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Define the mapping torus of a homeomorphism

  1. Jan 18, 2010 #1
    Define the mapping torus of a homeomorphism [itex]\phi:X \rightarrow X[/itex] to be the identification space

    [itex]T(\phi)= X \times I / \{ (x,0) \sim (\phi(x),1) | x \in X \}[/itex]

    I have to identify [itex]T(\phi)[/itex] with a standard space and prove that it is homotopy equivalent to [itex]S^1[/itex] by constructing explicit maps [itex]f:S^1 \rightarrow T(\phi), g: T(\phi) \rightarrow S^1[/itex] and explicit homotopies [itex]gf \simeq 1:S^1 \rightarrow S^1, fg \simeq 1:T(\phi) \rightarrow T(\phi)[/itex] in the two cases:

    (i) [itex]\phi(x)=x[/itex] for [itex]x \in X=I[/itex]
    (ii) [itex]\phi(x)=1-x[/itex] for [itex]x \in X=I[/itex]

    i found that since [itex]X=I[/itex], we have a square of side 1 to consider:

    in (i) we identify two opposite sides with one another, this gives us a cylinder.
    in (ii) we identify the point x with the point 1-x on the opposite side giving a kind of "twist" which i think leads to a Mobius strip.

    first of all, are my answers above correct? it says to identify them with a standard space. is there some sort of notation i can use for cylinders and Mobius strips? e.g. i can call a circle [itex]S^1[/itex], is there something like [itex]C^1[/itex] for a cylinder?

    then, how do i go about setting up the maps [itex]f[/itex] and [itex]g[/itex]?

  2. jcsd
  3. Jan 18, 2010 #2
    Re: Homotopies

    Your are correct: it's a cylinder and a Möbius strip; as far as I know, there isn't a specific abbreviation for these spaces, besides their names.

    Regarding the maps f and g, consider the "central" fibre of [tex]T\left(\phi \right)[/tex]: (1/2,y), [tex]y \in \left[0,1\right][/tex]; this is, in both cases, a homeomorphic image of [tex]S^1[/tex]; now f and g are very simple maps, and you can "contract" [tex]T\left(\phi \right)[/tex] to the central fibre; this gives you the homotopy.
  4. Jan 18, 2010 #3
    Re: Homotopies

    so [itex]g:T(\phi) \rightarrow S^1 ; (1/2,y) \mapsto ( \cos{ 2 \pi y}, \sin{ 2 \pi y } )[/itex] for [itex]y \in [0,1][/itex]. would that work? i don't understand how to contract down the x coordinate so that i can in fact only consider the central fibre as i have done above in ym function for g.

    assuming that's ok, [itex]f:S^1 \rightarrow T(\phi) ; ( \cos{ 2 \pi y}, \sin{ 2 \pi y } ) \mapsto (1/2,y)[/itex] for [itex]y \in [0,1][/itex], yes?

    pretty sure that's probably wrong but i can't see an alternative.
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