Fundamental group of $S^2 - \{N,S\}$

Question

I would like to calculate the fundamental group $\Pi_1$ of $S^2 - \{N,S\}$, where $N,S$ are the north respectively south pole of $S^2$, but do not see yet how to do this.

Answer 1

Using a stereographic projection shows that $S^2\setminus\{N,S\}$ is homeomorphic to $\mathbb{R}^2\setminus\{0\}$. What is the fundamental group of this space?

Answer 2

The map \begin{align}\phi:S^2\setminus \{N,S\}&\to S^1\times (-1,1)\\ \begin{pmatrix}x\\ y\\ z\end{pmatrix}&\mapsto \frac{1}{\sqrt{x^2+y^2}}\begin{pmatrix}x\\ y\\ z\sqrt{x^2+y^2}\end{pmatrix}\end{align} is a diffeomorphism.

Its inverse is $$\begin{pmatrix}a\\ b\\ c\end{pmatrix}\mapsto \begin{pmatrix}a\sqrt{1-c^2}\\ b\sqrt{1-c^2}\\ c\end{pmatrix}$$

$S^1$ is a deformation retract of $S^1\times(-1,1)$.

Answer 3

Yet another way to compute this is to note that $S^2\setminus \{N,S\}$ deformation retracts onto a graph. The fundamental group of a graph is not too hard to compute, once one can homotope an internal edge to a point.