$X,Y$ are locally path connected and path connected, with universal covers $tilde{X}, tilde{Y}$. If $X simeq...
$begingroup$
Suppose $X,Y$ are locally path connected and path connected, with universal covers $tilde{X}, tilde{Y}$. I'd like to prove that if $X simeq Y$ then $tilde{X} simeq tilde{Y}$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrm{id}_X$ and $fg simeq mathrm{id}_Y$, and let $p : tilde{X} to X$, $q : tilde{Y} to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tilde{Y}$. 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 $tilde{F} : C to B$ with $pi tilde{F} = F$ and $tilde{F}(c_0) = b_0$.
I'm going to use this Lemma with $C = tilde{Y}$ and $F = fq$. So pick points $x_0 in X$, $tilde{x_0} in tilde{X}$, $tilde{y_0} in Y$ such that $p(tilde{x_0}) = x_0 = fq(tilde{y_0})$. Then $fq$ has a unique lifting to a map $tilde{f}: tilde{Y} to tilde{X}$ such that $p tilde{f} = fq$ and $tilde{f}(tilde{y_0}) = tilde{x_0}$. Similarly, $gp$ has a unique lifting to a map $tilde{g} : tilde{X} to tilde{Y}$ such that $tilde{g} = gp$ and $tilde{g}(tilde{x_0}) = tilde{y_0}$.
I'd like it to be the case that $tilde{f}$ and $tilde{g}$ 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 $p tilde{f} tilde{g} simeq p$ and $q tilde{g} tilde{f} simeq q$ 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
MattMatt
448310
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $X,Y$ are locally path connected and path connected, with universal covers $tilde{X}, tilde{Y}$. I'd like to prove that if $X simeq Y$ then $tilde{X} simeq tilde{Y}$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrm{id}_X$ and $fg simeq mathrm{id}_Y$, and let $p : tilde{X} to X$, $q : tilde{Y} to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tilde{Y}$. 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 $tilde{F} : C to B$ with $pi tilde{F} = F$ and $tilde{F}(c_0) = b_0$.
I'm going to use this Lemma with $C = tilde{Y}$ and $F = fq$. So pick points $x_0 in X$, $tilde{x_0} in tilde{X}$, $tilde{y_0} in Y$ such that $p(tilde{x_0}) = x_0 = fq(tilde{y_0})$. Then $fq$ has a unique lifting to a map $tilde{f}: tilde{Y} to tilde{X}$ such that $p tilde{f} = fq$ and $tilde{f}(tilde{y_0}) = tilde{x_0}$. Similarly, $gp$ has a unique lifting to a map $tilde{g} : tilde{X} to tilde{Y}$ such that $tilde{g} = gp$ and $tilde{g}(tilde{x_0}) = tilde{y_0}$.
I'd like it to be the case that $tilde{f}$ and $tilde{g}$ 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 $p tilde{f} tilde{g} simeq p$ and $q tilde{g} tilde{f} simeq q$ 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
MattMatt
448310
$endgroup$
1
$begingroup$
See also here and here.
$endgroup$
– t.b.
May 22 '12 at 19:10
$begingroup$
@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 $p tilde{f} tilde{g} simeq p$ to $tilde{f} tilde{g} simeq mathrm{id}_{tilde{X}}$?
$endgroup$
– Matt
May 22 '12 at 19:15
$begingroup$
By construction $tilde{f}tilde{g}$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tilde{X} = 1$).
$endgroup$
– t.b.
May 22 '12 at 19:26
$begingroup$
@t.b. Actually, sorry. Why is $tilde{f} tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
$endgroup$
– Matt
May 22 '12 at 19:43
$begingroup$
You probably mean $pf = p$, no? That's what Clive verifies right before that.
$endgroup$
– t.b.
May 22 '12 at 19:56
|
show 1 more comment
$begingroup$
Suppose $X,Y$ are locally path connected and path connected, with universal covers $tilde{X}, tilde{Y}$. I'd like to prove that if $X simeq Y$ then $tilde{X} simeq tilde{Y}$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrm{id}_X$ and $fg simeq mathrm{id}_Y$, and let $p : tilde{X} to X$, $q : tilde{Y} to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tilde{Y}$. 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 $tilde{F} : C to B$ with $pi tilde{F} = F$ and $tilde{F}(c_0) = b_0$.
I'm going to use this Lemma with $C = tilde{Y}$ and $F = fq$. So pick points $x_0 in X$, $tilde{x_0} in tilde{X}$, $tilde{y_0} in Y$ such that $p(tilde{x_0}) = x_0 = fq(tilde{y_0})$. Then $fq$ has a unique lifting to a map $tilde{f}: tilde{Y} to tilde{X}$ such that $p tilde{f} = fq$ and $tilde{f}(tilde{y_0}) = tilde{x_0}$. Similarly, $gp$ has a unique lifting to a map $tilde{g} : tilde{X} to tilde{Y}$ such that $tilde{g} = gp$ and $tilde{g}(tilde{x_0}) = tilde{y_0}$.
