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.

  • 4
    $\begingroup$ 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. $\endgroup$ – BW. Nov 2 '14 at 19:52
  • 1
    $\begingroup$ 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

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 $\vee$.

  • $\begingroup$ 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

(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.


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 $\Bbb{R}^n,n\geq3$ $\pi_1(\Bbb{R}^n\backslash\{p_1,\ldots,p_k\})=\pi_1(\Bbb{R}^n)=\{0\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.