Constructing 6-fold cover of $S^1 vee S^1$ with deck transformation group $cong S_6$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Covering spaces Hatcher question 6.Universal Cover of wedge sums of spaces?Universal Cover of the Punctured TorusHow to find universal covering space?Deck transformations of universal cover are isomorphic to the fundamental group - explicitlyCover of the disk that behaves like a circleWhat does this action of $pi_1$ on the universal cover of $X$ induces a deck transformation?Does $S^1 vee S^1$ have a cover homotopically equivalent to the wedge sum of $k$ circles?For the space $X = T # T$ find its fundamental group and prove the universal cover is contractible.Deck transformation group isomorphic to $mathbbZ/mathbb6Z$
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Constructing 6-fold cover of $S^1 vee S^1$ with deck transformation group $cong S_6$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Covering spaces Hatcher question 6.Universal Cover of wedge sums of spaces?Universal Cover of the Punctured TorusHow to find universal covering space?Deck transformations of universal cover are isomorphic to the fundamental group - explicitlyCover of the disk that behaves like a circleWhat does this action of $pi_1$ on the universal cover of $X$ induces a deck transformation?Does $S^1 vee S^1$ have a cover homotopically equivalent to the wedge sum of $k$ circles?For the space $X = T # T$ find its fundamental group and prove the universal cover is contractible.Deck transformation group isomorphic to $mathbbZ/mathbb6Z$
$begingroup$
So i'm thinking that this will be a cover space of maximum possible symmetry. Will a "necklace" of 6 circles work? Any tips appreciated
algebraic-topology geometric-topology
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
$endgroup$
|
show 1 more comment
$begingroup$
So i'm thinking that this will be a cover space of maximum possible symmetry. Will a "necklace" of 6 circles work? Any tips appreciated
algebraic-topology geometric-topology
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
$endgroup$
3
$begingroup$
I don't know the answer immediately, but I don't think it's a necklace of $6$ circles. I think it's much bigger. The point is that the deck group acts freely, so the inverse image of a point must contain more than $6$ elements. In fact, by the orbit-stabilizer theorem (since the stabilizer must be trivial), the orbit must contain $|S_6| = 6! = 720$ elements. So you are looking for at least a 720-sheeted covering....(or I made a mistake. I usually just think about the deck group associated to the universal covering...)
$endgroup$
– Jason DeVito
Mar 26 at 14:27
2
$begingroup$
Supposing the cover is connected, it is impossible to have a 6 sheeted cover like that because the number of elements in the preimage is the index of the fundamental group under the induced map (call the image G). You might know that the deck transformations of a covering are isomorphic to the normalizer of G quotient G. This implies that index is at least 720.
$endgroup$
– Connor Malin
Mar 26 at 14:56
1
$begingroup$
Probably a better explanation (but in the same vein) is that a deck transformation is determined by where it sends a single point, so an n sheeted cover has at most n deck transformations.
$endgroup$
– Connor Malin
Mar 26 at 15:03
1
$begingroup$
It is a consequence of the lifting properties for covers. I think Hatcher proves it.
$endgroup$
– Connor Malin
Mar 26 at 15:22
3
$begingroup$
Your question doesn't mention anything about connectivity. If your cover is not required to be connected then you can take $6$ copies of $S^1 vee S^1$ and the deck transformation group just permutes the copies.
$endgroup$
– William
Mar 26 at 15:27
|
show 1 more comment
$begingroup$
So i'm thinking that this will be a cover space of maximum possible symmetry. Will a "necklace" of 6 circles work? Any tips appreciated
algebraic-topology geometric-topology
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
$endgroup$
So i'm thinking that this will be a cover space of maximum possible symmetry. Will a "necklace" of 6 circles work? Any tips appreciated
algebraic-topology geometric-topology
algebraic-topology geometric-topology
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
asked Mar 26 at 14:09
Mathematical MushroomMathematical Mushroom
1778
1778
3
$begingroup$
I don't know the answer immediately, but I don't think it's a necklace of $6$ circles. I think it's much bigger. The point is that the deck group acts freely, so the inverse image of a point must contain more than $6$ elements. In fact, by the orbit-stabilizer theorem (since the stabilizer must be trivial), the orbit must contain $|S_6| = 6! = 720$ elements. So you are looking for at least a 720-sheeted covering....(or I made a mistake. I usually just think about the deck group associated to the universal covering...)
