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Homotopy and Fundemental Group.

  1. Mar 5, 2008 #1
    I have a few questions on this topic, and i want to see if i got them partially right or wrong.
    1.[tex]\pi_1(X,x_0)[/tex] is the fundemental group of X based on x_0, i.e the groups of homotopy classes which are loops based in x0.
    Suppose that x1,x0 are points in a path connected space X, then the above group is abelian iff for every pair of paths: a,b from x0 to x1 we get that a'([f])=[a^-1]*[f]*[a]=[b^-1]*[f]*=b'([f]) for every [f] in the above group.

    2. Let A be a subset of X, suppose r:X->A is a continuous map s.t r(a)=a for each a in A. if a0 is in A, show that:
    [tex]r_*:\pi_1(X,a_0)->\pi_1(A,a_0)[/tex]
    is surjective where: [tex]r_*([f])=[rof][/tex]
    3. show that every star convex set is simply connected, i.e it's path connected and has the trivial fundemental group.

    here's what I did for 1:
    suppose the group is abelian, if I show that a'^-1ob' is the identity then it will follow that a'=b' ( or so me think).
    so here it goes:
    b'^-1oa'([f])=[a^-1][f][a][b^-1]
    but because the group is abelian and b*a^-1 (im using here the notation of my lecturer and munkres for pasting of paths) is a loop based in x_0, then [b*a^-1] commutes with [f] and it equals [a^-1] so the above term equals: [f][a^-1][a][b^-1]=[f]
    so it's indeed the identity function and thus a'=b'.
    to prove the next implication we simply notice that for every paths: a,b from x0 to x1,
    a'=b' so if [g] is in this group we can represnet it as:
    g=a*b^-1
    so if we operate this [f] we will get agian [f], i.e
    g'([f])=[g^-1][f][g]=[a^-1][f][a][b^-1]=b'^-1(a'([f]))=[f]
    so they commute and it's abelian, am i right here?


    now for two, if it acts only or pi_1(A,a_0) then obviously it's surjcetive but how to show it for other elements of pi_1(X,a_0), wait a minute I think that it's simple enough because every element of pi_1(A,a_0) is in the form of [rof] for f a loop based on a_0, cause if it's a loop, the we can paste it with another loop in A based in a_0, and thus every loop based in a_0 in pi_1(X,a_0).
    Not sure if i ogt it right i mean if [f] is in pi_1(A,a_0) then obviously r_*([f])=[rof]=[f] so in this case it's the identity but if for example [f] is in pi_1(X,a_0) and not in pi_1(A,a_0), then maybe because rof:[0,1]->A, and rof(0)=r(f(0))=r(a_0)=a_0=rof(1)=r(a_0) then also [rof] is homotopy loop class based in a_0, well now that I type this I think that this is rather rudimentary.

    Anyway for the third question I don't see a clear answer, I need to show that every loop based in the star point is the trivial one, i.e it goes and returns the same path but with different directions.
    Not sure how to show it, I mean can't we have a loop that passes through three distinct points and it passes by the star points twice and the other two only once (which is a non trivial loop).

    any hints?
     
  2. jcsd
  3. Mar 5, 2008 #2

    StatusX

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    Your first proof looks good. The second, you're right, is pretty simple, although you might also need to show the map is well defined (ie, r takes X-homotopic loops to A-homotopic loops).

    I'm not sure what you're saying here. You need to show every loop is homotopic to the constant loop. You know there's a line connecting every point to the basepoint, so try sending the points of the loop down along this line.
     
  4. Mar 5, 2008 #3

    mathwonk

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    assume the star point is the origin. let H(t,x) = tx, for t in [0,1]. restricted to the star set, it contracts the space to the origin.

    in particular it contracts any loop. but anyway, a contractible space has trivial pi one.
     
  5. Mar 6, 2008 #4
    But status I need to show that besides the constant loop there are no other loops based in the star point.

    So mathwonk, H(t,x)=tx is a homotopy between all paths connecting points with the star points and the constant loop, am I wrong here?
     
  6. Mar 6, 2008 #5

    JasonRox

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    The center star point has a straight path to any other point. You can use the standard straight-line homotopy to shrink all loops to the center point.

    Another question is to ask, why is it enough to find Pi_1 at the base point x0 rather than check for other base points?
     
  7. Mar 6, 2008 #6

    Hurkyl

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    You mean pi_1 ([itex]\pi_1[/itex]); Pi_1 ([itex]\Pi_1[/itex]) is the fundamental groupoid... which is very closely related to, but not the same as, the fundamental group.
     
  8. Mar 7, 2008 #7

    StatusX

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    No, you need to show all loops based at the star point are homotopic (to the constant loop), and I explained a way to do this.
     
  9. Mar 8, 2008 #8
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