Fundamental group of complement in $\mathbb{R}^3$ of $z$-axis and two circles

Question

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.

Answer 1

Your post seems to have two questions: Seifer-Van Kampen Thm and group presentation, I'll answer them separately...

Use Seifert-Van Kampen's Thm to compute $\pi_1(X)$:

For convenience, you can regard $X$ as $X'=B\setminus(\{(x,y,z)\mid x,y=0\}\cup (S^1\sqcup S^1))$, where $B$ is homeomorphic to a 3 dimensional ball (There is an obvious deformation retraction). Take $U=X'\cap\{(x,y,z)\mid z<2\}$ and $V=X'\cap\{(x,y,z)\mid z>1\}$, then $U\cap V\simeq\{\text{ punctured plane}\}\simeq S^1$, which means $\pi_1(U\cap V)\cong\Bbb{Z}=\langle \alpha\rangle$. Also, we have $U\approx V$ (you can easily see this by drawing a picture), and $U\simeq T^2=S^1\times S^1$, which means $\pi_1(U)\cong\Bbb{Z}^2=\langle a,b\rangle$ and $\pi_1(V)\cong\Bbb{Z}=\langle c,d\rangle$.

This conclusion doesn't seem to be obvious, but if you observe $U$, it has a vertical hole and a removed ring inside of it. By expanding the vertical hole and the tube inside of it, do you see the homotopy equivalence? If you feel confused, I can draw a picture to illustrate it. Alternatively, you can think of it as a punctured plane rotating with respect to $z$-axis, so now there is a homotopy equivalence between this space and $S^1\times S^1$ because the punctured plane deformation retracts onto $S^1$ (Hope it's clear now)

Now, consider the following mappings. $i_*:\pi_1(U\cap V)\to\pi_1(U)$ induced by the inclusion says $i_*(\alpha)=b$ (it is a loop that encloses the central vertical hole). Similarly, $j_*:\pi_1(U\cap V)\to\pi_1(V),\alpha\mapsto d$. Apply Seifert-Van Kampen's Thm, we get $$\pi_1(X)\cong\pi_1(X')\cong(\pi_1(U)*\pi_1(V))/N\cong(\Bbb{Z}^2(a,b)*\Bbb{Z}^2(c,d))/\langle b^{-1}d\rangle$$

Note that $a$ represents the basic loop that encloses the lower tube, whereas $c$ represents the loop that encloses the upper tube created by the removal of $S^1$.

Group presentation:

Claim: $(\Bbb{Z}^2(a,b)*\Bbb{Z}^2(c,d))/{\langle b^{-1}d\rangle}=(\Bbb{Z}*\Bbb{Z})\times\Bbb{Z}$

$$(\Bbb{Z}^2(a,b)*\Bbb{Z}^2(c,d))/\langle b^{-1}d\rangle=\langle a,b,c,d\mid b=d,ab=ba,cd=dc\rangle=\langle a,b,c\mid ab=ba,bc=cb\rangle$$ We see that $a$ and $c$ forms a free group of two generators and they're not commutative. So this is $(\Bbb{Z}*\Bbb{Z})\times\Bbb{Z}$, which agrees to your answer that solves it from a different perspective.

It's also possible to derive the same group presentation from your answer. Let $a$ be the loop that encloses the lower point of $A$ (resp. $c$ that encloses the upper one), and $b$ be the loop around the vertical hole. Then, $a$ and $c$ are the generators of the group $\Bbb{Z}*\Bbb{Z}$ and $b$ commutes with them. So, we have $\langle a,b,c\mid ab=ba,bc=cb\rangle$.

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Answer 2

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, $\pi_1(X) = \langle a, b, c | aba^{-1}b^{-1}, cbc^{-1}b^{-1} \rangle$.