Fundamental group of $\mathcal{B}^2/\mathcal{R}$ Let $\mathcal{B}^2$ be the unit closed disc in $\Bbb{R^2}.$
Now suppose we form the quotient $$\mathcal{B}^2/\mathcal{R}$$
where $\mathcal{R}$ is the equivalence relation generated by $z\sim z\exp{2i\pi /n}$ for $z\in S^1.$
How can I compute the fundamental group of such a space ?
Intuitively I would say it's homeomorphic to the cône of $S^1$ which it's contractile and therefore the fundamental group is trivial.
But how can I prove it? Does I need Van Kampen theorem ? 
 A: As noted in the comments, this space is not just the cone over $S^1$ if $n > 1$. (It is in fact the mapping cone of the map $S^1 \to S^1, z \mapsto z^n$.) For instance, $n=2$ yields a projective plane. We can use Van Kampen to determine $\pi_1(B^2/R)$: (Mouse over for progressive spoilers.)

Let $\pi$ be the projection onto the quotient. Take $U = \pi(\{z \in B^2: \lvert z \rvert < 3/4 \})$, $V = \pi(\{z \in B^2: \lvert z \rvert > 1/4 \})$. Now to apply Van Kampen, we have to determine the homotopy type of $U,V,U \cap V$, and the behavior of the maps induced by the inclusions $U \cap V \rightrightarrows U,V \rightrightarrows B^2/R$ at the homotopy level.



 $U$ is contractible, and $U \cap V \simeq S^1$. Note that $V$ deformation retracts onto a circle. Then the map $\mathbb{Z} \cong \pi_1(U\cap V) \to \pi_1(V) \cong \mathbb{Z}$ induced by the inclusion $U \cap V \to V$ is simply multiplication by $n$. (A loop going around the circle $U \cap V$ once goes around the circle $V$ $n$ times.) Therefore $\pi_1(B^2/R) \cong \mathbb{Z}_n$.

