4

What is the fundamental group of $\mathbb R^3 \setminus$ two linked circles? This is an example from Hatcher, he says that this space deformation retracts onto the wedge product of$S^2$ and a torus separating the two circles, but I have not really understood what exactly this deformation retract is.

I understand that the points inside of the torus can be pushed to the boundary of the torus and points outside of the sphere can be pushed to the sphere, but what about points outside the torus and inside the sphere.

Thanks much in advance!

amWhy
  • 209,954
Mel
  • 41
  • 1

1 Answers1

2

In the diagram enter image description here the equation $\color{blue}b = \color{red}c\color{green}a\color{red}{c^{-1}}$ holds because enter image description here therefore the group of the complement of enter image description here is $$\langle \color{red}x,\color{green} y | \color{green}y=\color{red}x \color{green}y \color{red}{x^{-1}}, \color{red}x = \color{green}y \color{red}x \color{green}{y^{-1}}\rangle \equiv \mathbb Z^2.$$

  • if the circles were not linked the group would be the free product $\mathbb Z*\mathbb Z$ because the order matters. –  Dec 06 '12 at 00:20
  • Hi thanks for your reply, but I still don't get what the deformation retract is. And can you please explain in words what you have shown? – Mel Dec 06 '12 at 16:23
  • @Mel, I just found the fundamental group of R^3 \ two linked circles. –  Dec 06 '12 at 16:28
  • So are you using Van Kampen's thm? – Mel Dec 06 '12 at 16:30
  • @Mel, yes. ---- –  Dec 06 '12 at 16:34