Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

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

share|improve this answer
    
if the circles were not linked the group would be the free product $\mathbb Z*\mathbb Z$ because the order matters. –  user51427 Dec 6 '12 at 0:20
    
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 6 '12 at 16:23
    
@Mel, I just found the fundamental group of R^3 \ two linked circles. –  user51427 Dec 6 '12 at 16:28
    
So are you using Van Kampen's thm? –  Mel Dec 6 '12 at 16:30
    
@Mel, yes. ---- –  user51427 Dec 6 '12 at 16:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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