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Suppose I define the mapping torus $M_f$ in the usual way by identifying $(x, 0)$ and $(f(x), 1)$. If I have a homeomorphism $f: X \rightarrow X$ and another homeomorphism $f': X \rightarrow X$ that are homotopic, are the spaces $M_f$ and $M_{f'}$ necessarily homeomorphic?

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  • $\begingroup$ Do you mean simply homotopic or homotopic through homeomorphisms? In the first case the answer is no, take for instance $X=[0,1]$ and $f=\mathrm{id}$ and $g=1-\mathrm{id}$. $\endgroup$ – Olivier Bégassat Jan 19 '15 at 3:07
  • $\begingroup$ For instance, in $S^1$ the identity and antipodal maps are homotopic and happen to give the same mapping torus. Is this coincidence? $\endgroup$ – Yon Kim Jan 19 '15 at 3:13
  • $\begingroup$ They are homotopic through homeomorphisms though. This is more restrictive than simply being homotopic in general. Do you see why the example I gave in my previous comment is a counter example? $\endgroup$ – Olivier Bégassat Jan 19 '15 at 3:14
  • $\begingroup$ Can you explain what you mean by homotopic through homeomorphisms? $\endgroup$ – Yon Kim Jan 19 '15 at 3:17
  • $\begingroup$ I mean a homotopy $H:X\times[0,1]\to X$ such that for all $t$, $H_t$ is a homeomorphism of $X$. In the case of the identity and the antipodal map of the sphere, you can take the homotopy $H(z,t)=e^{it\pi}z$, which is a homotopy through homeomorphisms. $\endgroup$ – Olivier Bégassat Jan 19 '15 at 3:19
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Asking for homotopic homeomorphisms isn't enough: if one takes $X=[0,1]$ and $f=\mathrm{id}$ and $g=1-f$, then $f$ and $g$ are homotopic homeomorphisms, but the mapping cylinder of $f$ is homeomorphic to $S^1\times[0,1]$ while the mapping cylinder of $g$ is homeomorphic to the Moebius strip. These spaces aren't homeomorphic.


Suppose $X$ locally compact Hausdorff and $Y$ Hausdorff, then a homotopy $$h:X\times [0,1]\to Y$$ is the same as a continuous path $$\widehat{h}:[0,1]\to C(X,Y)$$ (This is a special case of the exponential law for spaces where the set $C(X,Y)$ of continuous maps $X\to Y$ is equipped with the compact open topology. By a result of Arens, if $X$ is locally compact locally connected (or more trivially, if $X$ is compact), the subset of homeomorphisms $\mathrm{Aut}(X)\subset C(X,X)$ forms a topological group (composition is easily seen to be continuous, but the operation $\mathrm{inv}:f\mapsto f^{-1}$ may not be).


From now on $X$ is either compact or locally compact locally connected, so that the map $\mathrm{inv}:f\mapsto f^{-1}$ is a continuous one, and that the homotopy $H$ through homeomorphisms corresponds to a continuous path in $\mathrm{Aut}(X)$. Suppose $H$ is a homotopy through homeomorphisms from $f$ to $g$. Consider the map $$C:X\times [0,1]\to X\times[0,1],(x,t)\mapsto(H_t(x),t)$$ is a homeomorphism with inverse $$C':X\times [0,1]\to X\times[0,1],(x,t)\mapsto(\mathrm{inv}(H_t)(x),t)$$ then the map $D=C\circ(f^{-1}\times\mathrm{id}_{[0,1]})$ is a homeomorphism of $X\times[0,1]$, which satisfies for all $x\in X$, $D(x,0)=(x,0)$ and $D(f(x),1)=(g(x),1)$. This homeomorphism induces a homeomorphism between the mapping tori.

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  • $\begingroup$ Ok, there are some mistakes in the formulas, I'll get to them tomotopy. $\endgroup$ – Olivier Bégassat Jan 19 '15 at 4:24
  • $\begingroup$ Why is it $C'$ not $C$? $\endgroup$ – Yon Kim Jan 19 '15 at 5:22
  • $\begingroup$ @YonKim corrected the formulas. $\endgroup$ – Olivier Bégassat Jan 19 '15 at 10:34

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