TITLE: Fundamental group of complement in $\mathbb{R}^3$ of $z$-axis and two circles
QUESTION [2 upvotes]: Let $C_1$ be the circle in the plane $z=1$ centred at $(0,0,1)$ and $C_2$ be the circle in the plane $z=2$ centred at $(0,0,2)$, both of radius $1$.
I would like to compute $\pi_1(X)$, where $X=\mathbb{R}^3 \setminus \left(\{z\text{-axis}\} \cup C_1 \cup C_2\right)$, using Seifert-van Kampen.
 I know how to do it without Seifert-van Kampen. The set $X$ is homeomorphic to $A \times S^1$, where $A$ is a twice punctured plane, meaning that $X$ is obtained by rotating the twice punctured plane around the $z$-axis.
So we have
$$\pi_1(X)=\pi_1(A \times S^1)=\pi_1(A) \times \pi_1(S^1)=(\mathbb{Z}*\mathbb{Z}) \times \mathbb{Z}.$$
I would also like to obtain a presentation of this group.

REPLY [0 votes]: Kevin has already answered your question. Here I give you another argument for arriving that the same presentation as Kevin.
Firstly observe that your space deformation retracts to $\Bbb T^2 \cup_{S^1} \Bbb T^2$, that is, two torus stacked on top of one another such that they intersect each other in a circle.
Now, this space can be obtained by attaching a 2-cell to the space $\Bbb T^2 \vee S^1$ as follows: let $a$ be a meridian in $\Bbb T^2$, $b$ be a longitude in $\mathbb{T}^2$ and lastly, let $c$ denote the wedged $S^1$. We attach the 2-cell along $cbc^{-1}b^{-1}$, so, $b$ and $c$ commute. We already know that $a$ and $b$ commute. Thus, $\pi_1(X) = \langle a, b, c | aba^{-1}b^{-1}, cbc^{-1}b^{-1} \rangle$.
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