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I need help determining to what $S^n/${k points}--the n-dimensional sphere missing a finite k number of points-- is homotopy equivalent.

I tried envisioning the above for n=2:

$S^2\setminus${1 point} is homotopy equivalent to a point.

$S^2\setminus${2 points} is homotopy equivalent to $S^1$.

$S^2\setminus${3 points} is homotopy equivalent to $\vee_2S^1$, or the union of 2 $S^1$ joined at 1 point.

Assuming that the above are correct, I believe that I can generalize to $S^n\setminus${k points} is homotopy equivalent to $\vee_{k-1}S^{n-1}$.

Thus, I have 2 main questions.

  1. Is my generalization correct?

  2. And if so, how do I show this homotopy equivalence rigorously? I came up with the specific examples for $S^2$ by actually envisioning a sphere with k punctures.

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    $\begingroup$ The basic idea is to view $S^n\setminus\{k\text{ points}\}$ as a punctured sphere with $k-1$ other missing points. By stereographic projection (there are known formulas for this), this is homotopy equivalent to $\mathbb{R}^n$ minus $k-1$ points. The punctured plane is then homotopy equivalent to the wedge of $k-1$ copies of $S^{n-1}$. Geometrically, fix some base point, and then draw $k-1$ loops out from this point that encircle exactly one of the missing points. The rest of $\mathbb{R}^n$ deformation retracts onto these loops, and this is equivalent to the wedge sum. $\endgroup$ – Ben West Nov 2 '14 at 19:52
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    $\begingroup$ Right, first you should notice that $S^n$ minus a point is diffeomorphic to $\mathbb{R}^n$, and from there the visualization is easier; for example now you can visualize the case $n = 3$. $\endgroup$ – Qiaochu Yuan Nov 2 '14 at 19:54
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Your intuition is correct, though your notation for a few things is off.$^{\dagger}$

Without loss of generality, you can consider the $k$ missing points to be $(\cos \frac{2i\pi}{k}, \sin \frac{2i\pi}{k}, 0)$ for $0 \le i < k$. Then you can work out an explicit description of a homotopy equivalence with an $(k-1)$-fold wedge of circles by using the following intuition:

  • Expand the missing points towards the north and south poles of the sphere, so that the $i^{\text{th}}$ hole now looks like an open interval bent around the sphere from the north to the south pole passing through $(\cos \frac{2i\pi}{k}, \sin \frac{2i\pi}{k}, 0)$;
  • Fatten the holes so that the only remaining parts of the sphere are the north and south poles and $k$ curves joining them together;
  • Join the north and south poles together by contracting one of the $k$ curves to a point.

What you are left with is isomorphic to $\overbrace{S^1 \vee S^1 \vee \cdots \vee S^1}^{k-1\ \text{copies}}$.

This generalises to higher dimensions immediately: just stick in some more zeros.


$^{\dagger}$Regarding notation: $/$ typically means 'quotient' and $\setminus$ means 'remove', so the spaces you're considering are really $S^2 \setminus \{ k\ \text{points} \}$. Also, $\wedge$ typically refers to the smash product; what you want is the wedge, denoted $\vee$.

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  • $\begingroup$ Thank you so much for your help! I also really appreciate the note on the notation. I always get the notation confused, so that was very useful. $\endgroup$ – Jess Nov 2 '14 at 21:15
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(1) Yes

(2) The starting point is observing that a once-punctured $n$-sphere is homeomorphic to $\mathbb{R}^n$, so if you remove $k$ points from $S^n$, you get $\mathbb{R}^n$ with $(k-1)$ points removed.

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The stereographic projection gives us an homeomorphism between $\Bbb{S}^n\backslash\{p\}$ with $\Bbb{R}^n$, thus, for $n=2$ we have $$\Bbb{S}^2\backslash\{p\}\cong \Bbb{R}^2$$ and for high dimension we have $$\Bbb{S}^n\backslash\{p\}\cong \Bbb{R}^n$$ For $\Bbb{R}^n,n\geq3$ $\pi_1(\Bbb{R}^n\backslash\{p_1,\ldots,p_k\})=\pi_1(\Bbb{R}^n)=\{0\}$.

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