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    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 \}
    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 ) for \phi(x)=x for x \in X=I

    so then X \times I is a 1 by 1 square and then we identify two opposite sides meaning T( \phi ) is a cylinder.

    i then said: let
    f: S^1 \rightarrow T( \phi) ; ( \cos{\phi}, \sin{\phi}) \mapsto ( \cos{\phi}, \sin{\phi}, z) and
    g: T( \phi ) \rightarrow S^1 ; ( \cos{\phi}, \sin{\phi}, z) \mapsto ( \cos{\phi}, \sin{\phi})

    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.
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    Do you have the condition that X is contractible?
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    we have covered contractible spaces in lectures but it doesn't say anywhere in the question that X is contractible.
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    (Original post by latentcorpse)
    we have covered contractible spaces in lectures but it doesn't say anywhere in the question that X is contractible.
    Well then I don't think T(\phi) is necessarily homotopy equivalent to S^1. Eg, if X is disconnected and \phi is the identity, then T(\phi) isn't connected. Even if we take X=S^1 and \phi the identity, T(\phi) is a torus which isn't homotopy equivalent to a circle.

    Edit. Sorry, I didn't read the question properly. \phi 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.
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    well because i'm using f to map from a circle to a cylinder, i had z \in [0,1] 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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    (Original post by latentcorpse)
    well because i'm using f to map from a circle to a cylinder, i had z \in [0,1] so that each point on the circle owuld be mapped to a corresponding line on the side of the cylinder.
    You should write something like:
    f: S^1 \rightarrow T( \phi) ; ( \cos{\phi}, \sin{\phi}) \mapsto ( \cos{\phi}, \sin{\phi}, 0)
    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.
    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.
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    (Original post by SsEe)
    You should write something like:
    f: S^1 \rightarrow T( \phi) ; ( \cos{\phi}, \sin{\phi}) \mapsto ( \cos{\phi}, \sin{\phi}, 0)
    Otherwise it just looks like a general z.
    ok. true. what i wrote was a one-to-many 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.
    it says in the question that i want fg to be the identity itself though, doesn't that mean my actual functions are wrong?
    however, assuming we proceed this way, would the straight line homotopy work:

    h: T( \phi) \times I \rightarrow T(\phi); (s,t) \mapsto (1-t) gf(s) + \mathbb{I}_{T( \phi)} t where s=( \cos{\theta},\sin{\theta},z) \in T(\phi) and \mathbb{I}_{T(\phi)} is the identity map on the cylinder. since this then gives h(s,0)=(\cos{\theta},\sin{\theta  },0),h(s,1)=(\cos{\theta},\sin{\  theta},z)

    cheers
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    (Original post by latentcorpse)
    ok. true. what i wrote was a one-to-many 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?
    Indeed. The maps f and g when establishing homotopy equivalence don't have to be onto or 1-1.

    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
    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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    (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.
    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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    (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?
    You want fg, not gf in your definition of h.

    h: T( \phi) \times I \rightarrow T(\phi); (( \cos{\theta},\sin{\theta},z),t) \mapsto (1-t) ( \cos{\theta},\sin{\theta},0) + t ( \cos{\theta},\sin{\theta},z)
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    (Original post by SsEe)
    You want fg, not gf in your definition of h.

    h: T( \phi) \times I \rightarrow T(\phi); (( \cos{\theta},\sin{\theta},z),t) \mapsto (1-t) ( \cos{\theta},\sin{\theta},0) + t ( \cos{\theta},\sin{\theta},z)
    would it matter? surely if i had it the other way round, it would just be the reverse path h: T( \phi ) \times I \rightarrow T( \phi ) ; ( ( \cos{\theta} , \sin{\theta} , z ) , t ) \mapsto (1-t) ( \cos{\theta}, \sin{\theta}, z)+t ( \cos{\theta},\sin{\theta},0).

    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 T(\phi} should i be parameterising it with \phi instead of \theta or is that \phi referring to something else such as the map \phi: X \rightarrow X mentioned at the start of the question?

    thanks.
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    (Original post by latentcorpse)
    would it matter? surely if i had it the other way round, it would just be the reverse path h: T( \phi ) \times I \rightarrow T( \phi ) ; ( ( \cos{\theta} , \sin{\theta} , z ) , t ) \mapsto (1-t) ( \cos{\theta}, \sin{\theta}, z)+t ( \cos{\theta},\sin{\theta},0).

    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 T(\phi) should i be parameterising it with \phi instead of \theta or is that \phi referring to something else such as the map \phi: X \rightarrow X mentioned at the start of the question?

    thanks.
    Look at the maps. f maps from S^1 to T(\phi) so you can't plug s \in T(\phi) into it. You can't put them the other way around. It's probably best written as I did earlier.

    \phi refers to the map at the start. Not the parameter used to describe the space! So use \theta.
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    (Original post by SsEe)
    Look at the maps. f maps from S^1 to T(\phi) so you can't plug s \in T(\phi) into it. You can't put them the other way around. It's probably best written as I did earlier.

    \phi refers to the map at the start. Not the parameter used to describe the space! So use \theta.
    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 \phi(x)=1-x. i said that this would lead to T(\phi) 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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    (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?
    Yep. That's all obvious stuff. You shouldn't have to ask me if it's correct. Have some faith!

    i think i can have a bash at the next bit of the question now, which is essentially the same except that \phi(x)=1-x. i said that this would lead to T(\phi) 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.
    Just view it as a square with the top and bottom edges identified. Look up "homotopy retract" or "deformation retract". You want to retract the moebius strip onto the subspace \{ (\frac{1}{2}, t) : t \in [0,1] \} making sure the map you define is well defined at the top and bottom edge under the identification.

