cup product in cohomology ring of a suspensionLet $X = Sigma Y = Y wedge S^1$, cup product $tildeH^p(X) otimes tildeH^q(X) to tildeH^p+q(X)$ is the zero homomorphism?Triviality of relative cup product $H^2times H^2to H^4$ for spaces embeddable to $Bbb R^4$confusion about cup product in cohomology ringSuspension of a product - tricky homotopy equivalenceWhat map of ring spectra corresponds to a product in cohomology, especially the $cup$-product.Cup product on torusEasier proof about suspension of a manifoldSuspension of a CW complexhomology of suspensionThe property of suspensionCup product and suspensionCup Product: Why do we need to “consider cohomology with coefficients in a ring R”?
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cup product in cohomology ring of a suspension
Let $X = Sigma Y = Y wedge S^1$, cup product $tildeH^p(X) otimes tildeH^q(X) to tildeH^p+q(X)$ is the zero homomorphism?Triviality of relative cup product $H^2times H^2to H^4$ for spaces embeddable to $Bbb R^4$confusion about cup product in cohomology ringSuspension of a product - tricky homotopy equivalenceWhat map of ring spectra corresponds to a product in cohomology, especially the $cup$-product.Cup product on torusEasier proof about suspension of a manifoldSuspension of a CW complexhomology of suspensionThe property of suspensionCup product and suspensionCup Product: Why do we need to “consider cohomology with coefficients in a ring R”?
$begingroup$
Let $X$ be a CW-complex. Let $Sigma$ be suspension. Let $R$ be a commutative ring. Is the cup product of
$$
H^*(Sigma X;R)$$
trivial? How to prove? Where can I find the result?
algebraic-topology homology-cohomology homotopy-theory geometric-topology cw-complexes
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
$endgroup$
add a comment |
$begingroup$
Let $X$ be a CW-complex. Let $Sigma$ be suspension. Let $R$ be a commutative ring. Is the cup product of
$$
H^*(Sigma X;R)$$
trivial? How to prove? Where can I find the result?
algebraic-topology homology-cohomology homotopy-theory geometric-topology cw-complexes
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
$endgroup$
add a comment |
$begingroup$
Let $X$ be a CW-complex. Let $Sigma$ be suspension. Let $R$ be a commutative ring. Is the cup product of
$$
H^*(Sigma X;R)$$
trivial? How to prove? Where can I find the result?
algebraic-topology homology-cohomology homotopy-theory geometric-topology cw-complexes
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
$endgroup$
Let $X$ be a CW-complex. Let $Sigma$ be suspension. Let $R$ be a commutative ring. Is the cup product of
$$
H^*(Sigma X;R)$$
trivial? How to prove? Where can I find the result?
algebraic-topology homology-cohomology homotopy-theory geometric-topology cw-complexes
algebraic-topology homology-cohomology homotopy-theory geometric-topology cw-complexes
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
asked May 26 '15 at 1:44
ShiquanShiquan
3,70411237
3,70411237
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. Actually there is a more general result. If $X=bigcup_i=1^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $Sigma X$ is a union of two cones, each of which is contractible, the result follows.
To establish the general result, simply note that $H^*(X)cong H^*(X, A_i)$ as $A_i$ is contractible, and there is a cup product map begineqnarrayH^*(X, A_1)times cdots times H^*(X, A_n)to H^*(X, bigcup_i=1^n A_i)=H^*(X, X)=0endeqnarray
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
$endgroup$
add a comment |
$begingroup$
The answer depends on which homology theory you are using.
The statement fails for singular cohomology.
However there is a fairly easy way to show that the cup product is trivial on reduced cohomology $tildeH,^bullet(Sigma X)=H^bullet(Sigma X,textpt)$.
Write $Sigma X=textCone_+(X)cuptextCone_-(X)$ and let $iota:textptlongrightarrowtextCone(X)$ be the inclusion map. We get the following commutative diagram
$requireAMScd$
beginCD
H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt) @>smile>> H^p+q(Sigma X,textpt)\
@AH^p(iota)otimes H^q(iota)AA @AAH^p+q(iota)A\
H^p(Sigma X,textCone_+(X))otimes H^q(Sigma X,textCone_-(X)) @>>smile> H^p+q(Sigma X,Sigma X)
endCD
Now note that $H^bullet(iota):H^bullet(Sigma X,textCone(X))longrightarrow H^bullet(Sigma X,textpt)$ is an isomorphism because $textCone(X)$ is contractible.
