This is an exercise from Bredon:
Let $X$ be the quotient space of $\mathbb{D}^2$ obtained by identifying points on the boundary that are 120º apart. Find $\pi_1(X)$.
Could you give me any hints on how to proceed?
I tried to apply Van-Kampen's theorem to the sets $U = B(0,1/2)$ and $V = X \setminus \{0\}$, but what is the fundamental group of $V$?
On
Consider the above picture, Here the red, blue and purple arcs are identified.
You have $\Pi_{1}(U)=\{0\}$ and $\Pi_{1}(V)=\langle a\rangle\cong \Bbb{Z} $ where $a$ denotes the path $t\mapsto e^{2i\pi t}$ .
And $U\cap V$ deformation retracts to the path $a^{3}$ . $\Pi_{1}(U\cap V)=\langle a^{3}\rangle \cong \Bbb{Z}$ .
Thus By Van-Kampen Theorem, $\Pi_{1}(X)$ is the amalgamated product over $\Pi_{1}(U\cap V)$ . So $\Pi_{1}(X)= \langle a : a^{3}= 1\rangle = \frac{\Bbb{Z}}{\Bbb{3Z}}$
Three points in $\partial D^2 =S^1$ are same. Hence quotient space $\partial D^2/\sim$ is still $S^1$. Hence $\pi_1( \partial D^2/\sim )=\pi_1(S^1)=\mathbb{Z}=(a)$.
But in $X$, curve representing $a^3$ is contractible. Hence $\pi_1(X)=\mathbb{Z}_3$.