Mixing
Show that if B is simply-connected, then p is a homeomorphism.
$begingroup$
Let $p: E rightarrow B$ be a covering map with $E$ path-connected. Show that if $B$ is simply-connected, then $p$ is a homeomorphism.
I'm checking to see if my solution is flawed.
Since $p$ is a covering map it is a continuous, surjective and open map. That means that all I need to do to show that $p$ is a homeomorphism is to show that it is injective.
So for $p(a) = p(b)$ for $a,b in E$, we want to show that $a = b$.
Now, since $E$ is path-connected there exists a path from $a$ to $b$ denote it by $psi$.
Then $pcircpsi$ is a loop in $B$ since $p(a) = p(b)$.
But since $B$ is simply-connected $pcircpsi$ is homotopic to a point.
So $psi$ must be homotopic to a point when we lift it and therefore $a = b$.
So $p$ is injective and thus a homeomorphism.
algebraic-topology covering-spaces
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
$endgroup$
add a comment |
$begingroup$
Let $p: E rightarrow B$ be a covering map with $E$ path-connected. Show that if $B$ is simply-connected, then $p$ is a homeomorphism.
I'm checking to see if my solution is flawed.
Since $p$ is a covering map it is a continuous, surjective and open map. That means that all I need to do to show that $p$ is a homeomorphism is to show that it is injective.
So for $p(a) = p(b)$ for $a,b in E$, we want to show that $a = b$.
Now, since $E$ is path-connected there exists a path from $a$ to $b$ denote it by $psi$.
Then $pcircpsi$ is a loop in $B$ since $p(a) = p(b)$.
But since $B$ is simply-connected $pcircpsi$ is homotopic to a point.
So $psi$ must be homotopic to a point when we lift it and therefore $a = b$.
So $p$ is injective and thus a homeomorphism.
algebraic-topology covering-spaces
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
$endgroup$
$begingroup$
I think this works fine!
$endgroup$
– Daniele A
Apr 15 '14 at 20:51
1
$begingroup$
It works, but you'll want to be precise and say that $psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path.
$endgroup$
– Olivier Bégassat
Apr 15 '14 at 21:00
add a comment |
$begingroup$
Let $p: E rightarrow B$ be a covering map with $E$ path-connected. Show that if $B$ is simply-connected, then $p$ is a homeomorphism.
I'm checking to see if my solution is flawed.
Since $p$ is a covering map it is a continuous, surjective and open map. That means that all I need to do to show that $p$ is a homeomorphism is to show that it is injective.
So for $p(a) = p(b)$ for $a,b in E$, we want to show that $a = b$.
Now, since $E$ is path-connected there exists a path from $a$ to $b$ denote it by $psi$.
Then $pcircpsi$ is a loop in $B$ since $p(a) = p(b)$.
But since $B$ is simply-connected $pcircpsi$ is homotopic to a point.
So $psi$ must be homotopic to a point when we lift it and therefore $a = b$.
So $p$ is injective and thus a homeomorphism.
algebraic-topology covering-spaces
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
$endgroup$
Let $p: E rightarrow B$ be a covering map with $E$ path-connected. Show that if $B$ is simply-connected, then $p$ is a homeomorphism.
I'm checking to see if my solution is flawed.
Since $p$ is a covering map it is a continuous, surjective and open map. That means that all I need to do to show that $p$ is a homeomorphism is to show that it is injective.
So for $p(a) = p(b)$ for $a,b in E$, we want to show that $a = b$.
Now, since $E$ is path-connected there exists a path from $a$ to $b$ denote it by $psi$.
Then $pcircpsi$ is a loop in $B$ since $p(a) = p(b)$.
But since $B$ is simply-connected $pcircpsi$ is homotopic to a point.
So $psi$ must be homotopic to a point when we lift it and therefore $a = b$.
So $p$ is injective and thus a homeomorphism.
algebraic-topology covering-spaces
algebraic-topology covering-spaces
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
edited Apr 16 '14 at 12:45
Stefan Hamcke
21.9k42880
21.9k42880
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
asked Apr 15 '14 at 20:45
EgoKillaEgoKilla
1,207529
1,207529
$begingroup$
I think this works fine!
$endgroup$
– Daniele A
Apr 15 '14 at 20:51
1
$begingroup$
It works, but you'll want to be precise and say that $psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path.
$endgroup$
– Olivier Bégassat
Apr 15 '14 at 21:00
add a comment |
$begingroup$
I think this works fine!
