Homology computation for $T^2$
$begingroup$
Set $X=T^2$. Denote $E$ as the square and $partial E$ as the boundary of disk. Set $Z$ the image of $partial E$ under quotient of $X$. I am going to abuse notation for $H_star$ to denote both homology and reduced homology.
The book has shown $(E,partial E)to (X,Z)$ induces homology level isomorphism. It suffices to compute $H_star(E,partial E)$. I am looking for the mistake in my thought process. Consider long exact sequence $H_star(partial E)to H_star(E)to H_star(E,partial E)$. Since $E$ is a square/contractible, then $H_star(E)=0$. So $H_{star+1}(partial E)cong H_star(E,partial E)$. This computation is clearly wrong as $H_1(X)$ should have 2 generators.
$textbf{Q:}$ What is my mistake?
general-topology algebraic-topology
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
$endgroup$
add a comment |
$begingroup$
Set $X=T^2$. Denote $E$ as the square and $partial E$ as the boundary of disk. Set $Z$ the image of $partial E$ under quotient of $X$. I am going to abuse notation for $H_star$ to denote both homology and reduced homology.
The book has shown $(E,partial E)to (X,Z)$ induces homology level isomorphism. It suffices to compute $H_star(E,partial E)$. I am looking for the mistake in my thought process. Consider long exact sequence $H_star(partial E)to H_star(E)to H_star(E,partial E)$. Since $E$ is a square/contractible, then $H_star(E)=0$. So $H_{star+1}(partial E)cong H_star(E,partial E)$. This computation is clearly wrong as $H_1(X)$ should have 2 generators.
$textbf{Q:}$ What is my mistake?
general-topology algebraic-topology
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
$endgroup$
add a comment |
$begingroup$
Set $X=T^2$. Denote $E$ as the square and $partial E$ as the boundary of disk. Set $Z$ the image of $partial E$ under quotient of $X$. I am going to abuse notation for $H_star$ to denote both homology and reduced homology.
The book has shown $(E,partial E)to (X,Z)$ induces homology level isomorphism. It suffices to compute $H_star(E,partial E)$. I am looking for the mistake in my thought process. Consider long exact sequence $H_star(partial E)to H_star(E)to H_star(E,partial E)$. Since $E$ is a square/contractible, then $H_star(E)=0$. So $H_{star+1}(partial E)cong H_star(E,partial E)$. This computation is clearly wrong as $H_1(X)$ should have 2 generators.
$textbf{Q:}$ What is my mistake?
general-topology algebraic-topology
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
$endgroup$
Set $X=T^2$. Denote $E$ as the square and $partial E$ as the boundary of disk. Set $Z$ the image of $partial E$ under quotient of $X$. I am going to abuse notation for $H_star$ to denote both homology and reduced homology.
The book has shown $(E,partial E)to (X,Z)$ induces homology level isomorphism. It suffices to compute $H_star(E,partial E)$. I am looking for the mistake in my thought process. Consider long exact sequence $H_star(partial E)to H_star(E)to H_star(E,partial E)$. Since $E$ is a square/contractible, then $H_star(E)=0$. So $H_{star+1}(partial E)cong H_star(E,partial E)$. This computation is clearly wrong as $H_1(X)$ should have 2 generators.
$textbf{Q:}$ What is my mistake?
general-topology algebraic-topology
general-topology algebraic-topology
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
asked Dec 5 '18 at 0:14
user45765user45765
2,6942722
2,6942722
add a comment |
add a comment |
1 Answer
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$begingroup$
The relative homology group $H_star(E, partial E)$ should be giving the homology of the quotient space formed by collapsing $partial E$ to a point. But this quotient is actually homeomorphic to the sphere $S^2$ rather than a torus. To get a torus you must instead identify opposite sides.
The homology isomorphism is giving the first homology group of $X/Z$, which is not a torus but rather a torus with two circles collapsed to a point.
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
$endgroup$
$begingroup$
Ah my dumbness. Thanks a lot.
$endgroup$
– user45765
Dec 5 '18 at 0:39
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The relative homology group $H_star(E, partial E)$ should be giving the homology of the quotient space formed by collapsing $partial E$ to a point. But this quotient is actually homeomorphic to the sphere $S^2$ rather than a torus. To get a torus you must instead identify opposite sides.
The homology isomorphism is giving the first homology group of $X/Z$, which is not a torus but rather a torus with two circles collapsed to a point.
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
$endgroup$
$begingroup$
Ah my dumbness. Thanks a lot.
$endgroup$
– user45765
Dec 5 '18 at 0:39
add a comment |
$begingroup$
The relative homology group $H_star(E, partial E)$ should be giving the homology of the quotient space formed by collapsing $partial E$ to a point. But this quotient is actually homeomorphic to the sphere $S^2$ rather than a torus. To get a torus you must instead identify opposite sides.
The homology isomorphism is giving the first homology group of $X/Z$, which is not a torus but rather a torus with two circles collapsed to a point.
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
$endgroup$
$begingroup$
Ah my dumbness. Thanks a lot.
$endgroup$
– user45765
Dec 5 '18 at 0:39
add a comment |
$begingroup$
The relative homology group $H_star(E, partial E)$ should be giving the homology of the quotient space formed by collapsing $partial E$ to a point. But this quotient is actually homeomorphic to the sphere $S^2$ rather than a torus. To get a torus you must instead identify opposite sides.
The homology isomorphism is giving the first homology group of $X/Z$, which is not a torus but rather a torus with two circles collapsed to a point.
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
$endgroup$
The relative homology group $H_star(E, partial E)$ should be giving the homology of the quotient space formed by collapsing $partial E$ to a point. But this quotient is actually homeomorphic to the sphere $S^2$ rather than a torus. To get a torus you must instead identify opposite sides.
The homology isomorphism is giving the first homology group of $X/Z$, which is not a torus but rather a torus with two circles collapsed to a point.
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
answered Dec 5 '18 at 0:35
Rolf HoyerRolf Hoyer
11.2k31629
11.2k31629
$begingroup$
Ah my dumbness. Thanks a lot.
$endgroup$
– user45765
Dec 5 '18 at 0:39
add a comment |
$begingroup$
Ah my dumbness. Thanks a lot.
$endgroup$
– user45765
Dec 5 '18 at 0:39
$begingroup$
Ah my dumbness. Thanks a lot.
$endgroup$
– user45765
Dec 5 '18 at 0:39
$begingroup$
Ah my dumbness. Thanks a lot.
$endgroup$
– user45765
Dec 5 '18 at 0:39
add a comment |
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