Fundamental group via deck transformations considering rotation on sphere












2












$begingroup$


Let $Z_m$ act on $S^1$ by multiplication with $e^{2pi ki/m}$ for $k in Z_m$. Let $X = S^1 / Z_m$ be the orbit space of this action. Then we have a universal cover $q:S^1 rightarrow X$ given by the canonical projection. To determine the fundamental group of $X$ we can consider the group of deck transformations $Aut(q)$ of $q$, since we know that $Pi_1(X) = Aut(q)$.



$Aut(q)$ is by definition the group of all homeomorphisms $f: S^1 rightarrow S^1$ such that $qf=q$. We see that all rotations by degree $e^{2pi ki/m}$ for $k in Z_m$ are such homeos. But why are there no other ones?






algebraic-topology covering-spaces fundamental-groups







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $pi_1 (X) cong text{Aut}(q)$. In fact, it is correct that $text{Aut}(q) cong mathbb{Z}/mmathbb{Z}$, but $pi_1 (X) cong mathbb{Z}$, since $X$ is homeomorphic to $S^1$.
    $endgroup$
    – Sameer Kailasa
    Jan 2 at 19:43












  • $begingroup$
    yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$
    $endgroup$
    – CHwC
    Jan 2 at 19:52










  • $begingroup$
    I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations
    $endgroup$
    – CHwC
    Jan 2 at 19:53
















2












$begingroup$


Let $Z_m$ act on $S^1$ by multiplication with $e^{2pi ki/m}$ for $k in Z_m$. Let $X = S^1 / Z_m$ be the orbit space of this action. Then we have a universal cover $q:S^1 rightarrow X$ given by the canonical projection. To determine the fundamental group of $X$ we can consider the group of deck transformations $Aut(q)$ of $q$, since we know that $Pi_1(X) = Aut(q)$.



$Aut(q)$ is by definition the group of all homeomorphisms $f: S^1 rightarrow S^1$ such that $qf=q$. We see that all rotations by degree $e^{2pi ki/m}$ for $k in Z_m$ are such homeos. But why are there no other ones?






algebraic-topology covering-spaces fundamental-groups







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $pi_1 (X) cong text{Aut}(q)$. In fact, it is correct that $text{Aut}(q) cong mathbb{Z}/mmathbb{Z}$, but $pi_1 (X) cong mathbb{Z}$, since $X$ is homeomorphic to $S^1$.
    $endgroup$
    – Sameer Kailasa
    Jan 2 at 19:43












  • $begingroup$
    yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$
    $endgroup$
    – CHwC
    Jan 2 at 19:52










  • $begingroup$
    I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations
    $endgroup$
    – CHwC
    Jan 2 at 19:53














2












2








2





$begingroup$


Let $Z_m$ act on $S^1$ by multiplication with $e^{2pi ki/m}$ for $k in Z_m$. Let $X = S^1 / Z_m$ be the orbit space of this action. Then we have a universal cover $q:S^1 rightarrow X$ given by the canonical projection. To determine the fundamental group of $X$ we can consider the group of deck transformations $Aut(q)$ of $q$, since we know that $Pi_1(X) = Aut(q)$.



$Aut(q)$ is by definition the group of all homeomorphisms $f: S^1 rightarrow S^1$ such that $qf=q$. We see that all rotations by degree $e^{2pi ki/m}$ for $k in Z_m$ are such homeos. But why are there no other ones?






algebraic-topology covering-spaces fundamental-groups







share|cite|improve this question









$endgroup$




Let $Z_m$ act on $S^1$ by multiplication with $e^{2pi ki/m}$ for $k in Z_m$. Let $X = S^1 / Z_m$ be the orbit space of this action. Then we have a universal cover $q:S^1 rightarrow X$ given by the canonical projection. To determine the fundamental group of $X$ we can consider the group of deck transformations $Aut(q)$ of $q$, since we know that $Pi_1(X) = Aut(q)$.



