TITLE: Showing that the Mapping Torus is a topological manifold
QUESTION [7 upvotes]: Let $X$ be a conected topological $n$-manifold, and $f:X\rightarrow X$ an homeomorphism, the Mapping Torus $M_f$ is defined as,
$$M_f=X\times [0,1]/\sim$$
where $(x,0)\sim (f(x),1)$.
I am trying to prove that this is a manifold. Here is my approach:
This is obvious for every point $(x,a)$ with $a\neq 0,1$, so let's call $p\in M_f$ to $p=\pi(x_0,0)=\pi(f(x_0),1)$ where $\pi$ is the canonical projection. Let $\{\varphi, U\}$ be a chart of $X$ at $x_0$, and define,
$$\begin{matrix}
V_1&=&U\times [0,1/2)\\
V_2&=&U\times (1/2,1]
\end{matrix}$$
we have that, if $V=V_1\cup V_2$, $\pi(V)$ is an open set in $M_f$ and if we define,
$$\begin{matrix}
\phi:&V&\longrightarrow &\mathbb{R}^n\times (-1/2,1/2)\\
&(x,a)&\longmapsto&\left\{ \begin{matrix}(\varphi(x),a)&\text{if}&(x,a)\in V_1\\((\varphi\circ f^{-1})(x),a-1)&\text{if}&(x,a)\in V_2\end{matrix}\right.
\end{matrix}$$
from here I think it should be enough to show that $\phi$ is a quotient map, since the only points with same image are those related by $\sim$.
Is this argument valid?, and also is this the correct way to solve this problem or is there anything more elegant?

REPLY [2 votes]: You have the right idea, and this will work. There's nothing particularly more elegant, especially if (like me) you enjoy the elegance of the quotient topology. However, there are a couple of mistakes. 
First is a typo: the formulas for $V_1$ and $V_2$ should have Cartesian product symbols, not union symbols.
Second, the formula for $V_2$ is wrong. It should be
$$V_2 = f(U) \times (1/2,1]
$$