$X,Y$ are locally path connected and path connected, with universal covers $tildeX, tildeY$. If $X simeq Y$ then $tildeX simeq tildeY$
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Suppose $X,Y$ are locally path connected and path connected, with universal covers $tildeX, tildeY$. I'd like to prove that if $X simeq Y$ then $tildeX simeq tildeY$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrmid_X$ and $fg simeq mathrmid_Y$, and let $p : tildeX to X$, $q : tildeY to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tildeY$. There is a Lemma that says:
Suppose $pi : B to A$ is a covering projection, and $F: C to A$ is a continuous map where $C$ is simply connected and locally path connected. Suppose given base points $a_0, b_0, c_0$ of $A,B,C$ with $pi(b_0) = a_0 = F(c_0)$. Then there is a unique continuous $tildeF : C to B$ with $pi tildeF = F$ and $tildeF(c_0) = b_0$.
I'm going to use this Lemma with $C = tildeY$ and $F = fq$. So pick points $x_0 in X$, $tildex_0 in tildeX$, $tildey_0 in Y$ such that $p(tildex_0) = x_0 = fq(tildey_0)$. Then $fq$ has a unique lifting to a map $tildef: tildeY to tildeX$ such that $p tildef = fq$ and $tildef(tildey_0) = tildex_0$. Similarly, $gp$ has a unique lifting to a map $tildeg : tildeX to tildeY$ such that $tildeg = gp$ and $tildeg(tildex_0) = tildey_0$.
I'd like it to be the case that $tildef$ and $tildeg$ are homotopy equivalences. I don't know if this is true and, if it is, I don't know how to show it.
I can see that $p tildef tildeg simeq p$ and $q tildeg tildef simeq q$ which is quite close to what I want, but I don't know how to proceed.
I'm also concerned that I haven't use the "path connected" criterion anywhere. Hints would be great!
Thanks
algebraic-topology
asked May 22 '12 at 18:52
Matt
44839
 |Â
show 1 more comment
up vote
2
down vote
favorite
Suppose $X,Y$ are locally path connected and path connected, with universal covers $tildeX, tildeY$. I'd like to prove that if $X simeq Y$ then $tildeX simeq tildeY$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrmid_X$ and $fg simeq mathrmid_Y$, and let $p : tildeX to X$, $q : tildeY to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tildeY$. There is a Lemma that says:
Suppose $pi : B to A$ is a covering projection, and $F: C to A$ is a continuous map where $C$ is simply connected and locally path connected. Suppose given base points $a_0, b_0, c_0$ of $A,B,C$ with $pi(b_0) = a_0 = F(c_0)$. Then there is a unique continuous $tildeF : C to B$ with $pi tildeF = F$ and $tildeF(c_0) = b_0$.
I'm going to use this Lemma with $C = tildeY$ and $F = fq$. So pick points $x_0 in X$, $tildex_0 in tildeX$, $tildey_0 in Y$ such that $p(tildex_0) = x_0 = fq(tildey_0)$. Then $fq$ has a unique lifting to a map $tildef: tildeY to tildeX$ such that $p tildef = fq$ and $tildef(tildey_0) = tildex_0$. Similarly, $gp$ has a unique lifting to a map $tildeg : tildeX to tildeY$ such that $tildeg = gp$ and $tildeg(tildex_0) = tildey_0$.
I'd like it to be the case that $tildef$ and $tildeg$ are homotopy equivalences. I don't know if this is true and, if it is, I don't know how to show it.
I can see that $p tildef tildeg simeq p$ and $q tildeg tildef simeq q$ which is quite close to what I want, but I don't know how to proceed.
I'm also concerned that I haven't use the "path connected" criterion anywhere. Hints would be great!
Thanks
algebraic-topology
asked May 22 '12 at 18:52
Matt
44839
1
See also here and here.
– t.b.
May 22 '12 at 19:10
@t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p tildef tildeg simeq p$ to $tildef tildeg simeq mathrmid_tildeX$?
– Matt
May 22 '12 at 19:15
By construction $tildeftildeg$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tildeX = 1$).
– t.b.
May 22 '12 at 19:26
@t.b. Actually, sorry. Why is $tildef tildeg$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
– Matt
May 22 '12 at 19:43
You probably mean $pf = p$, no? That's what Clive verifies right before that.
– t.b.