$endgroup$
– Jason DeVito
Mar 26 at 14:27
2
$begingroup$
Supposing the cover is connected, it is impossible to have a 6 sheeted cover like that because the number of elements in the preimage is the index of the fundamental group under the induced map (call the image G). You might know that the deck transformations of a covering are isomorphic to the normalizer of G quotient G. This implies that index is at least 720.
$endgroup$
– Connor Malin
Mar 26 at 14:56
1
$begingroup$
Probably a better explanation (but in the same vein) is that a deck transformation is determined by where it sends a single point, so an n sheeted cover has at most n deck transformations.
$endgroup$
– Connor Malin
Mar 26 at 15:03
1
$begingroup$
It is a consequence of the lifting properties for covers. I think Hatcher proves it.
$endgroup$
– Connor Malin
Mar 26 at 15:22
3
$begingroup$
Your question doesn't mention anything about connectivity. If your cover is not required to be connected then you can take $6$ copies of $S^1 vee S^1$ and the deck transformation group just permutes the copies.
$endgroup$
– William
Mar 26 at 15:27
|
show 1 more comment
3
$begingroup$
I don't know the answer immediately, but I don't think it's a necklace of $6$ circles. I think it's much bigger. The point is that the deck group acts freely, so the inverse image of a point must contain more than $6$ elements. In fact, by the orbit-stabilizer theorem (since the stabilizer must be trivial), the orbit must contain $|S_6| = 6! = 720$ elements. So you are looking for at least a 720-sheeted covering....(or I made a mistake. I usually just think about the deck group associated to the universal covering...)
$endgroup$
– Jason DeVito
Mar 26 at 14:27
2
$begingroup$
Supposing the cover is connected, it is impossible to have a 6 sheeted cover like that because the number of elements in the preimage is the index of the fundamental group under the induced map (call the image G). You might know that the deck transformations of a covering are isomorphic to the normalizer of G quotient G. This implies that index is at least 720.
$endgroup$
– Connor Malin
Mar 26 at 14:56
1
$begingroup$
Probably a better explanation (but in the same vein) is that a deck transformation is determined by where it sends a single point, so an n sheeted cover has at most n deck transformations.
$endgroup$
– Connor Malin
Mar 26 at 15:03
1
$begingroup$
It is a consequence of the lifting properties for covers. I think Hatcher proves it.
$endgroup$
– Connor Malin
Mar 26 at 15:22
3
$begingroup$
Your question doesn't mention anything about connectivity. If your cover is not required to be connected then you can take $6$ copies of $S^1 vee S^1$ and the deck transformation group just permutes the copies.
$endgroup$
– William
Mar 26 at 15:27
3
3
$begingroup$
I don't know the answer immediately, but I don't think it's a necklace of $6$ circles. I think it's much bigger. The point is that the deck group acts freely, so the inverse image of a point must contain more than $6$ elements. In fact, by the orbit-stabilizer theorem (since the stabilizer must be trivial), the orbit must contain $|S_6| = 6! = 720$ elements. So you are looking for at least a 720-sheeted covering....(or I made a mistake. I usually just think about the deck group associated to the universal covering...)
$endgroup$
– Jason DeVito
Mar 26 at 14:27
$begingroup$
I don't know the answer immediately, but I don't think it's a necklace of $6$ circles. I think it's much bigger. The point is that the deck group acts freely, so the inverse image of a point must contain more than $6$ elements. In fact, by the orbit-stabilizer theorem (since the stabilizer must be trivial), the orbit must contain $|S_6| = 6! = 720$ elements. So you are looking for at least a 720-sheeted covering....(or I made a mistake. I usually just think about the deck group associated to the universal covering...)
