Let $X$ be a topological manifold, and $f:X\to X$ be a homeomorphism. The mapping cylinder is defined as $M_f:=(X\times[0,1]\sqcup X)/(x,1)\sim f(x)$. I am told somewhere that there exists an example of $(X,f)$ such that $M_f$ is not homeomorphic to $X\times[0,1]$. However, I don't know how to construct such an example.
I believe this statement is true since in the analogous mapping torus case, by taking $X=(0,1)$ and $f=1-id$, the mapping torus is not homeomorphic to $X\times S^1$. I even guess there exists a closed smooth manifold $X$, and diffeomorphism $f:X\to X$ such that $M_f$ is not homeomorphic to $X\times [0,1]$.
At first, I thought $X=S^1\subset \mathbb{C}$, and $f:z\mapsto \bar{z}$ can do the job. However, in this post and this post, people are claiming that for any homeomorphism $f:S^1\to S^1$, the mapping cylinder is homeomorphic to the product manifold $S^1\times [0,1]$. I can neither prove nor disprove their claim.
Any help is appreciated.
Best Answer
$M_f$ is formed as a quotient space of $X\times I$ and $X$ by the relation $(x, 1) \sim f(x)$, so a continuous function $g\colon M_f \to Y$ can be constructed by defining $g_{X\times I}\colon X\times I \to Y$ and $g_X \colon X \to Y$ in a way that respects the relation, that is we need $g_{X\times I} (x, 1) = g_X(f(x))$.
In particular we can define $h\colon M_f \to X\times I$ as follows: for $(x, t) \in X\times I$ define $h_{X\times I}(x, t) = (x, t)$, and for $x \in X$ define $h_X(x) = (f^{-1}(x), 1)$. Then $h_{X\times I}(x, 1)= (x, 1) = h_X(f(x))$.
I claim that $h$ is a homeomorphism, and that its inverse is given by the canonical map $X\times I \to M_f$ (but leave it to you to verify the details for yourself).