Find the fundamental group of a topological space X obtained from S3 by removing a figure 8

Find the fundamental group of $S^3$ = {$(x, y, z, w)$ $\in$ $\mathbb{R}^4$ : $x^2 + y^2 + z^2 + w^2 = 1$} by removing from it a figure eight, that is, the set:

$K$ = {$(x, y, $$\frac{1}{\sqrt2}$,$0)$ $\in$ $\mathbb{R}^4$ : $x^2 + y^2 = \frac{1}{2}$} $\cup$ {$(\frac{1}{\sqrt2}, 0, z, w)$ $\in$ $\mathbb{R}^4$ : $z^2 + w^2 = \frac{1}{2}$}

I am able to show that the fundamental group of the figure 8 is the free group on two generators, $\mathbb{Z}*\mathbb{Z}$.

But what I'm trying to do with this question is show that the above space, $S^3 -K$, is homotopy equivalent to the figure eight space, which I'm not sure how to do formally.

I guess one idea is to map this space homeomorphically to some subspace of $\mathbb{R}^3$ by using a stereographic projection, as defined in Munkres' Topology. But I'm having trouble defining such a homeopmorphism.

Can anyone assist?

Perhaps strangely, the result is just $S^{1}\vee S^{1}$ again. Maybe the easiest way to see this is to present $S^{3}$ as a $3$-disk with its boundary $S^{2}$ crushed to a point and recall the homotopy equivalence of the $3$-disk minus a “figure eight” to $S^{2}\vee S^{1}\vee S^{1}$.