Uniqueness of lift in covering space theory
$begingroup$
Let Y be a topological space and $pi: Xrightarrow Y$ a covering map. Take two lifts of the covering map $tilde {pi_1}$, $tilde {pi_2} : Xrightarrow X$ such that they agree on $x_0 in X$.
Under what hypothesis can I conclude that the two lifts coincide?
I have always seen proofs of existence and uniqueness together but I can't exactly single out what is needed for uniqueness only. I know the condition is connectedness, but I am not sure if I have to suppose it for $X$ or $Y$.
If you could also give a sketch of the proof, I would very much appreciate it.
algebraic-topology covering-spaces
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
$endgroup$
add a comment |
$begingroup$
Let Y be a topological space and $pi: Xrightarrow Y$ a covering map. Take two lifts of the covering map $tilde {pi_1}$, $tilde {pi_2} : Xrightarrow X$ such that they agree on $x_0 in X$.
Under what hypothesis can I conclude that the two lifts coincide?
I have always seen proofs of existence and uniqueness together but I can't exactly single out what is needed for uniqueness only. I know the condition is connectedness, but I am not sure if I have to suppose it for $X$ or $Y$.
If you could also give a sketch of the proof, I would very much appreciate it.
algebraic-topology covering-spaces
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
$endgroup$
add a comment |
$begingroup$
Let Y be a topological space and $pi: Xrightarrow Y$ a covering map. Take two lifts of the covering map $tilde {pi_1}$, $tilde {pi_2} : Xrightarrow X$ such that they agree on $x_0 in X$.
Under what hypothesis can I conclude that the two lifts coincide?
I have always seen proofs of existence and uniqueness together but I can't exactly single out what is needed for uniqueness only. I know the condition is connectedness, but I am not sure if I have to suppose it for $X$ or $Y$.
If you could also give a sketch of the proof, I would very much appreciate it.
algebraic-topology covering-spaces
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
$endgroup$
Let Y be a topological space and $pi: Xrightarrow Y$ a covering map. Take two lifts of the covering map $tilde {pi_1}$, $tilde {pi_2} : Xrightarrow X$ such that they agree on $x_0 in X$.
Under what hypothesis can I conclude that the two lifts coincide?
I have always seen proofs of existence and uniqueness together but I can't exactly single out what is needed for uniqueness only. I know the condition is connectedness, but I am not sure if I have to suppose it for $X$ or $Y$.
If you could also give a sketch of the proof, I would very much appreciate it.
algebraic-topology covering-spaces
algebraic-topology covering-spaces
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
edited Jan 1 at 19:39
Angelo Brillante Romeo
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
asked Jan 1 at 19:19
Angelo Brillante RomeoAngelo Brillante Romeo
1006
1006
add a comment |
add a comment |
1 Answer
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$begingroup$
More generally, let $f:Zto Y$ be a map, and $g,h : Zto X$ two lifts of $f$ that agree on $z_0in Z$.
Then ${xin Z mid g(x)=h(x)}$ is nonempty (it contains $z_0$). Moreover, it is open : if $h(x) = g(x)$, then consider a basic open set $U$, neighbourhood of $f(x)$ such that $pi^{-1}(U) =displaystylecoprod_{iin I}U_i$ with each $U_i$ mapped homeomorphically onto $U$.
Then $pi(h(x)) in U$ so $h(x) in U_i$ for some $i$. Let $pi_i$ be the restriction/corestriction of $pi$ to $U_ito U$. Then, restricted to $h^{-1}(U_i)$, $picirc h = f$, so $pi_icirc h_{mid h^{-1}(U_i)} = f_{mid h^{-1}(U_i)}$, and so $h= pi_i^{-1}circ f$ on $h^{-1}(U_i)$.
Similarly, on $g^{-1}(U_i)$, $g= pi_i^{-1}circ f$. So on $h^{-1}(U_i)cap g^{-1}(U_i)$ (which contains $x$), $g=h$. Therefore our set is open.
It is also closed. This either follows trivially if you're assuming $Y$ to be $T_2$; but even if it's not then if $g(x) neq h(x)$, then you can perform a similar argument to show that on a small enough neighbourhood of $x$, $g$ lands in $U_i$ while $h$ lands in $U_j, jneq i$, so $U_icap U_j = emptyset$, so that on this small neighbourhood, $g(y)neq h(y)$. If this isn't clear you should do this yourself.
Thus we have found a nonempty clopen subset of $Z$: if $Z$ is connected, it follows that it's $Z$.
So we have
If $Z$ is connected and $f:Zto Y$ has two lifts $g,h : Zto X$ that agree on a point $z_0in Z$, then $g=h$.
In your situation (after the edit) it means that if you assume $X$ to be connected, then the uniqueness follows.
If $X$ is not connected, then there are some counterexamples.
answered Jan 1 at 19:45
MaxMax
15.7k11143
$endgroup$
$begingroup$
For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $Xto X$ at all is a lift.
$endgroup$
– Eric Wofsey
Jan 1 at 19:49
$begingroup$
Thank you both!
