Algebraic Topology: Showing Cone(L(X,x)) is Homeomorphic to P(X,x)

[nughret]

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 couldn't 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

 

[yyat]

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.

 

[nughret]

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 knowledgeable 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

 

[nughret]

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?

 

[nughret]

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))/~'