$endgroup$
– Jason DeVito
Mar 26 at 14:27
2
2
$begingroup$
Supposing the cover is connected, it is impossible to have a 6 sheeted cover like that because the number of elements in the preimage is the index of the fundamental group under the induced map (call the image G). You might know that the deck transformations of a covering are isomorphic to the normalizer of G quotient G. This implies that index is at least 720.
$endgroup$
– Connor Malin
Mar 26 at 14:56
$begingroup$
Supposing the cover is connected, it is impossible to have a 6 sheeted cover like that because the number of elements in the preimage is the index of the fundamental group under the induced map (call the image G). You might know that the deck transformations of a covering are isomorphic to the normalizer of G quotient G. This implies that index is at least 720.
$endgroup$
– Connor Malin
Mar 26 at 14:56
1
1
$begingroup$
Probably a better explanation (but in the same vein) is that a deck transformation is determined by where it sends a single point, so an n sheeted cover has at most n deck transformations.
$endgroup$
– Connor Malin
Mar 26 at 15:03
$begingroup$
Probably a better explanation (but in the same vein) is that a deck transformation is determined by where it sends a single point, so an n sheeted cover has at most n deck transformations.
$endgroup$
– Connor Malin
Mar 26 at 15:03
1
1
$begingroup$
It is a consequence of the lifting properties for covers. I think Hatcher proves it.
$endgroup$
– Connor Malin
Mar 26 at 15:22
$begingroup$
It is a consequence of the lifting properties for covers. I think Hatcher proves it.
$endgroup$
– Connor Malin
Mar 26 at 15:22
3
3
$begingroup$
Your question doesn't mention anything about connectivity. If your cover is not required to be connected then you can take $6$ copies of $S^1 vee S^1$ and the deck transformation group just permutes the copies.
$endgroup$
– William
Mar 26 at 15:27
$begingroup$
Your question doesn't mention anything about connectivity. If your cover is not required to be connected then you can take $6$ copies of $S^1 vee S^1$ and the deck transformation group just permutes the copies.
$endgroup$
– William
Mar 26 at 15:27
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The comments are correct that this construction is impossible, however the reason is flawed.
First of all, the action of the deck transformation group $G$ of a connected covering map $f : X to Y$ a free action, meaning that the action of a nontrivial element of $G$ has no fixed points --- so no, a deck transformation cannot take $y_2$ to $y_3$ whilst fixing $y_1$.
Second, given $x in X$ and $y=f(x) in Y$, the action of $G$ on the subset $f^-1(y) subset X$ need not be transitive. In fact, that action is transitive if and only if the covering map $f : X to Y$ is regular, if and only if the image of the induced homomorphism $pi_1(X,x) to pi_1(Y,f(x))$ is a normal subgroup of $pi_1(Y,f(x))$. However, this only makes the construction more impossible (if that means anything), because it implies that the cardinality of $f^-1(x)$ is greater than or equal to the order of the group $G$.
So, in a covering map of degree $6$, the set $f^-1(x)$ has cardinality $6$, and there is no free action of a group of order $6! = 120$ on a set of cardinality $6$.
For proofs of the various things discussed in this answer, I suggest sitting down with a good solid course on the relationship between fundamental groups and covering spaces, as in Hatcher.
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
$endgroup$
add a comment |
$begingroup$
There is a $6!$-fold connected normal cover of $S^1vee S^1$ with deck transformation group $S_6$. Recall that $S_6$ has a presentation with two generators, hence there is a normal subgroup $Nsubset F_2$ of the free group on two generators. Let's identify $F_2=pi_1(S^1vee S^1)$. Hence, there is a homomorphism $f:pi_1(S^1vee S^1)to S_6$ through the quotient map $F_2to F_2/N$.
