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What is the fundamental group of the complement in $\mathbb R^3$ of a line and a circle. There are actually two cases to consider, one where the line goes through the interior of the circle and the other where it doesn?t. Are these two spaces homotopy equivalent?

I guess the answer is Z*Z. But I really have no idea about how to prove this rigorously.

  • in the two case it is Z. It is generated by the homotopy of a cosed path around the line – Tsemo Aristide Oct 18 '15 at 20:49
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    You can easily use van Kampen's thorem in the case where the circle does not encircle the line to see that the fundamental group is $\Bbb Z * \Bbb Z$. In the case where they're linked the fundamental group of the complement is $\Bbb Z^2$; you can either use van Kampen's more carefully or demonstrate that the space deformation retracts onto a torus. –  Oct 18 '15 at 20:52
  • @MikeMiller Thanks! Could you please explain more about the second case. What do you mean by a torus around the circle? –  Oct 18 '15 at 20:55
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    I deleted that last bit because it might be confusing. For ease of visualization take the line to be the $z$-axis and the circle to be the unit circle in the $xy$-plane. Then this space deformation retracts onto a torus; on one side of the torus is the circle, on the other side is the line. –  Oct 18 '15 at 20:56
  • @MikeMiller That is hard to imagine. Is that same with deformation retract the complement of two circles linked? –  Oct 18 '15 at 21:17
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    It's the same as the complement of the Hopf link in $S^3$. (The line, plus the point at infinity, is a circle.) If you search complement of Hopf link on MSE there are probably pictures of this. –  Oct 18 '15 at 21:40
  • @MikeMiller OK, thank you very much! –  Oct 18 '15 at 21:46

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