Fundamental Group Coset to preimage bijection

[PsychonautQQ]

Homework Statement

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.

Homework Equations

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 >.<

 

[andrewkirk]

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.

 

[PsychonautQQ]

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 $f$?

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 $B$?

 

[andrewkirk]

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 $B$?

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 $e_0$.

 

[PsychonautQQ]

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