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Fundamental Group Coset to preimage bijection

  1. Aug 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Let p: E-->B be a covering map, let p(e_0)=b_0 and let E be path connected. Show that there is a bijection between the collection of right cosets of p*F(E,e_0) in F(B,b_0) (where p* is the homomorphism of fundamental groups induced by p and F(E,e_0),F(B,b_0) are the fundamental groups of E and B based at e_0 and b_0 respectively) and the preimage of b_0 under the covering map p.

    2. Relevant equations


    3. The attempt at a solution
    I have already showed that the homomorphism induced by p injects F(E,e_0) into F(B,e_0). Now i'm trying to show why the cosets defined above have a bijective correspondence to the preimage of the covering map of b_0.... E being path connected definitely has something to do with it. Does anyone have any good hints/tips/insights? I seem to be going in circles >.<
     
  2. jcsd
  3. Aug 9, 2017 #2

    andrewkirk

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    A good strategy in trying to prove a bijection is to focus on the most natural map one can think of, between the two sets. Usually it turns out to be what one wants.

    What might be a nice natural map between the collection of right cosets and the pre-image ##H\triangleq p^{-1}(b_0)##?

    Every element of a right coset is a loop in B. If we could associate it with a path in E, we might reasonably expect that path to start and end on a point in ##H##.

    To try and head in that direction, we might start by defining A as the set of all paths in E that start at ##e_0## and end at a point in ##H##. Then partition that set based on the point at which the path ends. Then there is a bijection between the partition - call it R - and H.

    Now we need to try to relate R to the right coset ##Q\triangleq \pi_1(B,B_0)\ /\ p^*\left(\pi_1(E,e_0)\right)##.

    Pick a coset in Q and choose a representative element ##f##, which is a loop in B. Can we associate f with an element of R? We can't just take a path ##p^{-1}\circ f## in E because ##p^{-1}## is not a function. But since the domain of ##f## is ##[0,1]##, which is compact, maybe we can find an open cover of ##[0,1]## that, upon reducing to a finite open cover, gives us a series of homeomorphisms between a finite collection of open sets in B, whose union contains ##Im\ f##, and corresponding open sets in E. Since each sufficiently small open set ('evenly covered neighbourhood') in B will correspond to multiple homeomorphic images of it in E (the 'sheets' making up the 'fibre' above that nbd), we need a way of choosing one of those sheets. Since we want to relate ##f## to an element of R, which is a set of paths that start at ##e_0##, we can do that by requiring the open set from our cover that contains 0, to map to ##e_0##. Do you think you can go on from there, using the finite sub-cover of ##[0,1]##, to show that there is a unique path in E that corresponds to the loop ##f##?

    If we can do that then we have found a function, call it ##\phi##, from the set of loops in B, to R, by setting ##\phi(f)## to be the partition component containing the unique path (from ##A##) corresponding to ##f##.

    Next we'd like to show that if loops ##f_1,f_2## are in the same coset in Q then ##\phi(f_1)=\phi(f_2)##. If so then we can define a map ##\phi^*## from Q to R.

    Then all we have to do is show that ##\phi^*## is surjective and injective.

    PS This is a pretty first-principles approach. It is possible there are theorems that have already been given to you that could considerably short-cut this.
     
    Last edited: Aug 9, 2017
  4. Aug 9, 2017 #3
    The unique path in E that corresponds to that loop, you mean the unique lifting of ##f## that must exist because ##E## is a covering space of ##B##?
     
  5. Aug 9, 2017 #4

    andrewkirk

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    The terminology I am familiar with is similar, but slightly different. Under that terminology the path we are referring to is

    'the unique lift ##\gamma## of ##f## such that ##\gamma(0)=e_0##'

    There is a different 'lift' of ##f## for every point in the fibre over ##b_0##, but only one of those starts at ##e_0##.
     
  6. Aug 9, 2017 #5
    Btw thanks, i've made a lot of progress on this proof. I found a mapping and proved that it's well defined, my professor verified what I did. Now i'm trying to show injectiveness, he says surjective will be easy.

    I'm defining a map let's call it g such that g sends equivalence classes in the fundamental group of B to the end point of their unique lifting into E, thus it's a map that can send elements from the cosets of Q to preimages of b_0
     
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