By standard facts about covering spaces (as seen in Hatcher) there is then a connected normal cover of $S^1vee S^1$ associated to the normal subgroup $N$ for which $S_6$ is the group of deck transformations.
You might know this covering space as the Cayley graph of $S_6$ with the chosen generating set. In fact, the two-generator Cayley graphs of $S_6$ encompass the set of all connected normal covers of $S^1vee S^1$ with deck transformation group isomorphic to $S_6$.
(William in the comments points out that there is a $6$-fold disconnected cover of $S^1vee S^1$ with $S_6$ as its group of deck transformations: the disjoint union $coprod_i=1^6S^1vee S^1$.)
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The comments are correct that this construction is impossible, however the reason is flawed.
First of all, the action of the deck transformation group $G$ of a connected covering map $f : X to Y$ a free action, meaning that the action of a nontrivial element of $G$ has no fixed points --- so no, a deck transformation cannot take $y_2$ to $y_3$ whilst fixing $y_1$.
Second, given $x in X$ and $y=f(x) in Y$, the action of $G$ on the subset $f^-1(y) subset X$ need not be transitive. In fact, that action is transitive if and only if the covering map $f : X to Y$ is regular, if and only if the image of the induced homomorphism $pi_1(X,x) to pi_1(Y,f(x))$ is a normal subgroup of $pi_1(Y,f(x))$. However, this only makes the construction more impossible (if that means anything), because it implies that the cardinality of $f^-1(x)$ is greater than or equal to the order of the group $G$.
So, in a covering map of degree $6$, the set $f^-1(x)$ has cardinality $6$, and there is no free action of a group of order $6! = 120$ on a set of cardinality $6$.
For proofs of the various things discussed in this answer, I suggest sitting down with a good solid course on the relationship between fundamental groups and covering spaces, as in Hatcher.
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
$endgroup$
add a comment |
$begingroup$
The comments are correct that this construction is impossible, however the reason is flawed.
First of all, the action of the deck transformation group $G$ of a connected covering map $f : X to Y$ a free action, meaning that the action of a nontrivial element of $G$ has no fixed points --- so no, a deck transformation cannot take $y_2$ to $y_3$ whilst fixing $y_1$.
Second, given $x in X$ and $y=f(x) in Y$, the action of $G$ on the subset $f^-1(y) subset X$ need not be transitive. In fact, that action is transitive if and only if the covering map $f : X to Y$ is regular, if and only if the image of the induced homomorphism $pi_1(X,x) to pi_1(Y,f(x))$ is a normal subgroup of $pi_1(Y,f(x))$. However, this only makes the construction more impossible (if that means anything), because it implies that the cardinality of $f^-1(x)$ is greater than or equal to the order of the group $G$.
So, in a covering map of degree $6$, the set $f^-1(x)$ has cardinality $6$, and there is no free action of a group of order $6! = 120$ on a set of cardinality $6$.
For proofs of the various things discussed in this answer, I suggest sitting down with a good solid course on the relationship between fundamental groups and covering spaces, as in Hatcher.
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
$endgroup$
add a comment |
$begingroup$
The comments are correct that this construction is impossible, however the reason is flawed.
First of all, the action of the deck transformation group $G$ of a connected covering map $f : X to Y$ a free action, meaning that the action of a nontrivial element of $G$ has no fixed points --- so no, a deck transformation cannot take $y_2$ to $y_3$ whilst fixing $y_1$.
Second, given $x in X$ and $y=f(x) in Y$, the action of $G$ on the subset $f^-1(y) subset X$ need not be transitive. In fact, that action is transitive if and only if the covering map $f : X to Y$ is regular, if and only if the image of the induced homomorphism $pi_1(X,x) to pi_1(Y,f(x))$ is a normal subgroup of $pi_1(Y,f(x))$. However, this only makes the construction more impossible (if that means anything), because it implies that the cardinality of $f^-1(x)$ is greater than or equal to the order of the group $G$.
