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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!

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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 $$\langle \color{red}x,\color{green} y | \color{green}y=\color{red}x \color{green}y \color{red}{x^{-1}}, \color{red}x = \color{green}y \color{red}x \color{green}{y^{-1}}\rangle \equiv \mathbb Z^2.$$

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  • $\begingroup$ if the circles were not linked the group would be the free product $\mathbb Z*\mathbb Z$ because the order matters. $\endgroup$
    – user51427
    Dec 6, 2012 at 0:20
  • $\begingroup$ 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? $\endgroup$
    – Mel
    Dec 6, 2012 at 16:23
  • $\begingroup$ @Mel, I just found the fundamental group of R^3 \ two linked circles. $\endgroup$
    – user51427
    Dec 6, 2012 at 16:28
  • $\begingroup$ So are you using Van Kampen's thm? $\endgroup$
    – Mel
    Dec 6, 2012 at 16:30
  • $\begingroup$ @Mel, yes. ---- $\endgroup$
    – user51427
    Dec 6, 2012 at 16:34

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