We can assume that $m < n-1$.

$S^n \setminus S^m$ are the elements of $S^n$ such that the last $n-m$ entries are not all zero.

I would like to compute the fundamental group $\pi_1(S^n \setminus S^m)$

I do not know how to tackle this problem. For $n>2$ I cannot picture what is happening.

I know that $\pi_1(S^2 \setminus S^0)=\Bbb{Z}$, (corrected) because we have removed the points $(1,0,0)$ and $(-1, 0, 0)$.

If I were to calculate $\pi_1(S^2 \setminus S^1)$, it would be trivial, because we have cut the sphere in half by removing all points with $x_3=0$ and all loops on the half spheres can be contracted to trivial loops. I can see why this holds for all $S^n \setminus S^{n-1}$, but it is not asked.

I suspect that removing $S^0$ always leaves inconctractible loops and the rest only trivial loops. But I cannot imagine it at all and I know no formal process which would help me with this task. Any hints welcome.

  • $\begingroup$ If you remove a point from $S^2$, you get the plane thus $\pi_1(S^2 \setminus S^0)$ should be just $\mathbb Z$. $\endgroup$ – Roman Hric Oct 15 '19 at 19:10
  • $\begingroup$ @RomanHric We remove two points though, and a loop can circle each one of them. $\endgroup$ – 77and33is100 Oct 15 '19 at 19:13
  • $\begingroup$ @B.Swan You can slide one of those loops to another, if you are looking at the sphere. (If the loops have the same winding number wrt the poles, of course.) Another way to see why $\pi_1(S^2 \backslash S^0)=\mathbb{Z}$ is to realize that $S^2 \backslash S^0$ is homeomorphic to a cylinder, which is homotopy equivalent to a circle. $\endgroup$ – Aloizio Macedo Oct 15 '19 at 19:16
  • $\begingroup$ Roman's point is that when you remove one point you get the plane. When you remove another, you get the punctured plane, which has fundamental group $\mathbb{Z}$. $\endgroup$ – Aloizio Macedo Oct 15 '19 at 19:17
  • $\begingroup$ @AloizioMacedo Thank you, I understand what I got wrong. $\endgroup$ – 77and33is100 Oct 15 '19 at 19:19

Taking the stereographic projection originating from one of the points in $S^m$, we have an homeomorphism between $S^n- S^m$ and $\mathbb{R}^n-\mathbb{R}^m$. The latter is homotopy equivalent to $\mathbb{R}^{n-m} \backslash \{0\},$ which in turn is homotopy equivalent to $S^{n-m-1}$. Hence, $S^n-S^m \stackrel{hmtp}{\simeq} S^{n-m-1}$.

It follows that if $n-m=2$, then the fundamental group is $\mathbb{Z}$. If $n-m=1$, we have two (path)-connected components both with trivial fundamental group. (This is clear from the start: $S^{n+1}-S^{n}$ is just two disks.) In all other cases, we are left with a connected set with trivial fundamental group.


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