So, in a covering map of degree $6$, the set $f^-1(x)$ has cardinality $6$, and there is no free action of a group of order $6! = 120$ on a set of cardinality $6$.
For proofs of the various things discussed in this answer, I suggest sitting down with a good solid course on the relationship between fundamental groups and covering spaces, as in Hatcher.
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
$endgroup$
The comments are correct that this construction is impossible, however the reason is flawed.
First of all, the action of the deck transformation group $G$ of a connected covering map $f : X to Y$ a free action, meaning that the action of a nontrivial element of $G$ has no fixed points --- so no, a deck transformation cannot take $y_2$ to $y_3$ whilst fixing $y_1$.
Second, given $x in X$ and $y=f(x) in Y$, the action of $G$ on the subset $f^-1(y) subset X$ need not be transitive. In fact, that action is transitive if and only if the covering map $f : X to Y$ is regular, if and only if the image of the induced homomorphism $pi_1(X,x) to pi_1(Y,f(x))$ is a normal subgroup of $pi_1(Y,f(x))$. However, this only makes the construction more impossible (if that means anything), because it implies that the cardinality of $f^-1(x)$ is greater than or equal to the order of the group $G$.
So, in a covering map of degree $6$, the set $f^-1(x)$ has cardinality $6$, and there is no free action of a group of order $6! = 120$ on a set of cardinality $6$.
For proofs of the various things discussed in this answer, I suggest sitting down with a good solid course on the relationship between fundamental groups and covering spaces, as in Hatcher.
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
answered Apr 2 at 0:11
Lee MosherLee Mosher
52.5k33891
52.5k33891
add a comment |
add a comment |
$begingroup$
There is a $6!$-fold connected normal cover of $S^1vee S^1$ with deck transformation group $S_6$. Recall that $S_6$ has a presentation with two generators, hence there is a normal subgroup $Nsubset F_2$ of the free group on two generators. Let's identify $F_2=pi_1(S^1vee S^1)$. Hence, there is a homomorphism $f:pi_1(S^1vee S^1)to S_6$ through the quotient map $F_2to F_2/N$.
By standard facts about covering spaces (as seen in Hatcher) there is then a connected normal cover of $S^1vee S^1$ associated to the normal subgroup $N$ for which $S_6$ is the group of deck transformations.
You might know this covering space as the Cayley graph of $S_6$ with the chosen generating set. In fact, the two-generator Cayley graphs of $S_6$ encompass the set of all connected normal covers of $S^1vee S^1$ with deck transformation group isomorphic to $S_6$.
(William in the comments points out that there is a $6$-fold disconnected cover of $S^1vee S^1$ with $S_6$ as its group of deck transformations: the disjoint union $coprod_i=1^6S^1vee S^1$.)
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
$endgroup$
add a comment |
$begingroup$
There is a $6!$-fold connected normal cover of $S^1vee S^1$ with deck transformation group $S_6$. Recall that $S_6$ has a presentation with two generators, hence there is a normal subgroup $Nsubset F_2$ of the free group on two generators. Let's identify $F_2=pi_1(S^1vee S^1)$. Hence, there is a homomorphism $f:pi_1(S^1vee S^1)to S_6$ through the quotient map $F_2to F_2/N$.
By standard facts about covering spaces (as seen in Hatcher) there is then a connected normal cover of $S^1vee S^1$ associated to the normal subgroup $N$ for which $S_6$ is the group of deck transformations.
You might know this covering space as the Cayley graph of $S_6$ with the chosen generating set. In fact, the two-generator Cayley graphs of $S_6$ encompass the set of all connected normal covers of $S^1vee S^1$ with deck transformation group isomorphic to $S_6$.
(William in the comments points out that there is a $6$-fold disconnected cover of $S^1vee S^1$ with $S_6$ as its group of deck transformations: the disjoint union $coprod_i=1^6S^1vee S^1$.)
