Fundamental group and universal cover for this quotient space For a non-zero integer $p$ define the topological space $L_p$ by:
\begin{equation}
L_p=\mathbb{D}^2\sqcup\mathbb{S}^1\big{/}z\sim{z^p}
\end{equation}
Check that $L_1\cong\mathbb{D}^2$, and more generally, find the fundamental group $\pi_1(L_p)$ and a universal cover for $L_p$. 
This question comes from an optional assignment sheet given after the completion of my introductory course on algebraic topology, and in particular I have done next to nothing on universal covers, so any help on how to even start approaching that part would be much appreciated (my lecturer didn't do any examples). 
More importantly, I'm not even sure how to correctly interpret the quotient - should I be viewing $\mathbb{D}^2\sqcup\mathbb{S}^1$ sort of like a closed disc, and then imposing the equivalence relation on every point in this closed disc? This seems like it can't be the correct interpretation, since then $L_1$ is viewed as a closed disc without any equivalence relation, which is certainly not homeomorphic to the open disc? It seems that imposing a CW structure on $L_p$ would be useful in computing the fundamental groups, but again I haven't seen many examples of this, and I don't want to start on this until I've interpreted the question correctly!
If the interpretation is indeed ambiguous then I will ask my lecturer, but I think it's more likely I have misunderstood what's going on. If anyone can clear things up and explain how to proceed it would be much appreciated. 
 A: This question is answered in Example 1.47 of
Hatcher, Allen. "Algebraic topology. 2002." Cambridge UP, Cambridge 606.9 (2002).
See http://pi.math.cornell.edu/~hatcher/AT/AT.pdf.
The space $L_p$ is obtained by attaching $\mathbb{D}^2$ to $\mathbb{S}^1$ by the map $\mathbb{D}^2 \supset \mathbb{S}^1 \stackrel{f_p}{\rightarrow} \mathbb{S}^1$ where $f_p(z) = z^p$ wraps $\mathbb{S}^1$ $p$-times around itself. In other words, it is the quotient of $\mathbb{D}^2$ by the equivalence relation $\sim_p$ given by $z_1 \sim_p z_2$ if (a) $z_1, z_2 \in \mathbb{S}^1$ and $z_1^p = z_2^p$ or (b) $z_1, z_2 \notin \mathbb{S}^1$ and $z_1 = z_2$.
Let $U_r(\zeta) \subset \mathbb{R}^2$ denote the open disk with radius $r$ around $\zeta \in \mathbb{R}^2$. The quotient map $q : \mathbb{D}^2 \to L_p$ maps the interior $U_1(0)$ of $\mathbb{D}^2$ homeomorphically onto an open subspace $U'_1(0)$ of $L_p$, on the boundary $\mathbb{S}^1$ any two points differing by an angle of $2k\pi /p$ are mapped by $q$ to same point. Note that $\mathbb{S}^1/\sim_p$ is homeomorphic to $\mathbb{S}^1$.
Let us describe "nice" open neighborhoods of an equivalence class $[z] \in L_p$ with $z \in \mathbb{S}^1$. $[z]$ contains the $p$ points $\zeta_k(z) = z e^{2k \pi i / p} \in \mathbb{S}^1$, $k = 1,...,p$. For sufficiently small $\epsilon > 0$ the sets $V_\epsilon(\zeta_k(z)) = U_\epsilon(\zeta_k(z)) \cap \mathbb{D}^2$ (which are open in $\mathbb{D}^2$) are pairwise disjoint. Then $V_\epsilon([z]) = \bigcup_{k=1}^p q(V_\epsilon(\zeta_k(z)))$ is an open neighborhood of $[z]$ in $L_p$. The collection of all $V_\epsilon([z])$ for sufficiently small $\epsilon$ is in fact a basis of open neighborhoods of $[z]$ in $L_p$. Note that by construction $q^{-1}(V_\epsilon([z])) = \bigcup_{k=1}^p V_\epsilon(\zeta_k(z))$.
The universal covering of $L_p$ is $\tilde{L}_p = \mathbb{D}^2 \times \lbrace 1,...,p \rbrace / \sim$, where $\sim$ is generated by $(z,j) \sim (z,k)$ for $z \in \mathbb{S}^1$ and $j,k \in \lbrace 1,...,p \rbrace$; that is, in the disjoint union of $p$ copies of $\mathbb{D}^2$ the boundaries are identified in the obvious way. Let $\tilde{q} : \mathbb{D}^2 \times \lbrace 1,...,p \rbrace  \to \tilde{L}_p$ denote the quotient map.
The covering map $c : \tilde{L}_p \to L_p$ is given as follows:
Define $\gamma : \mathbb{D}^2 \times \lbrace 1,...,p \rbrace \to \mathbb{D}^2, \gamma(z,k) = z e^{2k\pi i / p}$. It is easy to verify that $\gamma(z,j) \sim_p \gamma(z,k)$. Therefore $\gamma$ induces a unique map $c : \tilde{L}_p \to L_p$ such that $c \circ \tilde{q} = q \circ \gamma$.
Note that $\tilde{q}^{-1}(c^{-1}(V_\epsilon([z]))) = \gamma^{-1}(q^{-1}(V_\epsilon([z]))) = \gamma^{-1}(\bigcup_{k=1}^p V_\epsilon(\zeta_k(z))) = $ $ \bigcup_{k=1}^p \gamma^{-1}(V_\epsilon(\zeta_k(z))) = \bigcup_{k=1}^p V_\epsilon(\zeta_k(z)) \times \lbrace 1,...,p \rbrace$.
Therefore $c^{-1}(V_\epsilon([z])) = \bigcup_{k=1}^p W_\epsilon(\zeta_k(z))$, where $W_\epsilon(\zeta_k(z))$ is obtained from $V_\epsilon(\zeta_k(z)) \times \lbrace 1,...,p \rbrace$  by identifying the corresponding boundary points $(z,j)$ with $z \in V_\epsilon(\zeta_k(z)) \cap \mathbb{S}^1$, $j = 1,...,p$. The $W_\epsilon(\zeta_k(z))$ are open in $\tilde{L}_p$ and pairwise disjoint. Each $W_\epsilon(\zeta_k(z))$ is mapped by $c$ homeomorphically onto $V_\epsilon([z]))$. This shows that $c$ is a $p$-sheeted covering (note that $c^{-1}(U'_1(0)) = \bigcup_{k=1}^p \tilde{q}(U_1(0) \times \lbrace k \rbrace)$, where each of the pairwise disjoint open sets $\tilde{q}(U_1(0) \times \lbrace k \rbrace)$ is mapped by $c$ homeomorphically onto $U'_1(0)$).
That $L_1 \cong \mathbb{D}^2$ is obvious -  on $\mathbb{D}^2$ we have $z_1 \sim_1 z_2$ if and only if $z_1 = z_2$.
