$X =\{z\in \mathbb{C} : 1 \leqslant |z| \leqslant 2\}$. Find a fundamental group of $X /_{\sim} $, where $z_{1} \sim z_{2}$ if and only if $[(|z_{1}| = |z_{2}| = 1$ and $(\frac{z_{1}}{z_{2}})^{3} = 1) $ or $( |z_{1}| = |z_{2}| = 2$ and $(\frac{z_{1}}{z_{2}})^{4} = 1)$ or ($z_{1}=z_{2})]$.
(It was edited - firstly I wrote an incorrect definition of X.)
I was trying to do it similarly as how you find a fundamental group of $\mathbb{R}P_{2}$ and it goes like this: $(|z_{1}| = |z_{2}| = 1$ and $(\frac{z_{1}}{z_{2}})^{3} = 1)$ if and only if $z_{1},z_{2} \in S(0,1)$ and the difference between their arguments equals $\frac{\pi}{3}$, whereas $ (|z_{1}| = |z_{2}| = 2$ and $(\frac{z_{1}}{z_{2}})^{4} = 1)$ if and only if $z_{1},z_{2} \in S(0,2)$ and the difference between their arguments equals $\frac{\pi}{4}$. So the only nontrivial equivalence classes are on $S(0,1)$ and $S(0,2)$. Images of those circles in $p: X \rightarrow X/_{\sim} $ are also circles, let name them $S_1$ and $S_2$ respectively. Now we can treat $X/_{\sim}$ as $X\cup_{f}(S_{1}\cup S_{2})$, where f is the restriction of p.
There is a lemma which says that if $Z=\bar{B^{2}} \cup_{f} Y$ where $f: S^{1} \rightarrow Y$ is a continuous map, then for $y=f(1)$, $i: Y \rightarrow Z$ enduces an epimorphism $i_{*}: \pi_{1}(Y,y) \rightarrow \pi_{1}(Z,y)$ with kernel $<<[\lambda_{f}]>>$, where $[\lambda_{f}]$ is a path such that for $t \in I=[0,1]$ we have $t \mapsto f(\exp(2 \pi i t)) \in Y$. Moreover, if $<S,R>$ is the presentation for $\pi_{1}(Y,y)$, then $<S|R\cup \{[\lambda_{f}]\}>$ is the presentation of $\pi_{1}(Z,y)$.
Is it possible to use those fact in some tricky way in my problem?