Fundamental group of the complement of an eight shape

Question

What is $\pi_1(\mathbb{R}^3 \setminus (S^1 \vee S^1))$? I would say that the space deformation retracts to a sphere $S^2$ surrounding the missing eight with two sticks stuck in the two loops of the eight. Those sticks can each be closed to form a loop, so that the space would be homotopy equivalent to $S^1 \vee S^2 \vee S^1$, yielding $\mathbb{Z} \ast \mathbb{Z}$ as the fundamental group. Is this correct?

Answer 1

You are essentially right. However (as noted in comments) you use $\wedge$ (meaning smash product) instead of $\vee$ (meaning wedge sum). And also the fundamental group of $S^1 \vee S^1$ is $\mathbb{Z} \ast \mathbb{Z}$ rather that $\mathbb{Z} × \mathbb{Z}$ (i.e. free sum rather than product (which is direct sum at the same time)).

So after your edits, it's correct.