Any group can be represented as the fundamental group of a 2-dimensional topological space
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I saw in the textbook affirmation that any group can be represented as the fundamental group of a $2$-dimensional topological space. Without proof. May be you can give some ideas, how can I proof that or where I can find the proof.
I know, that any group can be represent like quatient group of free group. And I know that the number of elements of generating set of group -- number of "holes" in space. But why only $2$-dimensional space in this affirmation?
group-theory algebraic-topology free-groups
asked 14 hours ago
Arsenii
975
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up vote
1
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favorite
I saw in the textbook affirmation that any group can be represented as the fundamental group of a $2$-dimensional topological space. Without proof. May be you can give some ideas, how can I proof that or where I can find the proof.
I know, that any group can be represent like quatient group of free group. And I know that the number of elements of generating set of group -- number of "holes" in space. But why only $2$-dimensional space in this affirmation?
group-theory algebraic-topology free-groups
asked 14 hours ago
Arsenii
975
Could you clarify which notion of "dimension" is used here? Do you mean a $2$-manifold, or a CW-complex with no cells of dimension higher than $2$, or something else?
– Tobias Kildetoft
14 hours ago
@TobiasKildetoft As I understood it, in the textbook was dealt with the second case
– Arsenii
14 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I saw in the textbook affirmation that any group can be represented as the fundamental group of a $2$-dimensional topological space. Without proof. May be you can give some ideas, how can I proof that or where I can find the proof.
I know, that any group can be represent like quatient group of free group. And I know that the number of elements of generating set of group -- number of "holes" in space. But why only $2$-dimensional space in this affirmation?
group-theory algebraic-topology free-groups
asked 14 hours ago
Arsenii
975
I saw in the textbook affirmation that any group can be represented as the fundamental group of a $2$-dimensional topological space. Without proof. May be you can give some ideas, how can I proof that or where I can find the proof.
I know, that any group can be represent like quatient group of free group. And I know that the number of elements of generating set of group -- number of "holes" in space. But why only $2$-dimensional space in this affirmation?
group-theory algebraic-topology free-groups
group-theory algebraic-topology free-groups
asked 14 hours ago
Arsenii
975
asked 14 hours ago
Arsenii
975
asked 14 hours ago
Arsenii
975
asked 14 hours ago
Arsenii
975
asked 14 hours ago
Arsenii
975
975
Could you clarify which notion of "dimension" is used here? Do you mean a $2$-manifold, or a CW-complex with no cells of dimension higher than $2$, or something else?
– Tobias Kildetoft
14 hours ago
@TobiasKildetoft As I understood it, in the textbook was dealt with the second case
– Arsenii
14 hours ago
add a comment |
Could you clarify which notion of "dimension" is used here? Do you mean a $2$-manifold, or a CW-complex with no cells of dimension higher than $2$, or something else?
– Tobias Kildetoft
14 hours ago
@TobiasKildetoft As I understood it, in the textbook was dealt with the second case
– Arsenii
14 hours ago
Could you clarify which notion of "dimension" is used here? Do you mean a $2$-manifold, or a CW-complex with no cells of dimension higher than $2$, or something else?
– Tobias Kildetoft
14 hours ago
Could you clarify which notion of "dimension" is used here? Do you mean a $2$-manifold, or a CW-complex with no cells of dimension higher than $2$, or something else?
– Tobias Kildetoft
14 hours ago
@TobiasKildetoft As I understood it, in the textbook was dealt with the second case
– Arsenii
14 hours ago
@TobiasKildetoft As I understood it, in the textbook was dealt with the second case
– Arsenii
14 hours ago
add a comment |
1 Answer
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Take a presentation $G = langle S mid R rangle$, i.e. $G$ is the quotient of the free group $langle S rangle$ generated by $S$, modulo the relations $R subset langle S rangle$. You can consider first the wedge sum of $S$-many circles, $X_1 = bigvee^S S^1$. Its fundamental group is $langle S rangle$ (immediate application of van Kampen's theorem). Then you attach a $2$-cell for each relation in $R$, along a path that represents the given element of $langle S rangle = pi_1(X_1)$. In this way you obtain a CW-complex $X = X_2$ of dimension $2$, and its fundamental group is $G$ by van Kampen's theorem (again).
answered 13 hours ago
Najib Idrissi
40.4k469136
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Take a presentation $G = langle S mid R rangle$, i.e. $G$ is the quotient of the free group $langle S rangle$ generated by $S$, modulo the relations $R subset langle S rangle$. You can consider first the wedge sum of $S$-many circles, $X_1 = bigvee^S S^1$. Its fundamental group is $langle S rangle$ (immediate application of van Kampen's theorem). Then you attach a $2$-cell for each relation in $R$, along a path that represents the given element of $langle S rangle = pi_1(X_1)$. In this way you obtain a CW-complex $X = X_2$ of dimension $2$, and its fundamental group is $G$ by van Kampen's theorem (again).
answered 13 hours ago
Najib Idrissi
40.4k469136
add a comment |
up vote
4
down vote
accepted
Take a presentation $G = langle S mid R rangle$, i.e. $G$ is the quotient of the free group $langle S rangle$ generated by $S$, modulo the relations $R subset langle S rangle$. You can consider first the wedge sum of $S$-many circles, $X_1 = bigvee^S S^1$. Its fundamental group is $langle S rangle$ (immediate application of van Kampen's theorem). Then you attach a $2$-cell for each relation in $R$, along a path that represents the given element of $langle S rangle = pi_1(X_1)$. In this way you obtain a CW-complex $X = X_2$ of dimension $2$, and its fundamental group is $G$ by van Kampen's theorem (again).
answered 13 hours ago
Najib Idrissi
40.4k469136
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Take a presentation $G = langle S mid R rangle$, i.e. $G$ is the quotient of the free group $langle S rangle$ generated by $S$, modulo the relations $R subset langle S rangle$. You can consider first the wedge sum of $S$-many circles, $X_1 = bigvee^S S^1$. Its fundamental group is $langle S rangle$ (immediate application of van Kampen's theorem). Then you attach a $2$-cell for each relation in $R$, along a path that represents the given element of $langle S rangle = pi_1(X_1)$. In this way you obtain a CW-complex $X = X_2$ of dimension $2$, and its fundamental group is $G$ by van Kampen's theorem (again).
answered 13 hours ago
Najib Idrissi
40.4k469136
Take a presentation $G = langle S mid R rangle$, i.e. $G$ is the quotient of the free group $langle S rangle$ generated by $S$, modulo the relations $R subset langle S rangle$. You can consider first the wedge sum of $S$-many circles, $X_1 = bigvee^S S^1$. Its fundamental group is $langle S rangle$ (immediate application of van Kampen's theorem). Then you attach a $2$-cell for each relation in $R$, along a path that represents the given element of $langle S rangle = pi_1(X_1)$. In this way you obtain a CW-complex $X = X_2$ of dimension $2$, and its fundamental group is $G$ by van Kampen's theorem (again).
answered 13 hours ago
Najib Idrissi
40.4k469136
answered 13 hours ago
Najib Idrissi
40.4k469136
answered 13 hours ago
Najib Idrissi
40.4k469136
answered 13 hours ago
Najib Idrissi
40.4k469136
40.4k469136
add a comment |
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Could you clarify which notion of "dimension" is used here? Do you mean a $2$-manifold, or a CW-complex with no cells of dimension higher than $2$, or something else?
– Tobias Kildetoft
14 hours ago
@TobiasKildetoft As I understood it, in the textbook was dealt with the second case
– Arsenii
14 hours ago