Define the mapping torus of a homeomorphism to be the identification space
Identify with a standard space and prove that it is homotopy equivalent to by constructing explicit maps and explicit homotopies for for
so then is a 1 by 1 square and then we identify two opposite sides meaning is a cylinder.
i then said: let
and
meaning gf=1 and fg=1
i don't feel like i've actually done anything at all here though. is this a sufficient answer?
thanks.

latentcorpse
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 24012010 12:50
Last edited by latentcorpse; 25012010 at 22:53. 
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 24012010 15:37
Do you have the condition that X is contractible?

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 24012010 15:48
we have covered contractible spaces in lectures but it doesn't say anywhere in the question that X is contractible.

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 24012010 16:09
(Original post by latentcorpse)
we have covered contractible spaces in lectures but it doesn't say anywhere in the question that X is contractible.
Edit. Sorry, I didn't read the question properly. is the identity and X is [0,1].
Your answer isn't correct. In the map f, what's z? gf=1 but fg is only homotopic to the identity.Last edited by SsEe; 24012010 at 16:20. 
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 24012010 17:29
well because i'm using f to map from a circle to a cylinder, i had so that each point on the circle owuld be mapped to a corresponding line on the side of the cylinder.
ok. i see fg isn't necessarily the identity as you won't always end up back at the same z value you started with.
any ideas on what i should try instead then?
thanks. 
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 24012010 17:35
Otherwise it just looks like a general z.
ok. i see fg isn't necessarily the identity as you won't always end up back at the same z value you started with.
any ideas on what i should try instead then?
thanks. 
latentcorpse
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 24012010 17:54
ok. true. what i wrote was a onetomany function which obviously can't exist. although i did it because i was trying to map from the circle to the cylinder. however, you're saying that f doesn't necessarily have to be surjective i.e. i can just map to the circle on the "base" of the cylinder. is that correct?
(Original post by SsEe)
fg isn't equal to the identity as it squashes the cylinder down to a circle (like crushing a coke can or something). You should be able to find a homotopy between fg and the identity.
however, assuming we proceed this way, would the straight line homotopy work:
where and is the identity map on the cylinder. since this then gives
cheersLast edited by latentcorpse; 24012010 at 18:02. 
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 24012010 18:00
(Original post by latentcorpse)
ok. true. what i wrote was a onetomany function which obviously can't exist. although i did it because i was trying to map from the circle to the cylinder. however, you're saying that f doesn't necessarily have to be surjective i.e. i can just map to the circle on the "base" of the cylinder. is that correct?
it says in the question that i want fg to be the identity itself though, doesn't that mean my actual functions are wrong?
cheers 
latentcorpse
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 24012010 19:06
(Original post by SsEe)
Nope. It wants fg and gf to be homotopic to the identity maps and it wants you to explicitly describe the homotopies between them. This is just the definition of spaces being homotopic. 
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 24012010 19:13
(Original post by latentcorpse)
ok. it appears you posted that just before i finished editing my last post. how does the straight line homotopy i wrote down look?

