Chain complexes, homology and homotopy type.Two CW complexes with isomorphic homotopy groups and homology,...
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Chain complexes, homology and homotopy type.
Two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalentHomotopy equivalent chain complexesHomology of Chain Complexes from Free ResolutionOn chain homotopy equivalenceAbstract homotopy invariance of homologyWhy is $mathbb{Z}/p^2mathbb{Z}$ indecomposable in the homotopy category of chain complexesChain Homotopy in abelian categoryInvert a quasi isomorphism of chain complexesComputing the Chain Homotopy Group of two chain complexes of free abelian groupschain homotopy equivalence and quasi-isomorphism
$begingroup$
Are they examples of easy chain complexes...
that have the same homotopy type but are not isomorphic?
That have the same homology groups but haven't got the same homotopy type?
homological-algebra homotopy-theory
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
$endgroup$
add a comment |
$begingroup$
Are they examples of easy chain complexes...
that have the same homotopy type but are not isomorphic?
That have the same homology groups but haven't got the same homotopy type?
homological-algebra homotopy-theory
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
$endgroup$
add a comment |
$begingroup$
Are they examples of easy chain complexes...
that have the same homotopy type but are not isomorphic?
That have the same homology groups but haven't got the same homotopy type?
homological-algebra homotopy-theory
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
$endgroup$
Are they examples of easy chain complexes...
that have the same homotopy type but are not isomorphic?
That have the same homology groups but haven't got the same homotopy type?
homological-algebra homotopy-theory
homological-algebra homotopy-theory
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
asked Mar 9 at 17:36
roi_saumonroi_saumon
60138
60138
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
For two homotopic non-isomorphic chain complex, take any simplicial complex with a simplicial contraction, e.g. $EG$ for any finite group $G$. Then the extra degeneracy will give a chain homotopy from $C^{text{simp}}(EG, mathbb{Z})$ to the chain complex with $mathbb{Z}$ in degree $0$ and zero everywhere else. On the other hand $C^{text{simp}}_1(EG, mathbb{Z})=mathbb{Z}[G]$.
For homologous chain complexes which are not homotopic, note that homotopic chain complexes are still homotopic after tensoring with any abelian group. Then the chain complexes
$require{AMScd}
begin{CD}
ldots @>>> 0 @>>> mathbb{Z} @>times 2>> mathbb{Z} @>>> 0 @>>> ldots\
@. @| @VV0V @VVtext{mod }2V @| @.\
ldots @>>> 0 @>>> 0 @>>> mathbb{Z}/2mathbb{Z} @>>> 0 @>>> ldots
end{CD}$
are homologous, but after tensoring with $mathbb{Z}/2mathbb{Z}$ they are not. Hence they cannot be homotopic.
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
For two homotopic non-isomorphic chain complex, take any simplicial complex with a simplicial contraction, e.g. $EG$ for any finite group $G$. Then the extra degeneracy will give a chain homotopy from $C^{text{simp}}(EG, mathbb{Z})$ to the chain complex with $mathbb{Z}$ in degree $0$ and zero everywhere else. On the other hand $C^{text{simp}}_1(EG, mathbb{Z})=mathbb{Z}[G]$.
For homologous chain complexes which are not homotopic, note that homotopic chain complexes are still homotopic after tensoring with any abelian group. Then the chain complexes
$require{AMScd}
begin{CD}
ldots @>>> 0 @>>> mathbb{Z} @>times 2>> mathbb{Z} @>>> 0 @>>> ldots\
@. @| @VV0V @VVtext{mod }2V @| @.\
ldots @>>> 0 @>>> 0 @>>> mathbb{Z}/2mathbb{Z} @>>> 0 @>>> ldots
end{CD}$
are homologous, but after tensoring with $mathbb{Z}/2mathbb{Z}$ they are not. Hence they cannot be homotopic.
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
$endgroup$
add a comment |
$begingroup$
For two homotopic non-isomorphic chain complex, take any simplicial complex with a simplicial contraction, e.g. $EG$ for any finite group $G$. Then the extra degeneracy will give a chain homotopy from $C^{text{simp}}(EG, mathbb{Z})$ to the chain complex with $mathbb{Z}$ in degree $0$ and zero everywhere else. On the other hand $C^{text{simp}}_1(EG, mathbb{Z})=mathbb{Z}[G]$.
For homologous chain complexes which are not homotopic, note that homotopic chain complexes are still homotopic after tensoring with any abelian group. Then the chain complexes
$require{AMScd}
begin{CD}
ldots @>>> 0 @>>> mathbb{Z} @>times 2>> mathbb{Z} @>>> 0 @>>> ldots\
@. @| @VV0V @VVtext{mod }2V @| @.\
ldots @>>> 0 @>>> 0 @>>> mathbb{Z}/2mathbb{Z} @>>> 0 @>>> ldots
end{CD}$
are homologous, but after tensoring with $mathbb{Z}/2mathbb{Z}$ they are not. Hence they cannot be homotopic.
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
$endgroup$
add a comment |
$begingroup$
For two homotopic non-isomorphic chain complex, take any simplicial complex with a simplicial contraction, e.g. $EG$ for any finite group $G$. Then the extra degeneracy will give a chain homotopy from $C^{text{simp}}(EG, mathbb{Z})$ to the chain complex with $mathbb{Z}$ in degree $0$ and zero everywhere else. On the other hand $C^{text{simp}}_1(EG, mathbb{Z})=mathbb{Z}[G]$.
For homologous chain complexes which are not homotopic, note that homotopic chain complexes are still homotopic after tensoring with any abelian group. Then the chain complexes
$require{AMScd}
begin{CD}
ldots @>>> 0 @>>> mathbb{Z} @>times 2>> mathbb{Z} @>>> 0 @>>> ldots\
@. @| @VV0V @VVtext{mod }2V @| @.\
ldots @>>> 0 @>>> 0 @>>> mathbb{Z}/2mathbb{Z} @>>> 0 @>>> ldots
end{CD}$
are homologous, but after tensoring with $mathbb{Z}/2mathbb{Z}$ they are not. Hence they cannot be homotopic.
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
$endgroup$
For two homotopic non-isomorphic chain complex, take any simplicial complex with a simplicial contraction, e.g. $EG$ for any finite group $G$. Then the extra degeneracy will give a chain homotopy from $C^{text{simp}}(EG, mathbb{Z})$ to the chain complex with $mathbb{Z}$ in degree $0$ and zero everywhere else. On the other hand $C^{text{simp}}_1(EG, mathbb{Z})=mathbb{Z}[G]$.
For homologous chain complexes which are not homotopic, note that homotopic chain complexes are still homotopic after tensoring with any abelian group. Then the chain complexes
$require{AMScd}
begin{CD}
ldots @>>> 0 @>>> mathbb{Z} @>times 2>> mathbb{Z} @>>> 0 @>>> ldots\
@. @| @VV0V @VVtext{mod }2V @| @.\
ldots @>>> 0 @>>> 0 @>>> mathbb{Z}/2mathbb{Z} @>>> 0 @>>> ldots
end{CD}$
are homologous, but after tensoring with $mathbb{Z}/2mathbb{Z}$ they are not. Hence they cannot be homotopic.
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
answered yesterday
Noah RiggenbachNoah Riggenbach
69028
69028
add a comment |
add a comment |
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