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
$endgroup$
add a comment |
$begingroup$
There is a $6!$-fold connected normal cover of $S^1vee S^1$ with deck transformation group $S_6$. Recall that $S_6$ has a presentation with two generators, hence there is a normal subgroup $Nsubset F_2$ of the free group on two generators. Let's identify $F_2=pi_1(S^1vee S^1)$. Hence, there is a homomorphism $f:pi_1(S^1vee S^1)to S_6$ through the quotient map $F_2to F_2/N$.
By standard facts about covering spaces (as seen in Hatcher) there is then a connected normal cover of $S^1vee S^1$ associated to the normal subgroup $N$ for which $S_6$ is the group of deck transformations.
You might know this covering space as the Cayley graph of $S_6$ with the chosen generating set. In fact, the two-generator Cayley graphs of $S_6$ encompass the set of all connected normal covers of $S^1vee S^1$ with deck transformation group isomorphic to $S_6$.
(William in the comments points out that there is a $6$-fold disconnected cover of $S^1vee S^1$ with $S_6$ as its group of deck transformations: the disjoint union $coprod_i=1^6S^1vee S^1$.)
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
$endgroup$
There is a $6!$-fold connected normal cover of $S^1vee S^1$ with deck transformation group $S_6$. Recall that $S_6$ has a presentation with two generators, hence there is a normal subgroup $Nsubset F_2$ of the free group on two generators. Let's identify $F_2=pi_1(S^1vee S^1)$. Hence, there is a homomorphism $f:pi_1(S^1vee S^1)to S_6$ through the quotient map $F_2to F_2/N$.
By standard facts about covering spaces (as seen in Hatcher) there is then a connected normal cover of $S^1vee S^1$ associated to the normal subgroup $N$ for which $S_6$ is the group of deck transformations.
You might know this covering space as the Cayley graph of $S_6$ with the chosen generating set. In fact, the two-generator Cayley graphs of $S_6$ encompass the set of all connected normal covers of $S^1vee S^1$ with deck transformation group isomorphic to $S_6$.
(William in the comments points out that there is a $6$-fold disconnected cover of $S^1vee S^1$ with $S_6$ as its group of deck transformations: the disjoint union $coprod_i=1^6S^1vee S^1$.)
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
answered Apr 2 at 1:50
Kyle MillerKyle Miller
10.1k930
10.1k930
add a comment |
add a comment |
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3
$begingroup$
I don't know the answer immediately, but I don't think it's a necklace of $6$ circles. I think it's much bigger. The point is that the deck group acts freely, so the inverse image of a point must contain more than $6$ elements. In fact, by the orbit-stabilizer theorem (since the stabilizer must be trivial), the orbit must contain $|S_6| = 6! = 720$ elements. So you are looking for at least a 720-sheeted covering....(or I made a mistake. I usually just think about the deck group associated to the universal covering...)
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– Jason DeVito
Mar 26 at 14:27
2
$begingroup$
Supposing the cover is connected, it is impossible to have a 6 sheeted cover like that because the number of elements in the preimage is the index of the fundamental group under the induced map (call the image G). You might know that the deck transformations of a covering are isomorphic to the normalizer of G quotient G. This implies that index is at least 720.
$endgroup$
– Connor Malin
Mar 26 at 14:56
1
$begingroup$
Probably a better explanation (but in the same vein) is that a deck transformation is determined by where it sends a single point, so an n sheeted cover has at most n deck transformations.
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– Connor Malin
Mar 26 at 15:03
1
$begingroup$
It is a consequence of the lifting properties for covers. I think Hatcher proves it.
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– Connor Malin
Mar 26 at 15:22
3
$begingroup$
Your question doesn't mention anything about connectivity. If your cover is not required to be connected then you can take $6$ copies of $S^1 vee S^1$ and the deck transformation group just permutes the copies.
$endgroup$
– William
Mar 26 at 15:27