Show that \(S^2 - \{p_0, p_1\} \simeq S^1\).
Solution
1. After a homeomorphism of \(S^2\), we may take \(p_0\) and \(p_1\) to be the north and south poles.
Any two distinct points of the sphere can be carried to the two poles by a homeomorphism of \(S^2\).
2. The complement of the two poles is homeomorphic to \(S^1\times\mathbb R\).
Write a non-polar point in spherical coordinates as \[ (\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi), \qquad \theta\in S^1,\quad 0<\phi<\pi. \] Thus the complement is \(S^1\times(0,\pi)\), and \((0,\pi)\cong\mathbb R\).
3. Hence\[ \boxed{S^2\setminus\{p_0,p_1\}\simeq S^1}. \]
The product \(S^1\times\mathbb R\) deformation-retracts onto \(S^1\times\{0\}\).