Is there a general way to calculate the fundamental group of a quotient space?
Suppose $X$ is a path-connected topological space, and $A$ is a path-connected subset of $X$. My question is, is there a way to calculate the fundamental group of the quotient space $X / A$ in terms of the fundamental groups of $X$ or $A$?
If there isn't a method in all cases, can you at least do it if $A$ has some special property?
general-topology algebraic-topology fundamental-groups quotient-spaces quotient-group
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
add a comment |
Suppose $X$ is a path-connected topological space, and $A$ is a path-connected subset of $X$. My question is, is there a way to calculate the fundamental group of the quotient space $X / A$ in terms of the fundamental groups of $X$ or $A$?
If there isn't a method in all cases, can you at least do it if $A$ has some special property?
general-topology algebraic-topology fundamental-groups quotient-spaces quotient-group
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
2
If the inclusion $i:Ahookrightarrow X$ is a cofibration then you get a cofibration sequence $Axrightarrow{i} Xrightarrow X/A$. You can use homology to study the connectivity of the map $i$, and in special cases says things about the low-dimensional homotopy groups of $X/A$.
– Tyrone
Nov 29 '18 at 23:21
add a comment |
Suppose $X$ is a path-connected topological space, and $A$ is a path-connected subset of $X$. My question is, is there a way to calculate the fundamental group of the quotient space $X / A$ in terms of the fundamental groups of $X$ or $A$?
If there isn't a method in all cases, can you at least do it if $A$ has some special property?
general-topology algebraic-topology fundamental-groups quotient-spaces quotient-group
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
Suppose $X$ is a path-connected topological space, and $A$ is a path-connected subset of $X$. My question is, is there a way to calculate the fundamental group of the quotient space $X / A$ in terms of the fundamental groups of $X$ or $A$?
If there isn't a method in all cases, can you at least do it if $A$ has some special property?
general-topology algebraic-topology fundamental-groups quotient-spaces quotient-group
general-topology algebraic-topology fundamental-groups quotient-spaces quotient-group
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
asked Nov 29 '18 at 20:49
Keshav SrinivasanKeshav Srinivasan
2,10111441
2,10111441
2
If the inclusion $i:Ahookrightarrow X$ is a cofibration then you get a cofibration sequence $Axrightarrow{i} Xrightarrow X/A$. You can use homology to study the connectivity of the map $i$, and in special cases says things about the low-dimensional homotopy groups of $X/A$.
– Tyrone
Nov 29 '18 at 23:21
add a comment |
2
If the inclusion $i:Ahookrightarrow X$ is a cofibration then you get a cofibration sequence $Axrightarrow{i} Xrightarrow X/A$. You can use homology to study the connectivity of the map $i$, and in special cases says things about the low-dimensional homotopy groups of $X/A$.
– Tyrone
Nov 29 '18 at 23:21
2
2
If the inclusion $i:Ahookrightarrow X$ is a cofibration then you get a cofibration sequence $Axrightarrow{i} Xrightarrow X/A$. You can use homology to study the connectivity of the map $i$, and in special cases says things about the low-dimensional homotopy groups of $X/A$.
– Tyrone
Nov 29 '18 at 23:21
If the inclusion $i:Ahookrightarrow X$ is a cofibration then you get a cofibration sequence $Axrightarrow{i} Xrightarrow X/A$. You can use homology to study the connectivity of the map $i$, and in special cases says things about the low-dimensional homotopy groups of $X/A$.
– Tyrone
Nov 29 '18 at 23:21
add a comment |
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I do not think that one can say anything about $X/A$. Consider $X = D^2 vee S^1$ and $A_1$ = boundary of $D^2$, $A_2$ = second component of $D^2 vee S^1$. Then $A_1$ and $A_2$ are homeomorphic. They are nicely embedded into $X$. In fact they are subpolyhedra, hence the inclusions $A_i to X$ are cofibrations. But $X/A_1 approx S^2 vee S^1$ has fundamental group $mathbb{Z}$ and $X/A_2 approx D^2$ has fundamental group $0$.
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
add a comment |
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1 Answer
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I do not think that one can say anything about $X/A$. Consider $X = D^2 vee S^1$ and $A_1$ = boundary of $D^2$, $A_2$ = second component of $D^2 vee S^1$. Then $A_1$ and $A_2$ are homeomorphic. They are nicely embedded into $X$. In fact they are subpolyhedra, hence the inclusions $A_i to X$ are cofibrations. But $X/A_1 approx S^2 vee S^1$ has fundamental group $mathbb{Z}$ and $X/A_2 approx D^2$ has fundamental group $0$.
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
add a comment |
I do not think that one can say anything about $X/A$. Consider $X = D^2 vee S^1$ and $A_1$ = boundary of $D^2$, $A_2$ = second component of $D^2 vee S^1$. Then $A_1$ and $A_2$ are homeomorphic. They are nicely embedded into $X$. In fact they are subpolyhedra, hence the inclusions $A_i to X$ are cofibrations. But $X/A_1 approx S^2 vee S^1$ has fundamental group $mathbb{Z}$ and $X/A_2 approx D^2$ has fundamental group $0$.
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
add a comment |
I do not think that one can say anything about $X/A$. Consider $X = D^2 vee S^1$ and $A_1$ = boundary of $D^2$, $A_2$ = second component of $D^2 vee S^1$. Then $A_1$ and $A_2$ are homeomorphic. They are nicely embedded into $X$. In fact they are subpolyhedra, hence the inclusions $A_i to X$ are cofibrations. But $X/A_1 approx S^2 vee S^1$ has fundamental group $mathbb{Z}$ and $X/A_2 approx D^2$ has fundamental group $0$.
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
I do not think that one can say anything about $X/A$. Consider $X = D^2 vee S^1$ and $A_1$ = boundary of $D^2$, $A_2$ = second component of $D^2 vee S^1$. Then $A_1$ and $A_2$ are homeomorphic. They are nicely embedded into $X$. In fact they are subpolyhedra, hence the inclusions $A_i to X$ are cofibrations. But $X/A_1 approx S^2 vee S^1$ has fundamental group $mathbb{Z}$ and $X/A_2 approx D^2$ has fundamental group $0$.
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
answered Nov 30 '18 at 10:35
Paul FrostPaul Frost
9,7053732
9,7053732
add a comment |
add a comment |
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If the inclusion $i:Ahookrightarrow X$ is a cofibration then you get a cofibration sequence $Axrightarrow{i} Xrightarrow X/A$. You can use homology to study the connectivity of the map $i$, and in special cases says things about the low-dimensional homotopy groups of $X/A$.
– Tyrone
Nov 29 '18 at 23:21