Fundamental group of $S^n\setminus S^k$

Question

I'm doing review questions for grad school examinations and I came across one that's stumped me for a while:

$S^n = \{(x_1, \dots , x_{n+1} \colon \Sigma x_i^2 = 1\}$ and $S^k = \{(x_1, \dots , x_{n+1} \in S^n \colon x_{k+2} = \cdots = x_{n+1} = 0\}$ What is $\pi_1(S^n\setminus S^k)$?

I've approached this with Seifert–van Kampen but can't get anywhere.

Answer 1

Hint: Stereographic projection. Pick a point on $S^k$.

Answer 2

$S^n \setminus S^k \simeq S^{n-k-1}$ which should answer your question.

Proof. Define

$j : S^{n-k-1} \to S^n \setminus S^k , j(y_1,\dots,y_{n-k}) = (0,\dots,0,y_1,\dots, y_{n-k}),$

$p: S^n \setminus S^k \to \mathbb R^{n-k} \setminus \{ 0\}, p(x_1,\dots,x_{n+1}) = (x_{k+2},\dots,x_{n+1}).$

$r : S^n \setminus S^k \to S^{n-k-1}, r(x) = p(x)/\lVert p(x) \rVert$.

We have $p(j(y)) = y$ and thus $r(j(y)) = y$. The latter means $r \circ j = id$.

For $x \in S^n \setminus S^k$ we have $j(r(x)) = (0,\dots,0,p(x)/\lVert p(x) \rVert)$. The linear path $l_x :[0,1] \to \mathbb R^{n+1}, l_x(t) = (1-t) x + t j(r(x))$, does not go through $0$:

For $t = 1$ we clearly have $l_x(t) \ne 0$ and for $t \ne 1$ the equation $l_x(t) = 0$ implies $x_1 = \dots = x_{k+1} = 0$. Hence we get $x = (0,\dots,0,p(x))$ which also shows $\lVert p(x) \rVert = 1$. We conclude $j(r(x)) = x$, thus $x = (1-t) x + t j(r(x)) = 0$ which is impossible.

This means that $l'_x: [0,1] \to S^n, l'_x(t) = l_x(t)/\lVert l_x(t) \rVert$, is well-defined.

We have $l'_x(t) \in S^n \setminus S^k$ (which is equivalent to $p(l'_x(t)) \ne 0$) for all $t$:

Since $p$ is linear, $p(l'_x(t)) = 0$ implies $p(l_x(t)) = 0$ and further

$$0 = (1-t) p(x) + t p(j(r(x)) = (1-t) p(x) + t r(x) = \\(1-t) p(x) + t p(x)/\lVert p(x) \rVert = \left(1-t+t/\lVert p(x) \rVert \right)p(x) .$$ Since $1-t+t/\lVert p(x) \rVert \ge 1-t+t = 1$, we conclude that $p(x) = 0$ which is impossible for $x \in S^n \setminus S^k$.

Therefore

$$H : (S^n \setminus S^k) \times [0,1] \to S^n \setminus S^k, H(x,t) = l'_x(t)$$ is well-defined and we have $H(x,0) = x, H(x,1) = j(r(x))$, i.e. $id \simeq j \circ r$.