The mapping torus $M_f$ of a homeomorphism $f$ of some topological space $X$ is a fiber bundle whose base is a circle and whose fiber is the original space $X$. If instead of a homeomorphism $f$ is just a homotopy equivalence of $X$, is $M_f$ a fibration over the circle with fiber homotopic to $X$?
3 Answers
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There's no reason to expect $M_f$ to be a fiber bundle, or even a Hurewicz or Serre fibration. (Think of $f: \mathbb{R} \to \mathbb{R}$ by $f(x)=0$. This is a homotopy equivalence, but $M_f$ is certainly not locally trivial, nor does $M_f \to S^1$ have any nice lifting properties.)
What is true is that the homotopy fiber of $M_f\to S^1$ is weakly equivalent to $X$, if $f$ is a homotopy equivalence (or even a weak equivalence). This often gets proved by using the theory of "quasi-fibrations".
answered Nov 12, 2009 at 19:55
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Check out
Homotopy equivalences and mapping torus projections D. S. Coram, P. F. Duvall Fund. Math. 109 (1980), 1-7
answered Dec 19, 2011 at 7:42
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This is closely related to a question of mine, which was motivated by wondering whether the mapping cylinder of a homotopy equivalence is a fibration over an interval. The counterexample given there (for the homotopy equivalence from I to a point) also gives a counterexample for the mapping torus, and makes it easy to see how it goes wrong.
answered Nov 12, 2009 at 22:58