Define the mapping torus of a homeomorphism

[latentcorpse]

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

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

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

(i) \phi(x)=x for x \in X=I
(ii) \phi(x)=1-x for x \in X=I

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

then, how do i go about setting up the maps f and g?

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

 

[latentcorpse]

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

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