$Aut(q)$ is by definition the group of all homeomorphisms $f: S^1 rightarrow S^1$ such that $qf=q$. We see that all rotations by degree $e^{2pi ki/m}$ for $k in Z_m$ are such homeos. But why are there no other ones?






algebraic-topology covering-spaces fundamental-groups




algebraic-topology covering-spaces fundamental-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 19:26









CHwCCHwC

309111




309111








  • 2




    $begingroup$
    Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $pi_1 (X) cong text{Aut}(q)$. In fact, it is correct that $text{Aut}(q) cong mathbb{Z}/mmathbb{Z}$, but $pi_1 (X) cong mathbb{Z}$, since $X$ is homeomorphic to $S^1$.
    $endgroup$
    – Sameer Kailasa
    Jan 2 at 19:43












  • $begingroup$
    yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$
    $endgroup$
    – CHwC
    Jan 2 at 19:52










  • $begingroup$
    I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations
    $endgroup$
    – CHwC
    Jan 2 at 19:53














  • 2




    $begingroup$
    Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $pi_1 (X) cong text{Aut}(q)$. In fact, it is correct that $text{Aut}(q) cong mathbb{Z}/mmathbb{Z}$, but $pi_1 (X) cong mathbb{Z}$, since $X$ is homeomorphic to $S^1$.
    $endgroup$
    – Sameer Kailasa
    Jan 2 at 19:43












  • $begingroup$
    yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$
    $endgroup$
    – CHwC
    Jan 2 at 19:52










  • $begingroup$
    I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations
    $endgroup$
    – CHwC
    Jan 2 at 19:53








2




2




$begingroup$
Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $pi_1 (X) cong text{Aut}(q)$. In fact, it is correct that $text{Aut}(q) cong mathbb{Z}/mmathbb{Z}$, but $pi_1 (X) cong mathbb{Z}$, since $X$ is homeomorphic to $S^1$.
$endgroup$
– Sameer Kailasa
Jan 2 at 19:43






$begingroup$
Since $S^1$ is not simply connected, your map $q$ is not a universal cover of $X$, so we can't say $pi_1 (X) cong text{Aut}(q)$. In fact, it is correct that $text{Aut}(q) cong mathbb{Z}/mmathbb{Z}$, but $pi_1 (X) cong mathbb{Z}$, since $X$ is homeomorphic to $S^1$.
$endgroup$
– Sameer Kailasa
Jan 2 at 19:43














$begingroup$
yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$
$endgroup$
– CHwC
Jan 2 at 19:52




$begingroup$
yes you're right I have been actually thinking about lens spaces which come from a group action on $S^3$ but I thought I could make things easier for my question by considering only $S^1$
$endgroup$
– CHwC
Jan 2 at 19:52












$begingroup$
I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations
$endgroup$
– CHwC
Jan 2 at 19:53




$begingroup$
I want to kind of understand why we cannot find other homeos of $S^1$ which permute the orbits than the rotations
$endgroup$
– CHwC
Jan 2 at 19:53










1 Answer
1






active

oldest

votes


















3












$begingroup$

$pi_1(X)$ is not meant to be equal to ${rm Aut}(q)$. That would be true if the covering was a universal cover, which is not the case here.



[An example of a universal cover of $X$ would be the map $q_{rm univ} : mathbb R to X$, sending $x mapsto x {rm mod } tfrac{2pi}{m}$. This is a universal cover because $mathbb R$ is simply connected. The deck transformation group for this universal cover is ${rm Aut}(q_{rm univ}) = mathbb Z$. Since $X$ itself is a small circle, $pi_1(X) = mathbb Z$, which agrees with ${rm Aut}(q_{rm univ})$.]



As for your original covering $q : S^1 to X$, we still have the following useful result (Hatcher 1.39): If $H : = q_star (pi_1(S^1)) $ is the image of $pi_1(S^1)$ under the group homomorphism $q_star : pi_1(S^1) to pi_1(X)$, then ${rm Aut}(q)$ is isomorphic to $N(H) / H$, where $N(H)$ is the normaliser of $q_star (pi_1(S^1))$ in $pi_1(X)$. Now $pi_1(X) = mathbb Z$, and $H$ is the subgroup $mmathbb Z$, which is a normal subgroup, so the theorem tells us that ${rm Aut}(q) = mathbb Z_m$. And this agrees perfectly with your counting.



