Fundamental Group of $S^3$ \ A , where A is a figure 8. I'm trying to figure out the fundamental group of a topological space X obtained from $\mathcal{S}^3$ by removing from it a "figure 8", i.e. the set $K=\{(x_1,x_2,0,0)\in\mathbb{R}^4 | x_1^2 + (x_2-1)^2 =1 \} \cup \{(x_1,x_2,0,0)\in\mathbb{R}^4 | x_1^2 + (x_2+1)^2 =1 \}$. My gut tells me it's going to be $\mathbb{Z}*\mathbb{Z}$, but I'm unsure how to prove it.

$S^3\setminus K$ is homotopy equivalent to a figure eight space itself! Here's how I visualize it. Removing the figure eight space is like first removing a ball and then adding two straight solid tubes to the inside of the ball you just removed. When you remove a ball in $S^3$, what's left is still a ball (think upper and lower hemisphere), and so by adding the solid tubes, you get a space which is a $3$-ball with two solid tubes added. This deformation retracts to a copy of $K$, which is a wedge of two circles. Now $\pi_1(K)\cong \mathbb Z*\mathbb Z$, the free group on $2$-generators, and we're done.
The fact that the complement of an unknotted wedge of two circles is itself homotopy equivalent to a wedge of two circles is related to the standard Heegard splittings of $S^3$ if you want some keywords to google.
