Hatcher's Algebraic Topology, Example 1.35 Hatcher considers the mapping cylinder A from $S^{1}$ to $S^{1}$ under the function $z \rightarrow z^m$. He claims without explanation that the universal cover of A is homeomorphic to a product $C_m \times \mathbb{R}$ where $C_m$ is the graph that is a cone on $m$ points. I don't understand where that came from. 
Here is a link to Hatcher's book, chapter 1. The example can be found on page 65.
http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf
 A: Work backwards. 
Consider $C_m$ as a subset of the complex plane (just so I can conviently write down the group action later), $$C_m = \{\lambda e^{2k\pi i/m} \mid 0 \leq \lambda \leq 1, k \in \Bbb Z\}.$$ Then $C_4$, for instance, is $+$; $C_6$ is $*$, $C_2$ is $|$.
There's a group action of $\Bbb Z$ on $C_m \times \Bbb R$, given by $n \cdot (z, t) = (e^{2\pi i n/m}z, t+n)$. This descends to an action of $\Bbb Z/m\Bbb Z$ on $C_m \times \Bbb R/m\Bbb Z = C_m \times S^1$. (Try to imagine what this looks like as a subset of $\Bbb R^3$, and the group action on it, in your mind.)  Now, why is this the mapping cylinder you describe? As a hint, the 'tendrils' of $C_m$ (times $S^1$) should end up corresponding to the 'cylinder' side of the mapping cylinder, and the center $0 \times S^1$) should correspond to the circle glued to by $z \mapsto z^m$.
A: Let's formalise this. I find it is very healthy to be explicit with our homeomorphisms and our maps in topology: visual intuition can be misleading.
Note: when Hatcher draws the picture of the cover of $A$, he fills in solid horizontal bars. These do not imply that the space is otherwise hollow (like rungs on a ladder): they are just example lines. The horiztonal lines he draws all share the same orbit under $\Bbb Z$ (explained below) but his picture is not to be literally interpreted. The cover is $C_m\times\Bbb R$ which shall be $m$ lots of solid 'wedges' ($\Bbb R\times I$) all joined together in a sheaf, a common line in the centre identifiable with $\Bbb R$.
Let $m\ge1$ be an integer. Let $S^1\times I\twoheadrightarrow A_m$ be the quotient map for the mapping cylinder of $S^1\to S^1,\,z\mapsto z^m$, i.e. $A_m=S^1\times I/(z,0)\sim(\zeta_m^k\cdot z,0)$ for all $z\in S^1$ and integer $k$. Here, $\zeta_m=\exp\left(2\pi i\cdot\frac{1}{m}\right)$.
As in the above answer, put: $$C_m:=\{\zeta^k_m\cdot t:k\in\Bbb Z,\,t\in I\}\subseteq\Bbb C$$
Let $\alpha:\bigsqcup_{j=0}^{m-1}\Bbb R\times I\twoheadrightarrow C_m\times\Bbb R$  be the quotient map that "glues the wedges together at zero", assembles the various copies of $\Bbb R\times I$ into a sheaf. That is: $$\alpha(x,t,j)=(\zeta^j_mt,x)$$For all $x,t,j$. For $\beta$, we need to wind $\Bbb R$ around $S^1$ in the manner corresponding to $S^1\cong\Bbb R/m\Bbb Z$ but we must stagger the windings. Explicitly: $$\beta(x,t,j)=\left(\exp\left(2\pi i\cdot\frac{x-j}{m}\right),t\right)$$For all $x,t,j$.
Finally, (as in the above answer) let $q:C_m\times\Bbb R\twoheadrightarrow C_m\times\Bbb R/\Bbb Z$ be the quotient corresponding to the natural covering space group action of $\Bbb Z$ on $C_m\times\Bbb R$, that points around a spiral staircase... Explicitly, this is the action generated by: $$C_m\times\Bbb R\to C_m\times\Bbb R,\,(z,x)\mapsto(\zeta_m\cdot z,x+1)$$

You can check (Hatcher discusses this) that $q$ is a covering map as well as a quotient. In the diagram below, there are quotient maps out of $\bigsqcup_{j=0}^{m-1}\Bbb R\times I$ that land in $C_m\times\Bbb R/\Bbb Z$ and $A_m$. One can check straightforwardly that these quotients enforce the same relation (of fibres) on their domain. That means that there is a canonical induced homeomorphism $C_m\times\Bbb R/\Bbb Z\cong A_m$. The composite $q:C_m\times\Bbb R\to C_m\times\Bbb R/\Bbb Z\cong A_m$ realises $C_m\times\Bbb R$ as a universal cover of $A_m$ since it is clearly simply connected.
Having an explicit but digestible map helps one in the battle to understand Hatcher's construction of a cover of the space $X_{m,n}$.