May 22 '12 at 19:56
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $X,Y$ are locally path connected and path connected, with universal covers $tildeX, tildeY$. I'd like to prove that if $X simeq Y$ then $tildeX simeq tildeY$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrmid_X$ and $fg simeq mathrmid_Y$, and let $p : tildeX to X$, $q : tildeY to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tildeY$. There is a Lemma that says:
Suppose $pi : B to A$ is a covering projection, and $F: C to A$ is a continuous map where $C$ is simply connected and locally path connected. Suppose given base points $a_0, b_0, c_0$ of $A,B,C$ with $pi(b_0) = a_0 = F(c_0)$. Then there is a unique continuous $tildeF : C to B$ with $pi tildeF = F$ and $tildeF(c_0) = b_0$.
I'm going to use this Lemma with $C = tildeY$ and $F = fq$. So pick points $x_0 in X$, $tildex_0 in tildeX$, $tildey_0 in Y$ such that $p(tildex_0) = x_0 = fq(tildey_0)$. Then $fq$ has a unique lifting to a map $tildef: tildeY to tildeX$ such that $p tildef = fq$ and $tildef(tildey_0) = tildex_0$. Similarly, $gp$ has a unique lifting to a map $tildeg : tildeX to tildeY$ such that $tildeg = gp$ and $tildeg(tildex_0) = tildey_0$.
I'd like it to be the case that $tildef$ and $tildeg$ are homotopy equivalences. I don't know if this is true and, if it is, I don't know how to show it.
I can see that $p tildef tildeg simeq p$ and $q tildeg tildef simeq q$ which is quite close to what I want, but I don't know how to proceed.
I'm also concerned that I haven't use the "path connected" criterion anywhere. Hints would be great!
Thanks
algebraic-topology
asked May 22 '12 at 18:52
Matt
44839
Suppose $X,Y$ are locally path connected and path connected, with universal covers $tildeX, tildeY$. I'd like to prove that if $X simeq Y$ then $tildeX simeq tildeY$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrmid_X$ and $fg simeq mathrmid_Y$, and let $p : tildeX to X$, $q : tildeY to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tildeY$. There is a Lemma that says:
Suppose $pi : B to A$ is a covering projection, and $F: C to A$ is a continuous map where $C$ is simply connected and locally path connected. Suppose given base points $a_0, b_0, c_0$ of $A,B,C$ with $pi(b_0) = a_0 = F(c_0)$. Then there is a unique continuous $tildeF : C to B$ with $pi tildeF = F$ and $tildeF(c_0) = b_0$.
I'm going to use this Lemma with $C = tildeY$ and $F = fq$. So pick points $x_0 in X$, $tildex_0 in tildeX$, $tildey_0 in Y$ such that $p(tildex_0) = x_0 = fq(tildey_0)$. Then $fq$ has a unique lifting to a map $tildef: tildeY to tildeX$ such that $p tildef = fq$ and $tildef(tildey_0) = tildex_0$. Similarly, $gp$ has a unique lifting to a map $tildeg : tildeX to tildeY$ such that $tildeg = gp$ and $tildeg(tildex_0) = tildey_0$.
I'd like it to be the case that $tildef$ and $tildeg$ are homotopy equivalences. I don't know if this is true and, if it is, I don't know how to show it.
I can see that $p tildef tildeg simeq p$ and $q tildeg tildef simeq q$ which is quite close to what I want, but I don't know how to proceed.
I'm also concerned that I haven't use the "path connected" criterion anywhere. Hints would be great!
Thanks
algebraic-topology
algebraic-topology
asked May 22 '12 at 18:52
Matt
44839
asked May 22 '12 at 18:52
Matt
44839
asked May 22 '12 at 18:52
Matt
44839
asked May 22 '12 at 18:52
Matt
44839
asked May 22 '12 at 18:52
Matt
44839
44839
1
See also here and here.
– t.b.
May 22 '12 at 19:10
@t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p tildef tildeg simeq p$ to $tildef tildeg simeq mathrmid_tildeX$?
– Matt
May 22 '12 at 19:15
By construction $tildeftildeg$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tildeX = 1$).
– t.b.
May 22 '12 at 19:26
@t.b. Actually, sorry. Why is $tildef tildeg$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
– Matt
May 22 '12 at 19:43
You probably mean $pf = p$, no? That's what Clive verifies right before that.
– t.b.