$endgroup$
– Daniele A
Apr 15 '14 at 20:51
1
$begingroup$
It works, but you'll want to be precise and say that $psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path.
$endgroup$
– Olivier Bégassat
Apr 15 '14 at 21:00
$begingroup$
I think this works fine!
$endgroup$
– Daniele A
Apr 15 '14 at 20:51
$begingroup$
I think this works fine!
$endgroup$
– Daniele A
Apr 15 '14 at 20:51
1
1
$begingroup$
It works, but you'll want to be precise and say that $psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path.
$endgroup$
– Olivier Bégassat
Apr 15 '14 at 21:00
$begingroup$
It works, but you'll want to be precise and say that $psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path.
$endgroup$
– Olivier Bégassat
Apr 15 '14 at 21:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your proof works fine, but you should say that $ppsi$ is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy, $ψ$ is path homotopic to a constant map, too, which implies $a=b$.
With little more effort, we can show that if a loop $phi$ at $b_0$ in $B$ is in the image
$p_*(pi_1(E,e_0))$ (which is a subgroup of $pi_1(B,b_0)$), then $ϕ$ lifts to a loop in $E$. For if $[ϕ]in p_*(pi_1(E,e_0))$, then there is a loop $lambda$ at $e_0$ such that $pλ$ is path homotopic to $ϕ$, and this homotopy lifts to a path homotopy $λsimeqψ$, where
$ψ$ is the lift of $ϕ$ at $e_0$, which is thus a loop.
In particular this implies that a null-homotopic $ϕ=pψ$ lifts to a loop $ψ$ since $[ϕ]=0$ is always in $p_*(pi_1(E,e_0))$. Therefore $ψ(0)=ψ(1)$.
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
$endgroup$
$begingroup$
"since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $psi$ is path homotopic to a constant map and not to some other arbitrary map?
$endgroup$
– Dean Gurvitz
Jan 26 at 18:48
$begingroup$
@DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:Itimes Ito B$ has a lifting $tilde H:Itimes Ito E$ such that $ptilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $tilde H$ satisfies $tilde H(0,t)=ψ(t)$ and $tilde H(1,t)=a$. Note that $tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $tilde H(s,0)$ must be fixed at $a$
$endgroup$
– Stefan Hamcke
Feb 2 at 17:59
$begingroup$
Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz
$endgroup$
– Stefan Hamcke
Feb 2 at 18:05
$begingroup$
I was aware but didn't put 1+1 together. However, I still don't understand why $bar{H}(1,t)$ is constant, since we only know $pbar{H}(1,t)$ is constant, which doesn't directly imply anything about $bar{H}$ on its own
$endgroup$
– Dean Gurvitz
Feb 2 at 19:55
1
$begingroup$
@DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path.
$endgroup$
– Stefan Hamcke
Feb 2 at 21:49
|
show 1 more comment
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$begingroup$
Your proof works fine, but you should say that $ppsi$ is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy, $ψ$ is path homotopic to a constant map, too, which implies $a=b$.
With little more effort, we can show that if a loop $phi$ at $b_0$ in $B$ is in the image
$p_*(pi_1(E,e_0))$ (which is a subgroup of $pi_1(B,b_0)$), then $ϕ$ lifts to a loop in $E$. For if $[ϕ]in p_*(pi_1(E,e_0))$, then there is a loop $lambda$ at $e_0$ such that $pλ$ is path homotopic to $ϕ$, and this homotopy lifts to a path homotopy $λsimeqψ$, where
$ψ$ is the lift of $ϕ$ at $e_0$, which is thus a loop.
In particular this implies that a null-homotopic $ϕ=pψ$ lifts to a loop $ψ$ since $[ϕ]=0$ is always in $p_*(pi_1(E,e_0))$. Therefore $ψ(0)=ψ(1)$.
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
$endgroup$
$begingroup$
"since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $psi$ is path homotopic to a constant map and not to some other arbitrary map?
$endgroup$
– Dean Gurvitz
Jan 26 at 18:48
$begingroup$
@DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:Itimes Ito B$ has a lifting $tilde H:Itimes Ito E$ such that $ptilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $tilde H$ satisfies $tilde H(0,t)=ψ(t)$ and $tilde H(1,t)=a$. Note that $tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $tilde H(s,0)$ must be fixed at $a$
$endgroup$
– Stefan Hamcke
Feb 2 at 17:59
$begingroup$
Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz
$endgroup$
– Stefan Hamcke
Feb 2 at 18:05
$begingroup$
I was aware but didn't put 1+1 together. However, I still don't understand why $bar{H}(1,t)$ is constant, since we only know $pbar{H}(1,t)$ is constant, which doesn't directly imply anything about $bar{H}$ on its own
$endgroup$
– Dean Gurvitz
Feb 2 at 19:55
1
$begingroup$
@DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path.
