Exercise 1.2.4

Proof. We will perform a series of deformation retractions. For any point $p\in {ℝ}^3-X$, we can retract it onto $S^2$ by moving it along the line joining $0$ and $p$. This is clearly continuous and is moreover a retraction since points already on $S^2$ do not move. The retract of ${ℝ}^3-X$ is then the $2n$-punctured sphere since each line that passes through the origin punctures the sphere in two locations; let ${\{x_i\}}_{1\le i\le 2n}$ be the set of missing points. Then, picking $x_{2n}$ as our pole, we can stereographically project $S^2-{\{x_i\}}_{1\le i\le 2n}$ onto ${ℝ}^2-{\{x_i^{\prime }\}}_{1\le i\le 2n-1}$, which as we recall is a homeomorphism. Finally, we can deformation retract the $(2n-1)$-punctured plane to ${\vee <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{2n-1}{S}^1$ in the same way we usually retract the doubly punctured plane to $S^1\vee S^1$, and so ${\pi }_1({ℝ}^3-X)={\ast <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{2n-1}\mathbb{Z}$ by Example $1.21$. □