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How can calculate the fundamental group of $\Bbb R^3\setminus (\text{axis }z \cup \text{circle})$ utilizing the Seifert-Van Kampen theorems. What opens sets I have to utilize ? I thought $A= \Bbb R^3 - (D^2 \times R)$ and other open sets?

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  • $\begingroup$ Yes..... I edited $\endgroup$ Jan 18 '17 at 22:04
  • $\begingroup$ To get better intuition you could fit that space into a compact cube which contains the circle and the axis. For example, if the circle has radius $1$ a cube of radius $2$ would fit. Why are you allowed to do this? Both spaces are homotopy equivalent. $\endgroup$
    – user60589
    Jan 19 '17 at 10:38
  • $\begingroup$ Then, if you cut the cube into slices, you get a cubes with $n$ lines removed. You can compute its fundamental group easily with Seifert-Van Kampen, which will be the free group on $n$ generators. - Using this, you can compute the fundamental group of your space using Seifert-Van Kampen again and taking care of the generators. $\endgroup$
    – user60589
    Jan 19 '17 at 10:47
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This can be a possible answer. Consider $A=${$(x,y) \in \Bbb R^2 | x>0 and (x,y)\neq (0,1)$} and $S^1$. Therefore $\Bbb R^3\setminus (\text{axis }z \cup \text{circle})$ can be obtained as $A\times S^1$ therefore $$\pi _1(\Bbb R^3\setminus (\text{axis }z \cup \text{circle}))= Z\times Z$$

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