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 24012010 23:40
would it matter? surely if i had it the other way round, it would just be the reverse path .
anyway. is that the question done now? the existance of these homotopies prove the spaces are homotopy equivalent, yes?
one minor detail...seeing as the space is should i be parameterising it with instead of or is that referring to something else such as the map mentioned at the start of the question?
thanks. 
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 24012010 23:47
(Original post by latentcorpse)
would it matter? surely if i had it the other way round, it would just be the reverse path .
anyway. is that the question done now? the existance of these homotopies prove the spaces are homotopy equivalent, yes?
one minor detail...seeing as the space is should i be parameterising it with instead of or is that referring to something else such as the map mentioned at the start of the question?
thanks.
refers to the map at the start. Not the parameter used to describe the space! So use .Last edited by SsEe; 25012010 at 13:49. 
latentcorpse
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 25012010 15:03
i think i can have a bash at the next bit of the question now, which is essentially the same except that . i said that this would lead to being a mobius strip but then i didn't know how to parameterise a mobius strip so i couldn't set up f or g. is there a standard parameterisation? i couldn't find one online. 
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 25012010 15:37
(Original post by latentcorpse)
ok thanks. so establishing the homotopy h proves fg is homotopic to the identity on T(phi) and since gf equals the identity on S1 it is obviously going to be homotopic to the identity on S1. is that right?
i think i can have a bash at the next bit of the question now, which is essentially the same except that . i said that this would lead to being a mobius strip but then i didn't know how to parameterise a mobius strip so i couldn't set up f or g. is there a standard parameterisation? i couldn't find one online.
This is a very common technique. What you did earlier was a retract of the cylinder onto its bottom edge. 
latentcorpse
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 25012010 15:50
a deformation retract is a subspace such that the inclusion defines a homotopy equivalence.
so if we let and we have and i need a map from X to Y so that i can show defines a homotopy equivalence. i thought about restricting the domain as follows but then not everything on the Mobius strip is going somewhere which can't be right.
am i right in thinking that i establish this homotopy equivalence in the deformation retract and then i can just use the same argument for the case but using Y instead of a Mobius strip? 
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 25012010 16:06
(Original post by latentcorpse)
a deformation retract is a subspace such that the inclusion defines a homotopy equivalence.
so if we let and we have and i need a map from X to Y so that i can show defines a homotopy equivalence. i thought about restricting the domain as follows but then not everything on the Mobius strip is going somewhere which can't be right.
am i right in thinking that i establish this homotopy equivalence in the deformation retract and then i can just use the same argument for the case but using Y instead of a Mobius strip?
The choice of r should be pretty obvious. I mean, there's one sensible choice. Just draw a picture of X, and mark Y on it. Then draw a picture of X being shrunk onto Y. That's what the map r must do. Make sure you show it's well defined at the top and bottom edge with the identification.
Then you must show that is homotopic to the identity on X. That's just a straight line homotopty. 
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 25012010 16:51
couple of things here:
(i) the definition of deformation retract in my notes seems different to yours. it says:
"A deformation retract of a topological space X is a subspace Y ⊆ X such that the inclusion i : Y → X is a homotopy equivalence."
(ii) if i use the method you're describing then surely
would do the trick.
although surely it would be more accurate to write
but then
and i need to show that is homotopic to the identity map on X
well i can take . this is from 1/2 to x and it's continuous as its the composition of continuous functions.
i'm not sure about checking is well defined though.
cheers. 
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 25012010 19:48
(Original post by latentcorpse)
couple of things here:
(i) the definition of deformation retract in my notes seems different to yours. it says:
"A deformation retract of a topological space X is a subspace Y ⊆ X such that the inclusion i : Y → X is a homotopy equivalence."
A homotopy retract is a map st r(y)=y on Y and . If such r exists then X and Y are homotopy equivalent.
The point is that the subspace is fixed and the rest of the space sort of shrink wrapped onto it.
(ii) if i use the method you're describing then surely
would do the trick.
although surely it would be more accurate to write
but then
and i need to show that is homotopic to the identity map on X
well i can take . this is from 1/2 to x and it's continuous as its the composition of continuous functions.
i'm not sure about checking is well defined though.
cheers.
I think the best way of saying it is probably:
gives a well defined map under the identification ~ in which (x,0)~(1x,1) because (x, 0) and (1x, 1) both map to the same point in Y, namely (0.5, 0) ~ (0.5, 1).
Then you define the homotopy
by . This is again well defined under the identification (exercise for reader...) 
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 25012010 21:59
(Original post by SsEe)
Hmm. That seems weaker than my definition:
A homotopy retract is a map st r(y)=y on Y and . If such r exists then X and Y are homotopy equivalent.
The point is that the subspace is fixed and the rest of the space sort of shrink wrapped onto it.
Basically. Need to keep an eye on details (both of us!! All this notation flying around is confusing...). isn't the moebius band. It's just a square. is the band. And Y is the subspace, which becomes a circle under the identification. is a cylinder.
I think the best way of saying it is probably:
gives a well defined map under the identification ~ in which (x,0)~(1x,1) because (x, 0) and (1x, 1) both map to the same point in Y, namely (0.5, 0) ~ (0.5, 1).
Then you define the homotopy
by . This is again well defined under the identification (exercise for reader...)
also to show the h you're using is well defined, i can't see how they match up because ?
you keep referring to r as a homotopy retract. is that somehow different from the deformation retract i'm reading about in my notes. i couldn't find anything about homotopy retracts.
cheers. 
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 25012010 23:05
(Original post by latentcorpse)
aargh. i understand the principles but i'm now confused by the notation. we're trying to show so shouldn't ?
also to show the h you're using is well defined, i can't see how they match up because ?
you keep referring to r as a homotopy retract. is that somehow different from the deformation retract i'm reading about in my notes. i couldn't find anything about homotopy retracts.
cheers.
To show h is well defined under the identification, you need to show that for all t and x in [0,1], h((x, 0), t) = h((1x, 1), t). Remember, (x,0) and (1x,1) are the same point. So h better take them to the same place!
http://en.wikipedia.org/wiki/Deformation_retract
It says what a retraction is and also: "An equivalent definition of deformation retraction is the following. A continuous map r: X → A is a deformation retraction if it is a retraction and its composition with the inclusion is homotopic to the identity map on X. In this formulation, a deformation retraction carries with it a homotopy between the identity map on X and itself."
I've seen all sorts of terminology but just keep in mind the idea of shrinking onto a subspace.
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