ for all $(x,y,t) \in (S^n\backslash S^m)\times I$. But

$$ (1-t)y + t\frac{y}{\|y\|} = 0 \qquad \Longleftrightarrow \qquad (1-t)y\|y\| +ty = 0 $$

Since $y\neq 0$, this would be the same as

$$ (1-t)\|y\| +t = 0 \ . $$

But, for $t \in [0,1]$, the left-hand side of the above equation represents points in the segment between $\|y\|$ and $1$. Since $y \neq 0$, this segment does not include the point $0$. Hence we are done.

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$\mathbb R^{n+1}$ is the direct sum

$$\mathbb R^{n+1} = \mathbb R^{m+1} \oplus \mathbb R^{n-m}$$

The unit sphere in $\mathbb R^{n+1}$ is $S^n$. The unit sphere in $\mathbb R^{m+1}$ is $S^m$.

Observe that a regular neighbourhood of $S^m$ in $S^n$ is a homeomorphic/diffeomorphic to a disc cross $S^n$. And the complement is a regular neighbourhood of the sphere in $\mathbb R^{n-m}$, which is a $S^{n-m-1}$.

Another way to make this argument is this way:

$\partial (I^{n+1}) \simeq S^n$. where $I$ is an interval.

But $I^{n+1} = I^{m+1} \times I^{n-m}$, so

$$ \partial (I^{n+1}) = (\partial I^{m+1}) \times I^{n-m} \cup I^{m+1} \times (\partial I^{n-m})$$


$$ S^n \simeq S^m \times D^{n-m} \cup D^{m+1} \times S^{n-m-1}$$

where I've identified the boundary of a cube with the sphere, and the cube with the disc. This is a union of sets, and their common intersection is $\partial (S^m \times D^{n-m}) = S^m \times S^{n-m-1} = \partial (D^{m+1} \times S^{n-m-1})$.

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