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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?

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  • $\begingroup$ As it stands, $V$ doesn't cover $\{\frac{1}{2}\}\times [0, 1]$. Instead, take $\left[0, \frac{1}{2} + \epsilon\right)$ and $\left(\frac{1}{2} - \epsilon, 1\right]$ in your definition of the $V_i$, and show that the transition map on the intersection is nicely behaved. $\endgroup$
    – anomaly
    Oct 12 '15 at 17:01
  • $\begingroup$ That's not (necessarily) a point of the space since $X$ is just a random manifold. $\endgroup$
    – Smurf
    Oct 12 '15 at 17:05
  • $\begingroup$ I messed up the notation: I meant $X\times \{\frac{1}{2}\}$. $\endgroup$
    – anomaly
    Oct 12 '15 at 17:12
  • $\begingroup$ Oh then you're right it doesn't cover any of those points, although I'm not trying to cover them, I just want a neighborhood of $p$. Actually I'm missing a lot more points, everyone from $X-U\times [0,1]$. $\endgroup$
    – Smurf
    Oct 12 '15 at 17:14
  • $\begingroup$ Sorry, you did explicitly state that you were just looking on a neighborhood of $p$, not working with a cover of the full manifold. $\endgroup$
    – anomaly
    Oct 12 '15 at 18:24
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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 $$V_2 = f(U) \times (1/2,1] $$

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  • $\begingroup$ I am not sure about that definition of $V_2$. I am trying $\pi(V)$ to be a neighborhood of $p$, so I need $V$ to be saturated, in particular $(f(x_0),1)$ must belong to it, which doesn't necessarily hold with that definition. $\endgroup$
    – Smurf
    Oct 12 '15 at 17:02
  • $\begingroup$ Ah, you are correct, I will fix my answer. I had not looked carefully but had assumed $(x,1) \sim (f(x),0)$ which is (somewhat) more standard. $\endgroup$
    – Lee Mosher
    Oct 12 '15 at 20:04
  • $\begingroup$ Although you were right about the elegance. After working this example I managed to get an intuitive idea of what a quotient map is, so the proof now looks pretty straightforward and intuitive. $\endgroup$
    – Smurf
    Oct 12 '15 at 20:16

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