Therefore we have found a factorisation of $asmile b=H^p+q(iota)left(H^p(iota)^-1(a)smile H^q(iota)^-1(b)right)$ over $H^p+q(Sigma X,Sigma X)cong0$, hence $asmile b=0$ for any $aotimes bin H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt)$.
answered Mar 16 at 19:47
LilalasLilalas
54
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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$begingroup$
Yes. Actually there is a more general result. If $X=bigcup_i=1^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $Sigma X$ is a union of two cones, each of which is contractible, the result follows.
To establish the general result, simply note that $H^*(X)cong H^*(X, A_i)$ as $A_i$ is contractible, and there is a cup product map begineqnarrayH^*(X, A_1)times cdots times H^*(X, A_n)to H^*(X, bigcup_i=1^n A_i)=H^*(X, X)=0endeqnarray
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
$endgroup$
add a comment |
$begingroup$
Yes. Actually there is a more general result. If $X=bigcup_i=1^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $Sigma X$ is a union of two cones, each of which is contractible, the result follows.
To establish the general result, simply note that $H^*(X)cong H^*(X, A_i)$ as $A_i$ is contractible, and there is a cup product map begineqnarrayH^*(X, A_1)times cdots times H^*(X, A_n)to H^*(X, bigcup_i=1^n A_i)=H^*(X, X)=0endeqnarray
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
$endgroup$
add a comment |
$begingroup$
Yes. Actually there is a more general result. If $X=bigcup_i=1^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $Sigma X$ is a union of two cones, each of which is contractible, the result follows.
To establish the general result, simply note that $H^*(X)cong H^*(X, A_i)$ as $A_i$ is contractible, and there is a cup product map begineqnarrayH^*(X, A_1)times cdots times H^*(X, A_n)to H^*(X, bigcup_i=1^n A_i)=H^*(X, X)=0endeqnarray
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
$endgroup$
Yes. Actually there is a more general result. If $X=bigcup_i=1^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $Sigma X$ is a union of two cones, each of which is contractible, the result follows.
To establish the general result, simply note that $H^*(X)cong H^*(X, A_i)$ as $A_i$ is contractible, and there is a cup product map begineqnarrayH^*(X, A_1)times cdots times H^*(X, A_n)to H^*(X, bigcup_i=1^n A_i)=H^*(X, X)=0endeqnarray
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
edited May 26 '15 at 3:29
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
answered May 26 '15 at 3:24
Alex FokAlex Fok
3,866818
3,866818
add a comment |
add a comment |
$begingroup$
The answer depends on which homology theory you are using.
The statement fails for singular cohomology.
However there is a fairly easy way to show that the cup product is trivial on reduced cohomology $tildeH,^bullet(Sigma X)=H^bullet(Sigma X,textpt)$.
Write $Sigma X=textCone_+(X)cuptextCone_-(X)$ and let $iota:textptlongrightarrowtextCone(X)$ be the inclusion map. We get the following commutative diagram
$requireAMScd$
beginCD
H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt) @>smile>> H^p+q(Sigma X,textpt)\
@AH^p(iota)otimes H^q(iota)AA @AAH^p+q(iota)A\
H^p(Sigma X,textCone_+(X))otimes H^q(Sigma X,textCone_-(X)) @>>smile> H^p+q(Sigma X,Sigma X)
endCD
Now note that $H^bullet(iota):H^bullet(Sigma X,textCone(X))longrightarrow H^bullet(Sigma X,textpt)$ is an isomorphism because $textCone(X)$ is contractible.
Therefore we have found a factorisation of $asmile b=H^p+q(iota)left(H^p(iota)^-1(a)smile H^q(iota)^-1(b)right)$ over $H^p+q(Sigma X,Sigma X)cong0$, hence $asmile b=0$ for any $aotimes bin H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt)$.
answered Mar 16 at 19:47
LilalasLilalas
54
$endgroup$
add a comment |
$begingroup$
The answer depends on which homology theory you are using.
The statement fails for singular cohomology.
