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
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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
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
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up vote
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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
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
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 |
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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).
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1 Answer
1
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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).
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).
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).
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
40.4k469136
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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