Mapping torus is an (m+1)-manifold

[noospace]

Homework Statement

Let X be an m-manifold. Let M(f) be the space obtained from $X\times [0,1]$ by gluing the ends together using $(x,0)\sim (f(x),1)$. Show that if M is an m-manifold then M(f) is an (m+1)-manifold.

The Attempt at a Solution

Since X has an atlas $\{ (U_\alpha,\varphi_\alpha) \}$, my first instinct was to define coordinate charts by $\psi_\alpha : U_\alpha \times [0,1] \to \mathbb{R}^{m+1}; (x,t) \to (\varphi_\alpha(x),t)$ but then we have to worry about the end-points. Right now I'm attempting to understand the solutions.

The idea behind the solutions is to map points with $1<t< 1$ in the usual way and to separately consider a point on the gluing edge $(x,0) \sim (f(x),1)$. There is a coordinate chart $\varphi : U \subset X \to \mathbb{R}^n$ where U is an open nbhd of x. This gives a coordinate chart at f(x) by $(f(U),\varphi\circ f^{-1})$. Now consider the subset of $X \times [0,1]$ given by $W = [0,\epsilon) \times U \cup (1-\epsilon,1]\times f(U)$. The claim is that this maps homeomorphically onto its image in M(f) but I don't see why. Can anyone help me understand this?

Thanks.

 

[mighty2000]

Sure, let's break it down step by step:

1. First, we define a map \psi: W \to M(f) by \psi(t,x) = (x,t).

2. Next, we show that \psi is injective. Suppose \psi(t_1,x_1) = \psi(t_2,x_2). Then, we have (x_1,t_1) = (x_2,t_2) in M(f), which means that either t_1 = t_2 or x_1 = x_2. If t_1 = t_2, then (x_1,t_1) = (x_2,t_2) implies that x_1 = x_2, so we have (x_1,t_1) = (x_2,t_2) in X \times [0,1]. But this contradicts the fact that (x_1,0) \sim (f(x_1),1) and (x_2,0) \sim (f(x_2),1) in M(f). Similarly, if x_1 = x_2, then (x_1,t_1) = (x_2,t_2) implies that t_1 = t_2, again contradicting the fact that (x_1,0) \sim (f(x_1),1) and (x_2,0) \sim (f(x_2),1) in M(f). Therefore, \psi is injective.

3. Next, we show that \psi is surjective. Let (y,s) \in M(f). Then, either s \neq 0 or y \notin f(U). If s \neq 0, then (y,s) \in (1-\epsilon,1]\times f(U) \subset W, so there exists some (x,t) \in W such that \psi(x,t) = (y,s). If y \notin f(U), then y \in X \setminus f(U), so (y,s) \in (0,\epsilon) \times X \setminus f(U) \subset W. Again, there exists some (x,t) \in W such that \psi(x,t) = (y,s). Therefore, \psi is surjective.

4. Finally, we show that \psi is a