Finally, I'll address the question in your comment: Can we see directly that there are no more than $m$ deck transformations for $q$? The answer is yes, because $q$ is an $m$-sheeted covering of $X$ (meaning that the preimage of $q^{-1}(x)$ for any point $x in X$ consists of $m $ points. Given a point $x in X$, a deck transformation $f$ is uniquely determined by where it sends $x$. But $f$ can only send $x$ to one of the points in $q^{-1}(x)$, and there are $m$ such points. Hence there can be at most $m$ deck transformations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations?
    $endgroup$
    – CHwC
    Jan 2 at 19:50










  • $begingroup$
    when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally
    $endgroup$
    – CHwC
    Jan 2 at 19:51










  • $begingroup$
    @CHwC See edit...
    $endgroup$
    – Kenny Wong
    Jan 2 at 20:02










  • $begingroup$
    why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing
    $endgroup$
    – CHwC
    Jan 2 at 21:35










  • $begingroup$
    @CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : widetilde X to X$ and a map $f : Y to X$ with two lifts $widetilde f_1, widetilde f_2 : Y to widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = widetilde X$ and $f = q$, then $widetilde f_1, widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere.
    $endgroup$
    – Kenny Wong
    Jan 2 at 21:40













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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$pi_1(X)$ is not meant to be equal to ${rm Aut}(q)$. That would be true if the covering was a universal cover, which is not the case here.



[An example of a universal cover of $X$ would be the map $q_{rm univ} : mathbb R to X$, sending $x mapsto x {rm mod } tfrac{2pi}{m}$. This is a universal cover because $mathbb R$ is simply connected. The deck transformation group for this universal cover is ${rm Aut}(q_{rm univ}) = mathbb Z$. Since $X$ itself is a small circle, $pi_1(X) = mathbb Z$, which agrees with ${rm Aut}(q_{rm univ})$.]



As for your original covering $q : S^1 to X$, we still have the following useful result (Hatcher 1.39): If $H : = q_star (pi_1(S^1)) $ is the image of $pi_1(S^1)$ under the group homomorphism $q_star : pi_1(S^1) to pi_1(X)$, then ${rm Aut}(q)$ is isomorphic to $N(H) / H$, where $N(H)$ is the normaliser of $q_star (pi_1(S^1))$ in $pi_1(X)$. Now $pi_1(X) = mathbb Z$, and $H$ is the subgroup $mmathbb Z$, which is a normal subgroup, so the theorem tells us that ${rm Aut}(q) = mathbb Z_m$. And this agrees perfectly with your counting.



Finally, I'll address the question in your comment: Can we see directly that there are no more than $m$ deck transformations for $q$? The answer is yes, because $q$ is an $m$-sheeted covering of $X$ (meaning that the preimage of $q^{-1}(x)$ for any point $x in X$ consists of $m $ points. Given a point $x in X$, a deck transformation $f$ is uniquely determined by where it sends $x$. But $f$ can only send $x$ to one of the points in $q^{-1}(x)$, and there are $m$ such points. Hence there can be at most $m$ deck transformations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations?
    $endgroup$
    – CHwC
    Jan 2 at 19:50










  • $begingroup$
    when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally
    $endgroup$
    – CHwC
    Jan 2 at 19:51










  • $begingroup$
    @CHwC See edit...
    $endgroup$
    – Kenny Wong
    Jan 2 at 20:02










  • $begingroup$
    why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing
    $endgroup$
    – CHwC
    Jan 2 at 21:35










  • $begingroup$
    @CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : widetilde X to X$ and a map $f : Y to X$ with two lifts $widetilde f_1, widetilde f_2 : Y to widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = widetilde X$ and $f = q$, then $widetilde f_1, widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere.
    $endgroup$
    – Kenny Wong
    Jan 2 at 21:40


















3












$begingroup$

$pi_1(X)$ is not meant to be equal to ${rm Aut}(q)$. That would be true if the covering was a universal cover, which is not the case here.



[An example of a universal cover of $X$ would be the map $q_{rm univ} : mathbb R to X$, sending $x mapsto x {rm mod } tfrac{2pi}{m}$. This is a universal cover because $mathbb R$ is simply connected. The deck transformation group for this universal cover is ${rm Aut}(q_{rm univ}) = mathbb Z$. Since $X$ itself is a small circle, $pi_1(X) = mathbb Z$, which agrees with ${rm Aut}(q_{rm univ})$.]



As for your original covering $q : S^1 to X$, we still have the following useful result (Hatcher 1.39): If $H : = q_star (pi_1(S^1)) $ is the image of $pi_1(S^1)$ under the group homomorphism $q_star : pi_1(S^1) to pi_1(X)$, then ${rm Aut}(q)$ is isomorphic to $N(H) / H$, where $N(H)$ is the normaliser of $q_star (pi_1(S^1))$ in $pi_1(X)$. Now $pi_1(X) = mathbb Z$, and $H$ is the subgroup $mmathbb Z$, which is a normal subgroup, so the theorem tells us that ${rm Aut}(q) = mathbb Z_m$. And this agrees perfectly with your counting.



