1

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.

Aliceyuenyuen
  • 502

1 Answers1

4

$M_f$ 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).

William
  • 9,310
  • 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? – Aliceyuenyuen Nov 29 '19 at 06:54
  • 1
    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? – William Nov 29 '19 at 15:32
  • 1
    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 $< 4$. – William Nov 29 '19 at 15:43