The foundamental group of $S^3/\mathbb{Z}_3$ expressed by Seifert-van Kampen theorem

Question

$S^3=\{(z_1,z_2); z_1,z_2\in\mathbb{C},|z_1|^2+|z_2|^2=1\}$, Then $S^3=S_1^3\cup S_2^3$, where $S_1^3=\{(z_1,z_2)\in\mathbb{C}^2; |z_1|^2\leqslant \frac{1}{2}, |z_2|^2=1-|z_1|^2\}$ and $S_2^3=\{(z_1,z_2)\in\mathbb{C}^2; |z_1|^2\geqslant \frac{1}{2}, |z_2|^2=1-|z_1|^2\}$. $S_1^3$ and $S_2^3$ are homeomorphic to solid tori. $S^3/\mathbb{Z}_3=S_1^3/\mathbb{Z}_3\cup S_2^3/\mathbb{Z}_3$, (where the action on $S^3$ in $\mathbb{Z}_3$ is generated by $v\mapsto e^{2\pi i/3}v$). How to use the Seifert-van Kampen theorem to express $\mathbb{Z}_3=\pi_1(S^3/\mathbb{Z}_3)$ as a quotient group of the free product of the fundamental group of $S_1^3/\mathbb{Z}_3$ and $S_2^3/\mathbb{Z}_3$. Is it true that $\pi_1(S_1^3/\mathbb{Z}_3)=\mathbb{Z}$ ?

Answer 1

To answer your last question first, yes, $\pi_1(S_1^3 / \mathbb Z_3) \approx \mathbb Z$. One simply deformation retracts $S_1^3 / \mathbb Z_3$ to the central circle $\{(z_1,z_2)\in \mathbb C^2 ; \, z_2 = 0, |z_1|^2 = 1\}/ \mathbb Z_3 $ using the deformation induced from deforming $S_1^3$ onto $\{(z_1,z_2)\in \mathbb C^2 ; \, z_2 = 0, |z_1|^2 = 1\}$. This is similar to the Moebius band whose boundary winds twice around the central circle. The real stumbling block for computing $\pi_1(S^3 / \mathbb Z_3)$ using Van Kampen Theorem is computation of the fundamental group of $(S_1^3 \cap S_2^3)/ \mathbb Z_3$, the boundary torus. For this reason I believe it is better to regard the covering space $p:S^3 \rightarrow S^3/ \mathbb Z_3$ and to use the following theorem from Hatcher's Algebraic Topology:

For an action $G$ on a path-connected, locally path-connected space $X$ satisfying the property for each $x \in X$ there is a neighborhood $V$ of $x$ such that all images $g(V)$ for varying $g \in G$ are disjoint, then $G \approx \pi_1(X/G)/p_*(\pi_1(X))$ where $p_*$ is the induced homomorphism of the quotient map $p: X \rightarrow X/G.$

It should be clear that the action $\mathbb Z_3$ on $S^3$ satisfies this condition(choose $V = B(x,\varepsilon)$ with $\varepsilon$ small). Additionally as $S^3$ is path-connected, locally path-connected we may apply the theorem. Thus $\mathbb Z_3 \approx \pi_1(S^3 / \mathbb Z_3)/ p_*(\pi_1(S^3))$. And as $S^3$ is simply connected, $p_*(\pi_1(S^3)) \approx 0$, and it follows $ \pi_1(S^3 / \mathbb Z_3) \approx \mathbb Z_3$.

EDIT(More directly addressing the question):

Considering the action of $\mathbb Z_3$ on the boundary torus, $\mathbb T$, we have the map $\gamma \mapsto \gamma^{-2}$ (similar to the antipodal map $x \mapsto -x$ being the action of $\mathbb Z_2$). We study this action of $\mathbb Z_3$ on the torus and notice that it has no fixed points so the map $p: \mathbb T \rightarrow \mathbb T/ \mathbb Z_3$ is a covering space, and $\mathbb T/ \mathbb Z_3$ must have a fundamental group with $\mathbb Z \times \mathbb Z$ as an index $3$ subgroup. Thus $\mathbb T/ \mathbb Z_3$ is the torus with a twisting action around both $S_1^3,S_2^3$ where a generator of the torus wraps once around one circle while twice around the other. So for the inclusion maps $i_1:\mathbb T/ \mathbb Z_3 \rightarrow \langle \alpha \rangle \approx S_1^3/ \mathbb Z_3, \,i_2:\mathbb T/ \mathbb Z_3 \rightarrow \langle \beta \rangle \approx S_2^3/ \mathbb Z_3 $ and for generators $a, b$ of $\pi_1(\mathbb T/ \mathbb Z_3)$ we have $N$ generated by $$i_1(a)i_2(a)^{-1}= \alpha \beta^{2}\\ i_1(b)i_2(b)^{-1}=\alpha^{-2}\beta^{-1}\\ i_1(ab)i_2(ab)^{-1}=\alpha^{-1}\beta$$ Thus we conclude $\pi_1(S^3/ \mathbb Z_3) \approx \mathbb Z * \mathbb Z/N$ and simplifying: $$\begin{split}\mathbb Z * \mathbb Z/N & = \langle \alpha,\beta \rangle / \langle \alpha \beta^{2},\alpha^{-2}\beta^{-1},\alpha^{-1}\beta \rangle\\ & = \langle \alpha \rangle / \langle \alpha^3 \rangle\\ & = \mathbb Z_3\end{split}$$