I'd like it to be the case that $tilde{f}$ and $tilde{g}$ 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 $p tilde{f} tilde{g} simeq p$ and $q tilde{g} tilde{f} simeq q$ 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
MattMatt
448310
$endgroup$
Suppose $X,Y$ are locally path connected and path connected, with universal covers $tilde{X}, tilde{Y}$. I'd like to prove that if $X simeq Y$ then $tilde{X} simeq tilde{Y}$.
I've had the following thoughts:
Let $f : X to Y$ and $g: Y to X$ be such that $gf simeq mathrm{id}_X$ and $fg simeq mathrm{id}_Y$, and let $p : tilde{X} to X$, $q : tilde{Y} to Y$ be the covering projections. Since $Y$ is locally path connected, so is $tilde{Y}$. 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 $tilde{F} : C to B$ with $pi tilde{F} = F$ and $tilde{F}(c_0) = b_0$.
I'm going to use this Lemma with $C = tilde{Y}$ and $F = fq$. So pick points $x_0 in X$, $tilde{x_0} in tilde{X}$, $tilde{y_0} in Y$ such that $p(tilde{x_0}) = x_0 = fq(tilde{y_0})$. Then $fq$ has a unique lifting to a map $tilde{f}: tilde{Y} to tilde{X}$ such that $p tilde{f} = fq$ and $tilde{f}(tilde{y_0}) = tilde{x_0}$. Similarly, $gp$ has a unique lifting to a map $tilde{g} : tilde{X} to tilde{Y}$ such that $tilde{g} = gp$ and $tilde{g}(tilde{x_0}) = tilde{y_0}$.
I'd like it to be the case that $tilde{f}$ and $tilde{g}$ 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 $p tilde{f} tilde{g} simeq p$ and $q tilde{g} tilde{f} simeq q$ 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
MattMatt
448310
asked May 22 '12 at 18:52
MattMatt
448310
asked May 22 '12 at 18:52
MattMatt
448310
asked May 22 '12 at 18:52
MattMatt
448310
asked May 22 '12 at 18:52
MattMatt
448310
448310
1
$begingroup$
See also here and here.
$endgroup$
– t.b.
May 22 '12 at 19:10
$begingroup$
@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 $p tilde{f} tilde{g} simeq p$ to $tilde{f} tilde{g} simeq mathrm{id}_{tilde{X}}$?
$endgroup$
– Matt
May 22 '12 at 19:15
$begingroup$
By construction $tilde{f}tilde{g}$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tilde{X} = 1$).
$endgroup$
– t.b.
May 22 '12 at 19:26
$begingroup$
@t.b. Actually, sorry. Why is $tilde{f} tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
$endgroup$
– Matt
May 22 '12 at 19:43
$begingroup$
You probably mean $pf = p$, no? That's what Clive verifies right before that.
$endgroup$
– t.b.
May 22 '12 at 19:56
|
show 1 more comment
1
$begingroup$
See also here and here.
$endgroup$
– t.b.
May 22 '12 at 19:10
$begingroup$
@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 $p tilde{f} tilde{g} simeq p$ to $tilde{f} tilde{g} simeq mathrm{id}_{tilde{X}}$?
$endgroup$
– Matt
May 22 '12 at 19:15
$begingroup$
By construction $tilde{f}tilde{g}$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tilde{X} = 1$).
$endgroup$
– t.b.
May 22 '12 at 19:26
$begingroup$
@t.b. Actually, sorry. Why is $tilde{f} tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
$endgroup$
– Matt
May 22 '12 at 19:43
$begingroup$
You probably mean $pf = p$, no? That's what Clive verifies right before that.
$endgroup$
– t.b.
May 22 '12 at 19:56
1
1
$begingroup$
See also here and here.
$endgroup$
– t.b.
May 22 '12 at 19:10
$begingroup$
See also here and here.
$endgroup$
– t.b.
May 22 '12 at 19:10
$begingroup$
@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 $p tilde{f} tilde{g} simeq p$ to $tilde{f} tilde{g} simeq mathrm{id}_{tilde{X}}$?
$endgroup$
– Matt
May 22 '12 at 19:15
$begingroup$
@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 $p tilde{f} tilde{g} simeq p$ to $tilde{f} tilde{g} simeq mathrm{id}_{tilde{X}}$?
$endgroup$
– Matt
May 22 '12 at 19:15
$begingroup$
By construction $tilde{f}tilde{g}$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tilde{X} = 1$).
$endgroup$
– t.b.
May 22 '12 at 19:26
$begingroup$
By construction $tilde{f}tilde{g}$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tilde{X} = 1$).
