1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

$M_f$ 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).

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Nov 29 '19 at 6:54
  • 1
    $\begingroup$ 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? $\endgroup$
    – William
    Nov 29 '19 at 15:32
  • 1
    $\begingroup$ 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 $< 4$. $\endgroup$
    – William
    Nov 29 '19 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.