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Algebraic Topology

  1. Mar 20, 2009 #1
    I am trying to show that the space Cone(L(X,x)) is homeomorphic to P(X,x)
    where L(X,x) = {loops in X base point x} and
    P(X,x) = {paths in X base point x}

    I firstly considered (L(X,x) x I) and tried to find a surjective map to P(X,x) that would quotient out right but i couldnt seem to find it. For example i considered
    (l,t) -> p where p is the path such that p(1)=l(t) and they agree naturally before
    i.e. F((l,t))(s) = l(ts)

    I was wondering if anyone could point me roughly the right way or just chip in with their own thoughts
  2. jcsd
  3. Mar 20, 2009 #2
    Why should this be true?

    If I take the simplest case X={x} then L(X,x)=P(X,x)={the constant path at x}. Cone(L(X,x)) is just [0,1] which is not homeomorphic to a point.
  4. Mar 20, 2009 #3
    Yes your example is clear and i to was running into similar problems when trying to consider where to map (c(x),t) for different values of t on a general space. The task was set by someone who is knowledgable about the subject so maybe if instead of the cone we consider

    (Cone(L(X,x)))/~ where (c(x),a) ~ (c(x),b), c(x) is the function with constat value x. This would then clear up the problem of how to map the constant function and we would have F(c(x),t) = F(l,0) = c(x).(l any loop)

    Then the problem remains the same, but if we consider the above identification made, it would appear natural to consider a method similar to my first post however i cannot manage to tweak it correctly
  5. Mar 24, 2009 #4
    Consider decomposing a path,p, into a loop,l, and a non looping part,n, and a time,t, at the point where we connect these paths: p = l (+t) n
    where for two paths a,b such that a(1)=b(0) we have (for t not equal 1,0)
    (a (+t) b) (s) = a(s/t) , 0<s<t
    b((s-t)/(1-t)) , t<s<1
    and (a (+0) b) (s) = a(0)
    (a (+1) b) (s) = a(s)

    Motivitated by this define the map from F: (L(X,x))x(N(X,x))xI -> P(X,x)
    where N(X,x) is the set of non looping maps n belongs to N(X,x) => n(t) is not equal to n(0)=x for all t>0 (strictly)
    Then F(l,t,n) = F(l',t',n') => l (+t) n = l' (+t') n' =>
    (t and t' not equal to 0)
    Then WlOG assume t is greater than or equal to t' then if t is strictly greater we would have that one path took the value x at this time whilst the other didn't a contradiction => t = t'
    and it then follows l = l' , n = n'
    Now let t = 0 , then F(l,0,n) = F(l',t',n') then it is clear t' =0 , we have no condition on l,n

    Hence we can find a bijection from Cone(L(X,x)xN(X,x)) to P(X,x) as i don't wish to consider the different possible topologies let us assume this is a continuous map.
    Then if the original statement (or its revised version) was true then we must have that
    Cone(L(X,x)xN(X,x)) is homeomorphic to (Cone(L(X,x)))/~
    Do the above constuctions appear correct?
  6. Mar 24, 2009 #5
    Re reading i see that we must also make the identification (l,1,n) ~' (l,1,n')
    But then we will just get ; Cone(L(X,x)xN(X,x))/~'
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