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

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 $z \mapsto z^m$.