$endgroup$
– Angelo Brillante Romeo
Jan 1 at 19:59
add a comment |
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$begingroup$
More generally, let $f:Zto Y$ be a map, and $g,h : Zto X$ two lifts of $f$ that agree on $z_0in Z$.
Then ${xin Z mid g(x)=h(x)}$ is nonempty (it contains $z_0$). Moreover, it is open : if $h(x) = g(x)$, then consider a basic open set $U$, neighbourhood of $f(x)$ such that $pi^{-1}(U) =displaystylecoprod_{iin I}U_i$ with each $U_i$ mapped homeomorphically onto $U$.
Then $pi(h(x)) in U$ so $h(x) in U_i$ for some $i$. Let $pi_i$ be the restriction/corestriction of $pi$ to $U_ito U$. Then, restricted to $h^{-1}(U_i)$, $picirc h = f$, so $pi_icirc h_{mid h^{-1}(U_i)} = f_{mid h^{-1}(U_i)}$, and so $h= pi_i^{-1}circ f$ on $h^{-1}(U_i)$.
Similarly, on $g^{-1}(U_i)$, $g= pi_i^{-1}circ f$. So on $h^{-1}(U_i)cap g^{-1}(U_i)$ (which contains $x$), $g=h$. Therefore our set is open.
It is also closed. This either follows trivially if you're assuming $Y$ to be $T_2$; but even if it's not then if $g(x) neq h(x)$, then you can perform a similar argument to show that on a small enough neighbourhood of $x$, $g$ lands in $U_i$ while $h$ lands in $U_j, jneq i$, so $U_icap U_j = emptyset$, so that on this small neighbourhood, $g(y)neq h(y)$. If this isn't clear you should do this yourself.
Thus we have found a nonempty clopen subset of $Z$: if $Z$ is connected, it follows that it's $Z$.
So we have
If $Z$ is connected and $f:Zto Y$ has two lifts $g,h : Zto X$ that agree on a point $z_0in Z$, then $g=h$.
In your situation (after the edit) it means that if you assume $X$ to be connected, then the uniqueness follows.
If $X$ is not connected, then there are some counterexamples.
answered Jan 1 at 19:45
MaxMax
15.7k11143
$endgroup$
$begingroup$
For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $Xto X$ at all is a lift.
$endgroup$
– Eric Wofsey
Jan 1 at 19:49
$begingroup$
Thank you both!
$endgroup$
– Angelo Brillante Romeo
Jan 1 at 19:59
add a comment |
$begingroup$
More generally, let $f:Zto Y$ be a map, and $g,h : Zto X$ two lifts of $f$ that agree on $z_0in Z$.
Then ${xin Z mid g(x)=h(x)}$ is nonempty (it contains $z_0$). Moreover, it is open : if $h(x) = g(x)$, then consider a basic open set $U$, neighbourhood of $f(x)$ such that $pi^{-1}(U) =displaystylecoprod_{iin I}U_i$ with each $U_i$ mapped homeomorphically onto $U$.
Then $pi(h(x)) in U$ so $h(x) in U_i$ for some $i$. Let $pi_i$ be the restriction/corestriction of $pi$ to $U_ito U$. Then, restricted to $h^{-1}(U_i)$, $picirc h = f$, so $pi_icirc h_{mid h^{-1}(U_i)} = f_{mid h^{-1}(U_i)}$, and so $h= pi_i^{-1}circ f$ on $h^{-1}(U_i)$.
Similarly, on $g^{-1}(U_i)$, $g= pi_i^{-1}circ f$. So on $h^{-1}(U_i)cap g^{-1}(U_i)$ (which contains $x$), $g=h$. Therefore our set is open.
It is also closed. This either follows trivially if you're assuming $Y$ to be $T_2$; but even if it's not then if $g(x) neq h(x)$, then you can perform a similar argument to show that on a small enough neighbourhood of $x$, $g$ lands in $U_i$ while $h$ lands in $U_j, jneq i$, so $U_icap U_j = emptyset$, so that on this small neighbourhood, $g(y)neq h(y)$. If this isn't clear you should do this yourself.
Thus we have found a nonempty clopen subset of $Z$: if $Z$ is connected, it follows that it's $Z$.
So we have
If $Z$ is connected and $f:Zto Y$ has two lifts $g,h : Zto X$ that agree on a point $z_0in Z$, then $g=h$.
In your situation (after the edit) it means that if you assume $X$ to be connected, then the uniqueness follows.
If $X$ is not connected, then there are some counterexamples.
answered Jan 1 at 19:45
MaxMax
15.7k11143
$endgroup$
$begingroup$
For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $Xto X$ at all is a lift.
$endgroup$
– Eric Wofsey
Jan 1 at 19:49
$begingroup$
Thank you both!
$endgroup$
– Angelo Brillante Romeo
Jan 1 at 19:59
add a comment |
$begingroup$
More generally, let $f:Zto Y$ be a map, and $g,h : Zto X$ two lifts of $f$ that agree on $z_0in Z$.