$endgroup$
– t.b.
May 22 '12 at 19:26
$begingroup$
@t.b. Actually, sorry. Why is $tilde{f} tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
$endgroup$
– Matt
May 22 '12 at 19:43
$begingroup$
@t.b. Actually, sorry. Why is $tilde{f} tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
$endgroup$
– Matt
May 22 '12 at 19:43
$begingroup$
You probably mean $pf = p$, no? That's what Clive verifies right before that.
$endgroup$
– t.b.
May 22 '12 at 19:56
$begingroup$
You probably mean $pf = p$, no? That's what Clive verifies right before that.
$endgroup$
– t.b.
May 22 '12 at 19:56
|
show 1 more comment
1 Answer
1
active
oldest
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$begingroup$
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tilde{X} longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tilde{X} longrightarrow tilde{Y}$. See the image.
Now do a similar map with $tilde{Y} longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tilde{X} longrightarrow tilde{Y}$ is a lift of $p: tilde{X} longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
$endgroup$
1
$begingroup$
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
$endgroup$
– Najib Idrissi
Mar 1 '18 at 8:58
$begingroup$
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.
$endgroup$
– Behnam Esmayli
Mar 2 '18 at 16:11
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tilde{X} longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tilde{X} longrightarrow tilde{Y}$. See the image.
Now do a similar map with $tilde{Y} longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tilde{X} longrightarrow tilde{Y}$ is a lift of $p: tilde{X} longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
$endgroup$
1
$begingroup$
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
$endgroup$
– Najib Idrissi
Mar 1 '18 at 8:58
$begingroup$
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.
$endgroup$
– Behnam Esmayli
Mar 2 '18 at 16:11
add a comment |
$begingroup$
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tilde{X} longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tilde{X} longrightarrow tilde{Y}$. See the image.
Now do a similar map with $tilde{Y} longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tilde{X} longrightarrow tilde{Y}$ is a lift of $p: tilde{X} longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
$endgroup$
1
$begingroup$
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
$endgroup$
– Najib Idrissi
Mar 1 '18 at 8:58
$begingroup$
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.
$endgroup$
– Behnam Esmayli
Mar 2 '18 at 16:11
add a comment |
$begingroup$
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tilde{X} longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tilde{X} longrightarrow tilde{Y}$. See the image.
Now do a similar map with $tilde{Y} longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tilde{X} longrightarrow tilde{Y}$ is a lift of $p: tilde{X} longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
$endgroup$
Assume that $phi:X longrightarrow Y$ is a homeomorphism. The map $ tilde{X} longrightarrow X longrightarrow Y$ can be (uniquely) lifted to a map $Phi:tilde{X} longrightarrow tilde{Y}$. See the image.
Now do a similar map with $tilde{Y} longrightarrow Y longrightarrow X$ (the vertical map.)
Notice that $Psi circ Phi : tilde{X} longrightarrow tilde{Y}$ is a lift of $p: tilde{X} longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
answered Jan 25 '18 at 3:26
Behnam EsmayliBehnam Esmayli
1,976515
1,976515
1
$begingroup$
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
$endgroup$
– Najib Idrissi
Mar 1 '18 at 8:58
$begingroup$
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.
$endgroup$
– Behnam Esmayli
Mar 2 '18 at 16:11
add a comment |
1
$begingroup$
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
$endgroup$
– Najib Idrissi
Mar 1 '18 at 8:58
$begingroup$
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.
$endgroup$
– Behnam Esmayli
Mar 2 '18 at 16:11
1
1
$begingroup$
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
$endgroup$
– Najib Idrissi
Mar 1 '18 at 8:58
$begingroup$
$X$ and $Y$ aren't homeomorphic, they're homotopy equivalent.
$endgroup$
– Najib Idrissi
Mar 1 '18 at 8:58
$begingroup$
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.
$endgroup$
– Behnam Esmayli
Mar 2 '18 at 16:11
$begingroup$
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.
$endgroup$
– Behnam Esmayli
Mar 2 '18 at 16:11
add a comment |
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1
$begingroup$
See also here and here.
$endgroup$
– t.b.
May 22 '12 at 19:10
$begingroup$
@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 $p tilde{f} tilde{g} simeq p$ to $tilde{f} tilde{g} simeq mathrm{id}_{tilde{X}}$?
$endgroup$
– Matt
May 22 '12 at 19:15
$begingroup$
By construction $tilde{f}tilde{g}$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($pi_1 tilde{X} = 1$).
$endgroup$
– t.b.
May 22 '12 at 19:26
$begingroup$
@t.b. Actually, sorry. Why is $tilde{f} tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
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– Matt
May 22 '12 at 19:43
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You probably mean $pf = p$, no? That's what Clive verifies right before that.
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– t.b.
May 22 '12 at 19:56