However there is a fairly easy way to show that the cup product is trivial on reduced cohomology $tildeH,^bullet(Sigma X)=H^bullet(Sigma X,textpt)$.
Write $Sigma X=textCone_+(X)cuptextCone_-(X)$ and let $iota:textptlongrightarrowtextCone(X)$ be the inclusion map. We get the following commutative diagram
$requireAMScd$
beginCD
H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt) @>smile>> H^p+q(Sigma X,textpt)\
@AH^p(iota)otimes H^q(iota)AA @AAH^p+q(iota)A\
H^p(Sigma X,textCone_+(X))otimes H^q(Sigma X,textCone_-(X)) @>>smile> H^p+q(Sigma X,Sigma X)
endCD
Now note that $H^bullet(iota):H^bullet(Sigma X,textCone(X))longrightarrow H^bullet(Sigma X,textpt)$ is an isomorphism because $textCone(X)$ is contractible.
Therefore we have found a factorisation of $asmile b=H^p+q(iota)left(H^p(iota)^-1(a)smile H^q(iota)^-1(b)right)$ over $H^p+q(Sigma X,Sigma X)cong0$, hence $asmile b=0$ for any $aotimes bin H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt)$.
answered Mar 16 at 19:47
LilalasLilalas
54
$endgroup$
add a comment |
$begingroup$
The answer depends on which homology theory you are using.
The statement fails for singular cohomology.
However there is a fairly easy way to show that the cup product is trivial on reduced cohomology $tildeH,^bullet(Sigma X)=H^bullet(Sigma X,textpt)$.
Write $Sigma X=textCone_+(X)cuptextCone_-(X)$ and let $iota:textptlongrightarrowtextCone(X)$ be the inclusion map. We get the following commutative diagram
$requireAMScd$
beginCD
H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt) @>smile>> H^p+q(Sigma X,textpt)\
@AH^p(iota)otimes H^q(iota)AA @AAH^p+q(iota)A\
H^p(Sigma X,textCone_+(X))otimes H^q(Sigma X,textCone_-(X)) @>>smile> H^p+q(Sigma X,Sigma X)
endCD
Now note that $H^bullet(iota):H^bullet(Sigma X,textCone(X))longrightarrow H^bullet(Sigma X,textpt)$ is an isomorphism because $textCone(X)$ is contractible.
Therefore we have found a factorisation of $asmile b=H^p+q(iota)left(H^p(iota)^-1(a)smile H^q(iota)^-1(b)right)$ over $H^p+q(Sigma X,Sigma X)cong0$, hence $asmile b=0$ for any $aotimes bin H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt)$.
answered Mar 16 at 19:47
LilalasLilalas
54
$endgroup$
The answer depends on which homology theory you are using.
The statement fails for singular cohomology.
However there is a fairly easy way to show that the cup product is trivial on reduced cohomology $tildeH,^bullet(Sigma X)=H^bullet(Sigma X,textpt)$.
Write $Sigma X=textCone_+(X)cuptextCone_-(X)$ and let $iota:textptlongrightarrowtextCone(X)$ be the inclusion map. We get the following commutative diagram
$requireAMScd$
beginCD
H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt) @>smile>> H^p+q(Sigma X,textpt)\
@AH^p(iota)otimes H^q(iota)AA @AAH^p+q(iota)A\
H^p(Sigma X,textCone_+(X))otimes H^q(Sigma X,textCone_-(X)) @>>smile> H^p+q(Sigma X,Sigma X)
endCD
Now note that $H^bullet(iota):H^bullet(Sigma X,textCone(X))longrightarrow H^bullet(Sigma X,textpt)$ is an isomorphism because $textCone(X)$ is contractible.
Therefore we have found a factorisation of $asmile b=H^p+q(iota)left(H^p(iota)^-1(a)smile H^q(iota)^-1(b)right)$ over $H^p+q(Sigma X,Sigma X)cong0$, hence $asmile b=0$ for any $aotimes bin H^p(Sigma X,textpt)otimes H^q(Sigma X,textpt)$.
answered Mar 16 at 19:47
LilalasLilalas
54
answered Mar 16 at 19:47
LilalasLilalas
54
answered Mar 16 at 19:47
LilalasLilalas
54
answered Mar 16 at 19:47
LilalasLilalas
54
54
add a comment |
add a comment |
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