May 22 '12 at 19:56
 |Â
show 1 more comment
1
See also here and here.
– t.b.
May 22 '12 at 19:10
@t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p tildef tildeg simeq p$ to $tildef tildeg simeq mathrmid_tildeX$?
– Matt
May 22 '12 at 19:15
By construction $tildeftildeg$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tildeX = 1$).
– t.b.
May 22 '12 at 19:26
@t.b. Actually, sorry. Why is $tildef tildeg$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
– Matt
May 22 '12 at 19:43
You probably mean $pf = p$, no? That's what Clive verifies right before that.
– t.b.
May 22 '12 at 19:56
1
1
See also here and here.
– t.b.
May 22 '12 at 19:10
See also here and here.
– t.b.
May 22 '12 at 19:10
@t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p tildef tildeg simeq p$ to $tildef tildeg simeq mathrmid_tildeX$?
– Matt
May 22 '12 at 19:15
@t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p tildef tildeg simeq p$ to $tildef tildeg simeq mathrmid_tildeX$?
– Matt
May 22 '12 at 19:15
By construction $tildeftildeg$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tildeX = 1$).
– t.b.
May 22 '12 at 19:26
By construction $tildeftildeg$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tildeX = 1$).
– t.b.
May 22 '12 at 19:26
@t.b. Actually, sorry. Why is $tildef tildeg$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
– Matt
May 22 '12 at 19:43
@t.b. Actually, sorry. Why is $tildef tildeg$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
– Matt
May 22 '12 at 19:43
You probably mean $pf = p$, no? That's what Clive verifies right before that.
– t.b.
May 22 '12 at 19:56
You probably mean $pf = p$, no? That's what Clive verifies right before that.
– t.b.
May 22 '12 at 19:56
 |Â
show 1 more comment
1 Answer
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Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tildeX longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tildeX longrightarrow tildeY$. See the image.
Now do a similar map with $tildeY longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tildeX longrightarrow tildeY$ is a lift of $p: tildeX longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
1
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
– Najib Idrissi
Mar 1 at 8:58
Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though.
– Behnam Esmayli
Mar 2 at 16:11
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tildeX longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tildeX longrightarrow tildeY$. See the image.
Now do a similar map with $tildeY longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tildeX longrightarrow tildeY$ is a lift of $p: tildeX longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
1
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
– Najib Idrissi
Mar 1 at 8:58
Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though.
– Behnam Esmayli
Mar 2 at 16:11
add a comment |Â
up vote
0
down vote
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tildeX longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tildeX longrightarrow tildeY$. See the image.
Now do a similar map with $tildeY longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tildeX longrightarrow tildeY$ is a lift of $p: tildeX longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
1
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
– Najib Idrissi
Mar 1 at 8:58
Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though.
– Behnam Esmayli
Mar 2 at 16:11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tildeX longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tildeX longrightarrow tildeY$. See the image.
Now do a similar map with $tildeY longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tildeX longrightarrow tildeY$ is a lift of $p: tildeX longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tildeX longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tildeX longrightarrow tildeY$. See the image.
Now do a similar map with $tildeY longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tildeX longrightarrow tildeY$ is a lift of $p: tildeX longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
answered Jan 25 at 3:26
Behnam Esmayli
1,865315
1,865315
1
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
– Najib Idrissi
Mar 1 at 8:58
Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though.
– Behnam Esmayli
Mar 2 at 16:11
add a comment |Â
1
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
– Najib Idrissi
Mar 1 at 8:58
Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though.
– Behnam Esmayli
Mar 2 at 16:11
1
1
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
– Najib Idrissi
Mar 1 at 8:58
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
– Najib Idrissi
Mar 1 at 8:58
Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though.
– Behnam Esmayli
Mar 2 at 16:11
Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though.
– Behnam Esmayli
Mar 2 at 16:11
add a comment |Â
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1
See also here and here.
– t.b.
May 22 '12 at 19:10
@t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p tildef tildeg simeq p$ to $tildef tildeg simeq mathrmid_tildeX$?
– Matt
May 22 '12 at 19:15
By construction $tildeftildeg$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tildeX = 1$).
– t.b.
May 22 '12 at 19:26
@t.b. Actually, sorry. Why is $tildef tildeg$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
– Matt
May 22 '12 at 19:43
You probably mean $pf = p$, no? That's what Clive verifies right before that.
– t.b.
May 22 '12 at 19:56