# Mapping cylinder of homeomorphism is not homeomorphic to product

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.

$$M_f$$ 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).
• The original version I am told is that "the mapping cylinder of a diffeomorphism on a smooth closed manifold $X$ is in general not diffeomorphic to $X\times I$". Do you think this statement is true? Nov 29 '19 at 6:54
• I think the map $h$ I constructed should be smooth if $f$ is a diffeomorphism (I'm not 100% on that, i have to think about it). I can see how if $f$ is just a homeomorphism then $M_f$ may not be diffeomorphic to $M\times I$. Where did you find this statement? Did they not have an example? Nov 29 '19 at 15:32
• I believe if $f$ is a diffeomorphism then $M_f$ is an h-cobordism between $X$ and itself, whose whitehead torsion vanishes because $f$ is a homeomorphism. Therefore the s-cobordism says that if $dim X > 4$ then $M_f \cong M\times I$. So if there is an example it would have to be in dimension $< 4$. Nov 29 '19 at 15:43