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Assume $M$ a topological space,$f\in Homeo (M)$ then torus bundle $M_f=M\times I/\{(x,0)\sim (f(x),1)|x\in M\}$ obiviouly $f$ plays a significant role in determing the the torus bundle. hence there should be some general result of the following type: $M_f$ and $M_g$ are bundle isomorphic(resp.diffeomorphic) iff "W" here "W "is a relation between $f$ and $g$.

can someone help give W and explain?Thank you!

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I'm not sure "torus bundle" is right, I've heard "mapping torus" used for this construction. Do you want your maps (between mapping tori or whatever) to necessarily preserve the I-coordinate, or at least take each M x {t_0} to some M x {t_1}? –  Aaron Mazel-Gee Apr 22 '10 at 8:07
    
@ Aaron Mazel-Gee.sorry,i made a mistake.as you point out,what i mean is mapping torus. –  sara Apr 22 '10 at 9:00
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@Sara - It would be nice if you edited your question to fix the title. –  Sam Nead Apr 22 '10 at 10:25
    
At least in 3-manifold topology, the term 'torus bundle' is usually used in the special case of this construction when M is a 2-torus. –  HJRW Apr 22 '10 at 18:25

2 Answers 2

Your $M_f$ is usually called the "mapping torus" of $f$, as Aaron points out. It comes with a map to the circle $S^1 = I/(0 \sim 1)$ and this map is a fibre bundle with fibres isomorphic to $M$.

First of all, the bundle isomorphism type of the bundle $M_f \to S^1$ only depends on the the isotopy class of $f$; i.e. if $f_0$ and $f_1$ are homotopic through homeomorphisms $f_t$ then $M_{f_0}$ and $M_{f_1}$ will be isomorphic bundles. This reduces you to working with the mapping class group, which you probably already knew since I see the mapping-class-groups tag included.

Now, it's an easy exercise to check that $M_f$ and $M_g$ are isomorphic bundles if and only if $f$ and $g$ are conjugate in the mapping class group of $M$.

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I wonder if there is a subtle point lurking here... if $f_0$ and $f_1$ are isotopic (homotopic through homeomorphisms) then the two bundles will be homeomorphic. However, the question asked about diffeomorphism. This is not my area, so I don't see that a diffeotopy (homotopic through diffeomorphisms?) gives a diffeomorphism of bundles... It feels like the "quality" of the derivative of the time direction will be important? –  Sam Nead Apr 22 '10 at 10:32
    
For a diffeotopy it is exactly the same short argument; topologically isotopic maps give homeomorphic bundles, and smoothly isotopic maps give diffeomorphic bundles. And in fact, reducing to the mapping class group is unnecessary. If f and g are isotopic then they are conjugate. –  Jeffrey Giansiracusa Apr 22 '10 at 11:31

Mapping tori come equipped with a projection to the circle: $p\colon M_f\to S^1$. Bundle isomorphism is isomorphism of the total space that commutes with the projection. A weaker notion of isomorphism for mapping tori is an isomorphism that commutes with the projection only up to homotopy. Pseudoisotopy is the name for the relation on isomorphisms of $M$ that replaces isotopy in this coarser notion of equivalence of mapping tori.

If $M$ is a simply connected manifold of high dimension, then pseudoisotopy implies isotopy and preserving the homotopy class of the map to the circle isn't a big deal, so isomorphism of mapping tori is pretty much isotopy of mapping classes. But if $M$ is not simply connected, there are two ways that mapping tori may be isomorphic without being bundle isomorphic: there may be more pseudoisotopies than isotopies and there may be isomorphisms that do not preserve the homotopy class of the map to the circle.

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