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?

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    $\begingroup$ Presumably, you want some "obvious" embedding of $S^1\wedge S^1$ into $\mathbb R^3$? $\endgroup$ – Thomas Andrews Sep 10 '14 at 8:41
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    $\begingroup$ Since you refer to an eight shape, I think you mean $S^1 \vee S^1$ (wedge sum), not $S^1 \wedge S^1$ (smash product). (The latex code for wedge-sums is \vee, and \wedge produces smash products.) $\endgroup$ – Martin Brandenburg Sep 10 '14 at 9:33
  • $\begingroup$ What Thomas is implying is that the answer (most likely) depends on exactly how you embed your $S^1$s into $\Bbb{R}^3$. The fundamental group of the complement of a single $S^1$ varies considerably depending on how knotty the ring is. For the simple loop (=the unknot) it is $\Bbb{Z}$ all right, but even for the trefoil embedding (=the snake on the surface of the torus here) the fundamental group of the complement is $\langle a,b |a^2=b^3\rangle$. $\endgroup$ – Jyrki Lahtonen Sep 10 '14 at 19:41

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.


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