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Thread: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

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

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

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

    [My question relates to the proof of the uniqueness of $\displaystyle \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 $\displaystyle s_0, s_1, .... s_n $ , such that for each i the set $\displaystyle f([s_i, s_{i+1} ])$ lies in an open set U (Use Lebesgue number lemma).

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

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

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

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

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

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

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

    But now focussing on the uniqueness of $\displaystyle \tilde{f} $ - couldn't we define $\displaystyle \tilde{f} (s) $ as belonging to any of the sets $\displaystyle \{ 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 $\displaystyle \tilde{f} (s) $ for each of the $\displaystyle \{ V_{\alpha} \} $ - but this is not may idea of a unique $\displaystyle \tilde{f} $.

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

    Peter
    Last edited by Bernhard; Jun 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 $\displaystyle \tilde{f}$ to "the next subinterval", we pick the ONLY pre-image of that U that contains $\displaystyle \tilde{f}(s_i)$.

    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:

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

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

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

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

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

    (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 $\displaystyle \tilde{h}(3/16) = 3\pi/4$,

    which is (0,π). again we see that $\displaystyle \ \tilde{h}(s) = 4\pi s$, 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 $\displaystyle \tilde{h}(15/16) = 15\pi/4$,

    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; Jun 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 $\displaystyle \ 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)

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

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

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


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

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

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

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

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

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

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


    Can anyone help?
    Attached Files Attached Files
    Last edited by Bernhard; Jun 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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