Finally, I'll address the question in your comment: Can we see directly that there are no more than $m$ deck transformations for $q$? The answer is yes, because $q$ is an $m$-sheeted covering of $X$ (meaning that the preimage of $q^{-1}(x)$ for any point $x in X$ consists of $m $ points. Given a point $x in X$, a deck transformation $f$ is uniquely determined by where it sends $x$. But $f$ can only send $x$ to one of the points in $q^{-1}(x)$, and there are $m$ such points. Hence there can be at most $m$ deck transformations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations?
    $endgroup$
    – CHwC
    Jan 2 at 19:50










  • $begingroup$
    when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally
    $endgroup$
    – CHwC
    Jan 2 at 19:51










  • $begingroup$
    @CHwC See edit...
    $endgroup$
    – Kenny Wong
    Jan 2 at 20:02










  • $begingroup$
    why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing
    $endgroup$
    – CHwC
    Jan 2 at 21:35










  • $begingroup$
    @CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : widetilde X to X$ and a map $f : Y to X$ with two lifts $widetilde f_1, widetilde f_2 : Y to widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = widetilde X$ and $f = q$, then $widetilde f_1, widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere.
    $endgroup$
    – Kenny Wong
    Jan 2 at 21:40
















3












3








3





$begingroup$

$pi_1(X)$ is not meant to be equal to ${rm Aut}(q)$. That would be true if the covering was a universal cover, which is not the case here.



[An example of a universal cover of $X$ would be the map $q_{rm univ} : mathbb R to X$, sending $x mapsto x {rm mod } tfrac{2pi}{m}$. This is a universal cover because $mathbb R$ is simply connected. The deck transformation group for this universal cover is ${rm Aut}(q_{rm univ}) = mathbb Z$. Since $X$ itself is a small circle, $pi_1(X) = mathbb Z$, which agrees with ${rm Aut}(q_{rm univ})$.]



As for your original covering $q : S^1 to X$, we still have the following useful result (Hatcher 1.39): If $H : = q_star (pi_1(S^1)) $ is the image of $pi_1(S^1)$ under the group homomorphism $q_star : pi_1(S^1) to pi_1(X)$, then ${rm Aut}(q)$ is isomorphic to $N(H) / H$, where $N(H)$ is the normaliser of $q_star (pi_1(S^1))$ in $pi_1(X)$. Now $pi_1(X) = mathbb Z$, and $H$ is the subgroup $mmathbb Z$, which is a normal subgroup, so the theorem tells us that ${rm Aut}(q) = mathbb Z_m$. And this agrees perfectly with your counting.



Finally, I'll address the question in your comment: Can we see directly that there are no more than $m$ deck transformations for $q$? The answer is yes, because $q$ is an $m$-sheeted covering of $X$ (meaning that the preimage of $q^{-1}(x)$ for any point $x in X$ consists of $m $ points. Given a point $x in X$, a deck transformation $f$ is uniquely determined by where it sends $x$. But $f$ can only send $x$ to one of the points in $q^{-1}(x)$, and there are $m$ such points. Hence there can be at most $m$ deck transformations.






share|cite|improve this answer











$endgroup$



$pi_1(X)$ is not meant to be equal to ${rm Aut}(q)$. That would be true if the covering was a universal cover, which is not the case here.



[An example of a universal cover of $X$ would be the map $q_{rm univ} : mathbb R to X$, sending $x mapsto x {rm mod } tfrac{2pi}{m}$. This is a universal cover because $mathbb R$ is simply connected. The deck transformation group for this universal cover is ${rm Aut}(q_{rm univ}) = mathbb Z$. Since $X$ itself is a small circle, $pi_1(X) = mathbb Z$, which agrees with ${rm Aut}(q_{rm univ})$.]



As for your original covering $q : S^1 to X$, we still have the following useful result (Hatcher 1.39): If $H : = q_star (pi_1(S^1)) $ is the image of $pi_1(S^1)$ under the group homomorphism $q_star : pi_1(S^1) to pi_1(X)$, then ${rm Aut}(q)$ is isomorphic to $N(H) / H$, where $N(H)$ is the normaliser of $q_star (pi_1(S^1))$ in $pi_1(X)$. Now $pi_1(X) = mathbb Z$, and $H$ is the subgroup $mmathbb Z$, which is a normal subgroup, so the theorem tells us that ${rm Aut}(q) = mathbb Z_m$. And this agrees perfectly with your counting.



