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Math Help - Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

  1. #1
    Super Member Bernhard's Avatar
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    Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

    In Section 54 of his book "Topology" on the Fundamental Group of the Circle, Munkres presents the following Lemma (Lemma 54.1 - see attachment)

    Lemma 54.1

    Let  p: E \rightarrow B be a covering map, let  p( e_0 ) = b_0 .

    Any path  f : [0.1] \rightarrow  B beginning at  b_0 has a unique lifting to a path  \tilde{f} in E beginning at  e_0

    ================================================== ====================================

    [My question relates to the proof of the uniqueness of  \tilde{f} ]

    Munkres begins the proof as follows:

    Proof:

    Cover B by open sets U each of which is evenly covered by p.

    Find a subdivision of [0.1], say  s_0, s_1, .... s_n , such that for each i the set  f([s_i, s_{i+1} ]) lies in an open set U (Use Lebesgue number lemma).

    We define the lifting  \tilde{f} step by step.

    First define  \tilde{f} (0) = e_0 .

    Then supposing  \tilde{f} (s) is defined for  0 \leq s \leq s_i we define  \tilde{f} ] on   [s_i , s_{i+1} as follows:

    The set  f([s_i, s_{i+1} ]) lies in some open set U that is evenly covered by p.

    Let  \{ V_{\alpha} \} be a partition of   p^{-1} (U) into slices; each  \{ V_{\alpha} \} is mapped homeomorphically onto U by p.

    Now  \tilde{f} (s_i) lies in one of these sets. ... ... etc etc ... see attachement

    ================================================== ==========================

    But now focussing on the uniqueness of  \tilde{f} - couldn't we define  \tilde{f} (s) as belonging to any of the sets  \{ V_{\alpha} \} and make this work?

    So any of the sets would do since the covering is even.

    Then the argument for uniqueness would follow (see page 342 of attachment) - so we would only have one  \tilde{f} (s) for each of the  \{ V_{\alpha} \} - but this is not may idea of a unique  \tilde{f}  .

    Can someone please help clarify Munkres argument regarding the uniqueness of  \tilde{f}

    Peter
    Last edited by Bernhard; June 22nd 2012 at 07:20 PM.
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    Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

    the uniqueness derives from the fact that we are considering "evenly-covered" open neighborhoods U in B. this means the pre-images p-1(U) are disjoint.

    in particular, e0 lies in only one of them. so when we pick the "slice" Vα to restrict p to (for the interval [0,s1]), we pick the one containing e0. so not just any of the slices will do.

    when we extend the lifted map \tilde{f} to "the next subinterval", we pick the ONLY pre-image of that U that contains \tilde{f}(s_i).

    for example, given the path h(s) = (cos(4πs),sin(4πs)) in B = S1, suppose we choose e0 = 0 in E = R, under the covering map:

    p(x) = (cos(x),sin(x)).

    for our open cover of the circle, all we need to do is choose "arc-lengths" less than 2pi radians. so let's use:

    U1 = p((-π/2, π/2))
    U2 = p((0,π))
    U3 = p((π/2,3π/2))
    U4 = p((π,2π))

    now we need to pick our subintervals of [0,1].

    we can pick s1 = 3/16, so that h([0,3/16]) ⊂ U1.
    we can pick s2 = 7/16, so that h([3/16,7/16]) ⊂ U2.
    we can pick s3 = 11/16, so that h([7/16,11/16]) ⊂ U3.
    we can pick s4 = 15/16, so that h([11/16,15/16]) ⊂ U4.

    finally, we see that h([15/16,1]) ⊂ U1.

    now let's look at p-1(U1), which is:

    \bigcup_{k \in \Mathbb{Z}} (-\pi/2+2k\pi,\pi/2+2k\pi)

    only one of these intervals contains the real number 0, namely the one with k = 0. so that's where we start, and since

    p \circ \tilde{h} = h we have:

    (\cos(\tilde{h}(s)),\sin(\tilde{h}(s))) = (\cos(4\pi s),\sin(4\pi s))

    on (-π/2,π/2), p is injective, so we see that on [0,3/16], that \ \tilde{h}(s) = 4\pi s

    (it's the injectivity of p on (-π/2,π/2) that allows us to uniquely "undo" the trig functions).

    so that gives us the lift of h on [0,3/16]. to extend this to the next sub-interval [3/16,7/16],

    we want to find the interval in the pre-image of U2 that contains \tilde{h}(3/16) = 3\pi/4,

    which is (0,π). again we see that \ \tilde{h}(s) = 4\pi s, on this interval as well.

    we continue in this way, obtaining a path in R that goes from 0 to 4π. the only interesting part is when we pick the slice for the last pre-image.

    note that although we want a pre-image of the same open set U1, now we want the interval in p-1(U1) that contains \tilde{h}(15/16) = 15\pi/4,

    which is no longer (-π/2,π/2) but rather (7π/2,9π/2) instead (k = 2, instead of k = 0).

    yes, there are other paths in R that p would map to h, but this is the only continuous one.
    Last edited by Deveno; June 23rd 2012 at 12:33 AM.
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    Super Member Bernhard's Avatar
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    Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

    Thanks for the help and guidance, Deveno

    Will now work through the post

    Peter
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    Super Member Bernhard's Avatar
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    Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

    I am working through Deveno's example - but I have a problem (which may well be an error I am making)

    This is annoying since the example illustrates the theorem very well - can anyone help with this possible error?

    Problem: It does not appear to me that  \ h( [0, 3/16] ) \subset U_1 as is required

    My working is as follows: (See attachment Example for Munkres - Topology - Lemma 54.1)

    To start we have (see Deveno's post above)

     h(s) = (cos \ 4 \pi s, sin \ 4 \pi s )

    p(x) = ( cos x, sin x )

    U_1 = p( \ ( - \pi / 2 , \pi / 2 ) )


    Now  p( - \pi /2 ) = (cos ( - \pi / 2, sin ( - \pi /2 ) = (0, -1)

     p( \pi / 2) = (cos \pi / 2 , sin \pi /2) = (0, 1)

     p (0) = (cos \ 0, sin \ 0 ) = (1,0)

    Now check that  h ( [s_1, s_2] ) = h( [0, 3/16] ) \subset U_1

     h(s_1) = h(0) = (cos 0, sin 0) = (1,0) \in U_1

     h(s_2) = h(3/16) = (cos 3 \pi / 4, sin 3 \pi /4 ) = (- 0.7071, + 0.7071) which is not in  U_1 --- Problem!!!

    For diagrams see attachment Example for Munkres - Topology - Lemma 54.1


    Can anyone help?
    Attached Files Attached Files
    Last edited by Bernhard; June 23rd 2012 at 07:27 PM.
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    Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

    oh my goodness, you are correct! we need, of course p(h(s1)) to be in U1, so my subinterval is "way too big" (i forgot about the factor of 4).

    changing the first to: [0,1/16] we then have: h([0,1/16]) lies in U1.

    we should be able to continue then with s2 = 3/16, because h([1/16,3/16]) then lies in U2.

    it appears the subintervals (and corresponding U's) should go as follows:

    [3/16,5/16] <--> U3
    [5/16,7/16] <--> U4
    [7/16,9/16] <--> U1
    [9/16,11/16] <--> U2
    [11/16,13/16] <--> U3
    [13/16,15/16] <--> U4
    [15/16,1] <--> U1, as before.

    which wraps around the circle twice, as i intended.
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    Super Member Bernhard's Avatar
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    Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

    Thanks so much for the revised set of subintervals and Ui

    will now rework the problem

    So helpful to have an example like this ... I wish the various texts had more such examples or that there was supplementary texts with such examples ... Would make it so much easier for those working alone ... I wish algebraic topology had the equivalent of Project Crazy Project

    thanks again

    peter
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