$endgroup$
– Stefan Hamcke
Feb 2 at 21:49
|
show 1 more comment
$begingroup$
Your proof works fine, but you should say that $ppsi$ is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy, $ψ$ is path homotopic to a constant map, too, which implies $a=b$.
With little more effort, we can show that if a loop $phi$ at $b_0$ in $B$ is in the image
$p_*(pi_1(E,e_0))$ (which is a subgroup of $pi_1(B,b_0)$), then $ϕ$ lifts to a loop in $E$. For if $[ϕ]in p_*(pi_1(E,e_0))$, then there is a loop $lambda$ at $e_0$ such that $pλ$ is path homotopic to $ϕ$, and this homotopy lifts to a path homotopy $λsimeqψ$, where
$ψ$ is the lift of $ϕ$ at $e_0$, which is thus a loop.
In particular this implies that a null-homotopic $ϕ=pψ$ lifts to a loop $ψ$ since $[ϕ]=0$ is always in $p_*(pi_1(E,e_0))$. Therefore $ψ(0)=ψ(1)$.
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
$endgroup$
$begingroup$
"since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $psi$ is path homotopic to a constant map and not to some other arbitrary map?
$endgroup$
– Dean Gurvitz
Jan 26 at 18:48
$begingroup$
@DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:Itimes Ito B$ has a lifting $tilde H:Itimes Ito E$ such that $ptilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $tilde H$ satisfies $tilde H(0,t)=ψ(t)$ and $tilde H(1,t)=a$. Note that $tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $tilde H(s,0)$ must be fixed at $a$
$endgroup$
– Stefan Hamcke
Feb 2 at 17:59
$begingroup$
Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz
$endgroup$
– Stefan Hamcke
Feb 2 at 18:05
$begingroup$
I was aware but didn't put 1+1 together. However, I still don't understand why $bar{H}(1,t)$ is constant, since we only know $pbar{H}(1,t)$ is constant, which doesn't directly imply anything about $bar{H}$ on its own
$endgroup$
– Dean Gurvitz
Feb 2 at 19:55
1
$begingroup$
@DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path.
$endgroup$
– Stefan Hamcke
Feb 2 at 21:49
|
show 1 more comment
$begingroup$
Your proof works fine, but you should say that $ppsi$ is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy, $ψ$ is path homotopic to a constant map, too, which implies $a=b$.
With little more effort, we can show that if a loop $phi$ at $b_0$ in $B$ is in the image
$p_*(pi_1(E,e_0))$ (which is a subgroup of $pi_1(B,b_0)$), then $ϕ$ lifts to a loop in $E$. For if $[ϕ]in p_*(pi_1(E,e_0))$, then there is a loop $lambda$ at $e_0$ such that $pλ$ is path homotopic to $ϕ$, and this homotopy lifts to a path homotopy $λsimeqψ$, where
$ψ$ is the lift of $ϕ$ at $e_0$, which is thus a loop.
In particular this implies that a null-homotopic $ϕ=pψ$ lifts to a loop $ψ$ since $[ϕ]=0$ is always in $p_*(pi_1(E,e_0))$. Therefore $ψ(0)=ψ(1)$.
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
$endgroup$
Your proof works fine, but you should say that $ppsi$ is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy, $ψ$ is path homotopic to a constant map, too, which implies $a=b$.
With little more effort, we can show that if a loop $phi$ at $b_0$ in $B$ is in the image
$p_*(pi_1(E,e_0))$ (which is a subgroup of $pi_1(B,b_0)$), then $ϕ$ lifts to a loop in $E$. For if $[ϕ]in p_*(pi_1(E,e_0))$, then there is a loop $lambda$ at $e_0$ such that $pλ$ is path homotopic to $ϕ$, and this homotopy lifts to a path homotopy $λsimeqψ$, where
$ψ$ is the lift of $ϕ$ at $e_0$, which is thus a loop.
In particular this implies that a null-homotopic $ϕ=pψ$ lifts to a loop $ψ$ since $[ϕ]=0$ is always in $p_*(pi_1(E,e_0))$. Therefore $ψ(0)=ψ(1)$.