Then ${xin Z mid g(x)=h(x)}$ is nonempty (it contains $z_0$). Moreover, it is open : if $h(x) = g(x)$, then consider a basic open set $U$, neighbourhood of $f(x)$ such that $pi^{-1}(U) =displaystylecoprod_{iin I}U_i$ with each $U_i$ mapped homeomorphically onto $U$.
Then $pi(h(x)) in U$ so $h(x) in U_i$ for some $i$. Let $pi_i$ be the restriction/corestriction of $pi$ to $U_ito U$. Then, restricted to $h^{-1}(U_i)$, $picirc h = f$, so $pi_icirc h_{mid h^{-1}(U_i)} = f_{mid h^{-1}(U_i)}$, and so $h= pi_i^{-1}circ f$ on $h^{-1}(U_i)$.
Similarly, on $g^{-1}(U_i)$, $g= pi_i^{-1}circ f$. So on $h^{-1}(U_i)cap g^{-1}(U_i)$ (which contains $x$), $g=h$. Therefore our set is open.
It is also closed. This either follows trivially if you're assuming $Y$ to be $T_2$; but even if it's not then if $g(x) neq h(x)$, then you can perform a similar argument to show that on a small enough neighbourhood of $x$, $g$ lands in $U_i$ while $h$ lands in $U_j, jneq i$, so $U_icap U_j = emptyset$, so that on this small neighbourhood, $g(y)neq h(y)$. If this isn't clear you should do this yourself.
Thus we have found a nonempty clopen subset of $Z$: if $Z$ is connected, it follows that it's $Z$.
So we have
If $Z$ is connected and $f:Zto Y$ has two lifts $g,h : Zto X$ that agree on a point $z_0in Z$, then $g=h$.
In your situation (after the edit) it means that if you assume $X$ to be connected, then the uniqueness follows.
If $X$ is not connected, then there are some counterexamples.
answered Jan 1 at 19:45
MaxMax
15.7k11143
$endgroup$
More generally, let $f:Zto Y$ be a map, and $g,h : Zto X$ two lifts of $f$ that agree on $z_0in Z$.
Then ${xin Z mid g(x)=h(x)}$ is nonempty (it contains $z_0$). Moreover, it is open : if $h(x) = g(x)$, then consider a basic open set $U$, neighbourhood of $f(x)$ such that $pi^{-1}(U) =displaystylecoprod_{iin I}U_i$ with each $U_i$ mapped homeomorphically onto $U$.
Then $pi(h(x)) in U$ so $h(x) in U_i$ for some $i$. Let $pi_i$ be the restriction/corestriction of $pi$ to $U_ito U$. Then, restricted to $h^{-1}(U_i)$, $picirc h = f$, so $pi_icirc h_{mid h^{-1}(U_i)} = f_{mid h^{-1}(U_i)}$, and so $h= pi_i^{-1}circ f$ on $h^{-1}(U_i)$.
Similarly, on $g^{-1}(U_i)$, $g= pi_i^{-1}circ f$. So on $h^{-1}(U_i)cap g^{-1}(U_i)$ (which contains $x$), $g=h$. Therefore our set is open.
It is also closed. This either follows trivially if you're assuming $Y$ to be $T_2$; but even if it's not then if $g(x) neq h(x)$, then you can perform a similar argument to show that on a small enough neighbourhood of $x$, $g$ lands in $U_i$ while $h$ lands in $U_j, jneq i$, so $U_icap U_j = emptyset$, so that on this small neighbourhood, $g(y)neq h(y)$. If this isn't clear you should do this yourself.
Thus we have found a nonempty clopen subset of $Z$: if $Z$ is connected, it follows that it's $Z$.
So we have
If $Z$ is connected and $f:Zto Y$ has two lifts $g,h : Zto X$ that agree on a point $z_0in Z$, then $g=h$.
In your situation (after the edit) it means that if you assume $X$ to be connected, then the uniqueness follows.
If $X$ is not connected, then there are some counterexamples.
answered Jan 1 at 19:45
MaxMax
15.7k11143
answered Jan 1 at 19:45
MaxMax
15.7k11143
answered Jan 1 at 19:45
MaxMax
15.7k11143
answered Jan 1 at 19:45
MaxMax
15.7k11143
15.7k11143
$begingroup$
For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $Xto X$ at all is a lift.
$endgroup$
– Eric Wofsey
Jan 1 at 19:49
$begingroup$
Thank you both!
$endgroup$
– Angelo Brillante Romeo
Jan 1 at 19:59
add a comment |
$begingroup$
For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $Xto X$ at all is a lift.
$endgroup$
– Eric Wofsey
Jan 1 at 19:49
$begingroup$
Thank you both!
$endgroup$
– Angelo Brillante Romeo
Jan 1 at 19:59
$begingroup$
For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $Xto X$ at all is a lift.
$endgroup$
– Eric Wofsey
Jan 1 at 19:49
$begingroup$
For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $Xto X$ at all is a lift.
$endgroup$
– Eric Wofsey
Jan 1 at 19:49
$begingroup$
Thank you both!
$endgroup$
– Angelo Brillante Romeo
Jan 1 at 19:59
$begingroup$
Thank you both!
$endgroup$
– Angelo Brillante Romeo
Jan 1 at 19:59
add a comment |
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