I am trying to find the fundamental group of the space $X=\mathbb{R}^3\setminus S$ where $S$ is the union of the $z$-axis, and two circles of radius 1 on the $xy$-plane, centered at $(0,0,1)$ and $(0,0,-1)$.
My approach: We know that the fundamental group of $\mathbb{R}^3$ minus circle+$z$-axis is that of a torus, which is $\mathbb{Z}_2$. Split the given space into two, namely, into $U_1=X\cap \{(x,y,z-\epsilon):(x,y,z)\in\text{upper half plane}, 0<\epsilon<1\}$ and $U_2=X\cap \{(x,y,z+\epsilon):(x,y,z)\in\text{lower half plane}, 0<\epsilon<1\}$. Then $U_1\cup U_2=X$, and $U_1$, $U_2$ are panth connected and open. The intersection then deformation retracts to $\mathbb{R}^2-\{0,0\}$, whose fundamental group is $\mathbb{Z}$.
We now have to figure out the induced homomorphism $i^*:\mathbb{Z}\to\mathbb{Z}_2$. This is where I am most uncertain. I thought, since the generating loop of $U_1\cap U_2$ simply gets included to one of the generating loops of the torus (the one going around the center hole), we get the presentation \begin{equation} \pi(X) \cong <a,b,c,d:ab=ba,cd=dc,a=c>? \end{equation} (What is this group, by the way?)
Is what I did correct?
Many thanks!
Maybe it's easier to think of $X$ as a plane with two holes that rotates around the $z$ axis. If the plane with two holes is $Y$,then $X$ should be homeomorphic to $Y \times S^1$. And thus the fundamental group is $(\mathbb{Z}*\mathbb{Z})\times\mathbb{Z}$. I hope it helps.
And, by the way, the fundamental group of the torus is $\mathbb{Z}\times\mathbb{Z}$, how did you get $\mathbb{Z_{2}}$?