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
edited Feb 2 at 17:36
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
answered Apr 16 '14 at 13:35
Stefan HamckeStefan Hamcke
21.9k42880
21.9k42880
$begingroup$
"since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $psi$ is path homotopic to a constant map and not to some other arbitrary map?
$endgroup$
– Dean Gurvitz
Jan 26 at 18:48
$begingroup$
@DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:Itimes Ito B$ has a lifting $tilde H:Itimes Ito E$ such that $ptilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $tilde H$ satisfies $tilde H(0,t)=ψ(t)$ and $tilde H(1,t)=a$. Note that $tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $tilde H(s,0)$ must be fixed at $a$
$endgroup$
– Stefan Hamcke
Feb 2 at 17:59
$begingroup$
Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz
$endgroup$
– Stefan Hamcke
Feb 2 at 18:05
$begingroup$
I was aware but didn't put 1+1 together. However, I still don't understand why $bar{H}(1,t)$ is constant, since we only know $pbar{H}(1,t)$ is constant, which doesn't directly imply anything about $bar{H}$ on its own
$endgroup$
– Dean Gurvitz
Feb 2 at 19:55
1
$begingroup$
@DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path.
$endgroup$
– Stefan Hamcke
Feb 2 at 21:49
|
show 1 more comment
$begingroup$
"since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $psi$ is path homotopic to a constant map and not to some other arbitrary map?
$endgroup$
– Dean Gurvitz
Jan 26 at 18:48
$begingroup$
@DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:Itimes Ito B$ has a lifting $tilde H:Itimes Ito E$ such that $ptilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $tilde H$ satisfies $tilde H(0,t)=ψ(t)$ and $tilde H(1,t)=a$. Note that $tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $tilde H(s,0)$ must be fixed at $a$
$endgroup$
– Stefan Hamcke
Feb 2 at 17:59
$begingroup$
Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz
$endgroup$
– Stefan Hamcke
Feb 2 at 18:05
$begingroup$
I was aware but didn't put 1+1 together. However, I still don't understand why $bar{H}(1,t)$ is constant, since we only know $pbar{H}(1,t)$ is constant, which doesn't directly imply anything about $bar{H}$ on its own
$endgroup$
– Dean Gurvitz
Feb 2 at 19:55
1
$begingroup$
@DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path.
$endgroup$
– Stefan Hamcke
Feb 2 at 21:49
$begingroup$
"since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $psi$ is path homotopic to a constant map and not to some other arbitrary map?
$endgroup$
– Dean Gurvitz
Jan 26 at 18:48
$begingroup$
"since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $psi$ is path homotopic to a constant map and not to some other arbitrary map?
$endgroup$
– Dean Gurvitz
Jan 26 at 18:48
$begingroup$
@DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:Itimes Ito B$ has a lifting $tilde H:Itimes Ito E$ such that $ptilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $tilde H$ satisfies $tilde H(0,t)=ψ(t)$ and $tilde H(1,t)=a$. Note that $tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $tilde H(s,0)$ must be fixed at $a$
$endgroup$
– Stefan Hamcke
Feb 2 at 17:59
$begingroup$
@DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:Itimes Ito B$ has a lifting $tilde H:Itimes Ito E$ such that $ptilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $tilde H$ satisfies $tilde H(0,t)=ψ(t)$ and $tilde H(1,t)=a$. Note that $tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $tilde H(s,0)$ must be fixed at $a$
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– Stefan Hamcke
Feb 2 at 17:59
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Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz
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– Stefan Hamcke
Feb 2 at 18:05
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Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz
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– Stefan Hamcke
Feb 2 at 18:05
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I was aware but didn't put 1+1 together. However, I still don't understand why $bar{H}(1,t)$ is constant, since we only know $pbar{H}(1,t)$ is constant, which doesn't directly imply anything about $bar{H}$ on its own
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– Dean Gurvitz
Feb 2 at 19:55
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I was aware but didn't put 1+1 together. However, I still don't understand why $bar{H}(1,t)$ is constant, since we only know $pbar{H}(1,t)$ is constant, which doesn't directly imply anything about $bar{H}$ on its own
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– Dean Gurvitz
Feb 2 at 19:55
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1
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@DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path.
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– Stefan Hamcke
Feb 2 at 21:49
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@DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path.
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– Stefan Hamcke
Feb 2 at 21:49
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I think this works fine!
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– Daniele A
Apr 15 '14 at 20:51
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It works, but you'll want to be precise and say that $psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path.
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– Olivier Bégassat
Apr 15 '14 at 21:00