Equivalence of surjections from a surface group to a free group
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Let $g geq 2$. Let $S = langle a_1,b_2,...,a_g,b_g | [a_1,b_1] cdots [a_g,b_g] rangle$ be the fundamental group of a genus $g$ surface and let $F_g$ be a free group with $g$ generators. Given two surjections $f_1,f_2 : S to F_g$ is there a way to determine if there are automophisms $phi: S to S$ and $psi: F_g to F_g$ so that $f_1 = phi circ f_2 circ psi$?
Is there an example of two surjections $f_1,f_2$ that are not equivalent in the above way?
I asked the question on MSE before but didn't get much.
gr.group-theory gt.geometric-topology free-groups
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Let $g geq 2$. Let $S = langle a_1,b_2,...,a_g,b_g | [a_1,b_1] cdots [a_g,b_g] rangle$ be the fundamental group of a genus $g$ surface and let $F_g$ be a free group with $g$ generators. Given two surjections $f_1,f_2 : S to F_g$ is there a way to determine if there are automophisms $phi: S to S$ and $psi: F_g to F_g$ so that $f_1 = phi circ f_2 circ psi$?
Is there an example of two surjections $f_1,f_2$ that are not equivalent in the above way?
I asked the question on MSE before but didn't get much.
gr.group-theory gt.geometric-topology free-groups
A naive question: is it clear such a surjection exists?
â PseudoNeo
3 hours ago
1
@PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles.
â Mike Miller
3 hours ago
Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_2g$) and was very confused.
â PseudoNeo
3 hours ago
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Let $g geq 2$. Let $S = langle a_1,b_2,...,a_g,b_g | [a_1,b_1] cdots [a_g,b_g] rangle$ be the fundamental group of a genus $g$ surface and let $F_g$ be a free group with $g$ generators. Given two surjections $f_1,f_2 : S to F_g$ is there a way to determine if there are automophisms $phi: S to S$ and $psi: F_g to F_g$ so that $f_1 = phi circ f_2 circ psi$?
Is there an example of two surjections $f_1,f_2$ that are not equivalent in the above way?
I asked the question on MSE before but didn't get much.
gr.group-theory gt.geometric-topology free-groups
Let $g geq 2$. Let $S = langle a_1,b_2,...,a_g,b_g | [a_1,b_1] cdots [a_g,b_g] rangle$ be the fundamental group of a genus $g$ surface and let $F_g$ be a free group with $g$ generators. Given two surjections $f_1,f_2 : S to F_g$ is there a way to determine if there are automophisms $phi: S to S$ and $psi: F_g to F_g$ so that $f_1 = phi circ f_2 circ psi$?
Is there an example of two surjections $f_1,f_2$ that are not equivalent in the above way?
I asked the question on MSE before but didn't get much.
gr.group-theory gt.geometric-topology free-groups
gr.group-theory gt.geometric-topology free-groups
asked 5 hours ago
user101010
892210
892210
A naive question: is it clear such a surjection exists?
â PseudoNeo
3 hours ago
1
@PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles.
â Mike Miller
3 hours ago
Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_2g$) and was very confused.
â PseudoNeo
3 hours ago
add a comment |Â
A naive question: is it clear such a surjection exists?
â PseudoNeo
3 hours ago
1
@PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles.
â Mike Miller
3 hours ago
Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_2g$) and was very confused.
â PseudoNeo
3 hours ago
A naive question: is it clear such a surjection exists?
â PseudoNeo
3 hours ago
A naive question: is it clear such a surjection exists?
â PseudoNeo
3 hours ago
1
1
@PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles.
â Mike Miller
3 hours ago
@PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles.
â Mike Miller
3 hours ago
Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_2g$) and was very confused.
â PseudoNeo
3 hours ago
Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_2g$) and was very confused.
â PseudoNeo
3 hours ago
add a comment |Â
1 Answer
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This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $g$ handlebody.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $g$ handlebody.
add a comment |Â
up vote
3
down vote
This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $g$ handlebody.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $g$ handlebody.
This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $g$ handlebody.
answered 16 mins ago
Jean Raimbault
1,683819
1,683819
add a comment |Â
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A naive question: is it clear such a surjection exists?
â PseudoNeo
3 hours ago
1
@PseudoNeo Algebraically, yes: kill all of the $b_i$. Geometrically, yes: the surface is the boundary of a handlebody, equivalent to a wedge of $g$ circles.
â Mike Miller
3 hours ago
Oh, thank you, I was misreading the question (I mixed up $F_g$ and $F_2g$) and was very confused.
â PseudoNeo
3 hours ago