Homotopies

latentcorpse

Define the mapping torus of a homeomorphism [latex]\phi:X \rightarrow X[/latex] to be the identification space

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

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

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

i found that since [latex]X=I[/latex], 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 [latex]S^1[/latex], is there something like [latex]C^1[/latex] for a cylinder?

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

thanks

JSuarez

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.

latentcorpse

so [latex]g:T(\phi) \rightarrow S^1 ; (1/2,y) \mapsto ( \cos{ 2 \pi y}, \sin{ 2 \pi y } )[/latex] for [latex]y \in [0,1][/latex]. 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, [latex]f:S^1 \rightarrow T(\phi) ; ( \cos{ 2 \pi y}, \sin{ 2 \pi y } ) \mapsto (1/2,y)[/latex] for [latex]y \in [0,1][/latex], yes?

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