    This is a very common technique. What you did earlier was a retract of the cylinder onto its bottom edge.
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    a deformation retract is a subspace Y \subseteq X such that the inclusion \iota: Y \rightarrow X defines a homotopy equivalence.

    so if we let X= \text{Mobius strip},Y= \{ ( \frac{1}{2},t) | t \in [0,1] \} and we have \iota : Y \rightarrow X ; y \mapsto y and i need a map from X to Y so that i can show \iota defines a homotopy equivalence. i thought about restricting the domain as follows \rho : \iota(Y) \rightarrow Y ; x \mapsto x 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 \phi(x)=x case but using Y instead of a Mobius strip?
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    (Original post by latentcorpse)
    a deformation retract is a subspace Y \subseteq X such that the inclusion \iota: Y \rightarrow X defines a homotopy equivalence.

    so if we let X= \text{Mobius strip},Y= \{ ( \frac{1}{2},t) | t \in [0,1] \} and we have \iota : Y \rightarrow X ; y \mapsto y and i need a map from X to Y so that i can show \iota defines a homotopy equivalence. i thought about restricting the domain as follows \rho : \iota(Y) \rightarrow Y ; x \mapsto x 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 \phi(x)=x case but using Y instead of a Mobius strip?
    The retract is map r:X \rightarrow Y such that r(y)=y \forall y \in Y, and such that \iota \circ r is homotopic to the identity on X.

    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 \iota \circ r is homotopic to the identity on X. That's just a straight line homotopty.
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    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
    r: X \rightarrow Y ; x \mapsto \frac{1}{2} would do the trick.
    although surely it would be more accurate to write
    r: X \times I \rightarrow Y \times I ; (x,t) \mapsto (\frac{1}{2},t)
    but then \iota \circ r : X \rightarrow X ; x \mapsto \frac{1}{2}
    and i need to show that is homotopic to the identity map on X

    well i can take h: X \times I \rightarrow X ; (x,t) \mapsto \frac{1}{2}(1-t)+tx. this is from 1/2 to x and it's continuous as its the composition of continuous functions.

    i'm not sure about checking r is well defined though.

    cheers.
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    (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."
    Hmm. That seems weaker than my definition:

    A homotopy retract is a map r:X \rightarrow Y st r(y)=y on Y and \iota \circ r ~ id_X. 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
    r: X \rightarrow Y ; x \mapsto \frac{1}{2} would do the trick.
    although surely it would be more accurate to write
    r: X \times I \rightarrow Y \times I ; (x,t) \mapsto (\frac{1}{2},t)
    but then \iota \circ r : X \rightarrow X ; x \mapsto \frac{1}{2}
    and i need to show that is homotopic to the identity map on X

    well i can take h: X \times I \rightarrow X ; (x,t) \mapsto \frac{1}{2}(1-t)+tx. this is from 1/2 to x and it's continuous as its the composition of continuous functions.

    i'm not sure about checking r is well defined though.

    cheers.
    Basically. Need to keep an eye on details (both of us!! All this notation flying around is confusing...). X \times I isn't the moebius band. It's just a square. T(\phi) is the band. And Y is the subspace, which becomes a circle under the identification. Y \times I is a cylinder.

    I think the best way of saying it is probably:
    r: X \times I \rightarrow Y; (x,t) \mapsto (\frac{1}{2},t) gives a well defined map under the identification ~ in which (x,0)~(1-x,1) because (x, 0) and (1-x, 1) both map to the same point in Y, namely (0.5, 0) ~ (0.5, 1).
    Then you define the homotopy
    h: ( X \times I  ) \times I \rightarrow X \times I by ((x,s), t) \mapsto t(x,s) + (1-t)(\frac{1}{2}, s). This is again well defined under the identification (exercise for reader...)
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    (Original post by SsEe)
    Hmm. That seems weaker than my definition:

    A homotopy retract is a map r:X \rightarrow Y st r(y)=y on Y and \iota \circ r ~ id_X. 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...). X \times I isn't the moebius band. It's just a square. T(\phi) is the band. And Y is the subspace, which becomes a circle under the identification. Y \times I is a cylinder.

    I think the best way of saying it is probably:
    r: X \times I \rightarrow Y; (x,t) \mapsto (\frac{1}{2},t) gives a well defined map under the identification ~ in which (x,0)~(1-x,1) because (x, 0) and (1-x, 1) both map to the same point in Y, namely (0.5, 0) ~ (0.5, 1).
    Then you define the homotopy
    h: ( X \times I  ) \times I \rightarrow X \times I by ((x,s), t) \mapsto t(x,s) + (1-t)(\frac{1}{2}, s). This is again well defined under the identification (exercise for reader...)
    aargh. i understand the principles but i'm now confused by the notation. we're trying to show T(\phi) \simeq S^1 so shouldn't h: T(\phi) \rightarrow S^1?

    also to show the h you're using is well defined, i can't see how they match up because h((x,s),0)=(\frac{1}{2},s) \neq h((1-x,s),1)=(1-x,s)?

    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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    (Original post by latentcorpse)
    aargh. i understand the principles but i'm now confused by the notation. we're trying to show T(\phi) \simeq S^1 so shouldn't h: T(\phi) \rightarrow S^1?

    also to show the h you're using is well defined, i can't see how they match up because h((x,s),0)=(\frac{1}{2},s) \neq h((1-x,s),1)=(1-x,s)?

    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.
    h is showing the homotopy equivalence of two maps T(\phi) \rightarrow T(\phi) (namely \iota \circ r and id_{T(\phi)}. So it should be a map T(\phi) \times I \rightarrow T(\phi)

    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((1-x, 1), t). Remember, (x,0) and (1-x,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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