Define the mapping torus of a homeomorphism

1. Jan 18, 2010

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

2. Jan 18, 2010

JSuarez

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 $$T\left(\phi \right)$$: (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.

3. Jan 18, 2010

latentcorpse

Re: Homotopies

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.