5
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.