Finally, I'll address the question in your comment: Can we see directly that there are no more than $m$ deck transformations for $q$? The answer is yes, because $q$ is an $m$-sheeted covering of $X$ (meaning that the preimage of $q^{-1}(x)$ for any point $x in X$ consists of $m $ points. Given a point $x in X$, a deck transformation $f$ is uniquely determined by where it sends $x$. But $f$ can only send $x$ to one of the points in $q^{-1}(x)$, and there are $m$ such points. Hence there can be at most $m$ deck transformations.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 21:03

























answered Jan 2 at 19:48









Kenny WongKenny Wong

19.1k21440




19.1k21440












  • $begingroup$
    okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations?
    $endgroup$
    – CHwC
    Jan 2 at 19:50










  • $begingroup$
    when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally
    $endgroup$
    – CHwC
    Jan 2 at 19:51










  • $begingroup$
    @CHwC See edit...
    $endgroup$
    – Kenny Wong
    Jan 2 at 20:02










  • $begingroup$
    why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing
    $endgroup$
    – CHwC
    Jan 2 at 21:35










  • $begingroup$
    @CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : widetilde X to X$ and a map $f : Y to X$ with two lifts $widetilde f_1, widetilde f_2 : Y to widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = widetilde X$ and $f = q$, then $widetilde f_1, widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere.
    $endgroup$
    – Kenny Wong
    Jan 2 at 21:40




















  • $begingroup$
    okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations?
    $endgroup$
    – CHwC
    Jan 2 at 19:50










  • $begingroup$
    when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally
    $endgroup$
    – CHwC
    Jan 2 at 19:51










  • $begingroup$
    @CHwC See edit...
    $endgroup$
    – Kenny Wong
    Jan 2 at 20:02










  • $begingroup$
    why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing
    $endgroup$
    – CHwC
    Jan 2 at 21:35










  • $begingroup$
    @CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : widetilde X to X$ and a map $f : Y to X$ with two lifts $widetilde f_1, widetilde f_2 : Y to widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = widetilde X$ and $f = q$, then $widetilde f_1, widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere.
    $endgroup$
    – Kenny Wong
    Jan 2 at 21:40


















$begingroup$
okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations?
$endgroup$
– CHwC
Jan 2 at 19:50




$begingroup$
okay but is there any way to see "directly" why there can be no other homeos which satisfy $qf=q$ then the rotations?
$endgroup$
– CHwC
Jan 2 at 19:50












$begingroup$
when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally
$endgroup$
– CHwC
Jan 2 at 19:51




$begingroup$
when I try to imagine such other homeos then I have to twist things around so much that I feel like the continuity of these maps is lost somewhere, but I dont see how to express this formally
$endgroup$
– CHwC
Jan 2 at 19:51












$begingroup$
@CHwC See edit...
$endgroup$
– Kenny Wong
Jan 2 at 20:02




$begingroup$
@CHwC See edit...
$endgroup$
– Kenny Wong
Jan 2 at 20:02












$begingroup$
why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing
$endgroup$
– CHwC
Jan 2 at 21:35




$begingroup$
why is a deck transformation uniquely determined by the value it takes on an arbitrary point x? i guess thats the point that i am missing
$endgroup$
– CHwC
Jan 2 at 21:35












$begingroup$
@CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : widetilde X to X$ and a map $f : Y to X$ with two lifts $widetilde f_1, widetilde f_2 : Y to widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = widetilde X$ and $f = q$, then $widetilde f_1, widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere.
$endgroup$
– Kenny Wong
Jan 2 at 21:40






$begingroup$
@CHwC This is a consequence of the Unique Lifting Property (Hatcher 1.34): Given a covering space $q : widetilde X to X$ and a map $f : Y to X$ with two lifts $widetilde f_1, widetilde f_2 : Y to widetilde X$ that agree at one point of $Y$, then if $Y$ is connected, these two lifts must agree on all of $Y$. If you apply this with $Y = widetilde X$ and $f = q$, then $widetilde f_1, widetilde f_2$ are deck transformations, so the theorem tells you that if two deck transformations agree at one point, then they agree everywhere.
$endgroup$
– Kenny Wong
Jan 2 at 21:40




















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