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What is the fundamental group of the complement of the closed disk in $\mathbb R^{3}$ ?

i.e $X = \{(x,y,z)\in \mathbb R^{3} \ | \ z=0, \ x^{2}+y^{2} \leq 1\}$

what is $\pi_{1}(\mathbb R^{3}-X)$ ?

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Your space can be contracted to $S^2$, so the fund. group is trivial. –  user8268 Jun 3 '11 at 19:40
    
thanks, I was thinking that way we can contract the closed disk to a point then our space has homotopy type of $R^{3}-{0}$. Then this means that the complement of the disk can be contracted to complement of the point, is that right ? –  hebele Jun 3 '11 at 20:05
    
Yes, but that's also homotopy-equivalent to $S^2$. –  Zhen Lin Jun 3 '11 at 20:31
    
okay, thanks for help, by the way this is NOT a homework question, I am gonna a take Top prelim, and it is past top prelim question –  hebele Jun 3 '11 at 20:35
    
"Then this means that the complement of the disk can be contracted to complement of the point, is that right ?" $S^2$ is not contractible, but it is simply connected. –  JSchlather Jun 4 '11 at 3:29

1 Answer 1

This question has been answered in comments. The space is homotopy equivalent to $S^2$, so the fundamental group is trivial.

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