Looking for reference on cup and cap product without invoking acyclic model theorem
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I am looking for reference on cup and cap product without invoking acyclic model theorem. To me, acyclic model theorem is very strange phenomena though I could understand it but I do not see direct construction.
$textbf{Q:}$ Is there a reference on cup and cap product construction without invoking acyclic model theorem(relative and $C(X)otimes_Z C(Y)cong C(Xtimes Y)$ as quasi isomorphism)? I would like to see a direct (computable) construction which will demonstrate non-commuativity of cup, associativity of both cap and cup whenever they are well defined. I am having trouble to see "obviously" $ucup v=(-1)^{deg(v)deg(u)}vcup u$ as well.(Note here I should not have written in this way as $uin H^i(X), vin H^j(X)$ but I have identified $H^{i+j}(Xtimes Y)=H^{i+j}(Ytimes X)$ in the image. This is indicating diagram is commutative up to a sign.) Most of time, the book proves this by acyclic model via producing homotopy to a chain map with a sign.
abstract-algebra general-topology reference-request algebraic-topology
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
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add a comment |
$begingroup$
I am looking for reference on cup and cap product without invoking acyclic model theorem. To me, acyclic model theorem is very strange phenomena though I could understand it but I do not see direct construction.
$textbf{Q:}$ Is there a reference on cup and cap product construction without invoking acyclic model theorem(relative and $C(X)otimes_Z C(Y)cong C(Xtimes Y)$ as quasi isomorphism)? I would like to see a direct (computable) construction which will demonstrate non-commuativity of cup, associativity of both cap and cup whenever they are well defined. I am having trouble to see "obviously" $ucup v=(-1)^{deg(v)deg(u)}vcup u$ as well.(Note here I should not have written in this way as $uin H^i(X), vin H^j(X)$ but I have identified $H^{i+j}(Xtimes Y)=H^{i+j}(Ytimes X)$ in the image. This is indicating diagram is commutative up to a sign.) Most of time, the book proves this by acyclic model via producing homotopy to a chain map with a sign.
abstract-algebra general-topology reference-request algebraic-topology
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
$endgroup$
$begingroup$
You could try GW Whitehead's paper 'Generalized Homology Theories' for a stable homotopy theoretic approach to the whole business, including cup, cap, slant, etc.. products.
$endgroup$
– Tyrone
Dec 13 '18 at 10:09
$begingroup$
The cup product can be defined on the chain level by the Alexander-Whitney map and the diagonal. These maps are given by explicit formulas. If you want to go deeper, there are $E_infty$ operads acting on cochains, with a huge number of explicit operations spelling out the higher homotopy commutativity of cochains.
$endgroup$
– Justin Young
Dec 13 '18 at 17:14
add a comment |
$begingroup$
I am looking for reference on cup and cap product without invoking acyclic model theorem. To me, acyclic model theorem is very strange phenomena though I could understand it but I do not see direct construction.
$textbf{Q:}$ Is there a reference on cup and cap product construction without invoking acyclic model theorem(relative and $C(X)otimes_Z C(Y)cong C(Xtimes Y)$ as quasi isomorphism)? I would like to see a direct (computable) construction which will demonstrate non-commuativity of cup, associativity of both cap and cup whenever they are well defined. I am having trouble to see "obviously" $ucup v=(-1)^{deg(v)deg(u)}vcup u$ as well.(Note here I should not have written in this way as $uin H^i(X), vin H^j(X)$ but I have identified $H^{i+j}(Xtimes Y)=H^{i+j}(Ytimes X)$ in the image. This is indicating diagram is commutative up to a sign.) Most of time, the book proves this by acyclic model via producing homotopy to a chain map with a sign.
abstract-algebra general-topology reference-request algebraic-topology
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
$endgroup$
I am looking for reference on cup and cap product without invoking acyclic model theorem. To me, acyclic model theorem is very strange phenomena though I could understand it but I do not see direct construction.
$textbf{Q:}$ Is there a reference on cup and cap product construction without invoking acyclic model theorem(relative and $C(X)otimes_Z C(Y)cong C(Xtimes Y)$ as quasi isomorphism)? I would like to see a direct (computable) construction which will demonstrate non-commuativity of cup, associativity of both cap and cup whenever they are well defined. I am having trouble to see "obviously" $ucup v=(-1)^{deg(v)deg(u)}vcup u$ as well.(Note here I should not have written in this way as $uin H^i(X), vin H^j(X)$ but I have identified $H^{i+j}(Xtimes Y)=H^{i+j}(Ytimes X)$ in the image. This is indicating diagram is commutative up to a sign.) Most of time, the book proves this by acyclic model via producing homotopy to a chain map with a sign.
abstract-algebra general-topology reference-request algebraic-topology
abstract-algebra general-topology reference-request algebraic-topology
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
asked Dec 13 '18 at 4:59
user45765user45765
2,7282722
2,7282722
$begingroup$
You could try GW Whitehead's paper 'Generalized Homology Theories' for a stable homotopy theoretic approach to the whole business, including cup, cap, slant, etc.. products.
$endgroup$
– Tyrone
Dec 13 '18 at 10:09
$begingroup$
The cup product can be defined on the chain level by the Alexander-Whitney map and the diagonal. These maps are given by explicit formulas. If you want to go deeper, there are $E_infty$ operads acting on cochains, with a huge number of explicit operations spelling out the higher homotopy commutativity of cochains.
$endgroup$
– Justin Young
Dec 13 '18 at 17:14
add a comment |
$begingroup$
You could try GW Whitehead's paper 'Generalized Homology Theories' for a stable homotopy theoretic approach to the whole business, including cup, cap, slant, etc.. products.
$endgroup$
– Tyrone
Dec 13 '18 at 10:09
$begingroup$
The cup product can be defined on the chain level by the Alexander-Whitney map and the diagonal. These maps are given by explicit formulas. If you want to go deeper, there are $E_infty$ operads acting on cochains, with a huge number of explicit operations spelling out the higher homotopy commutativity of cochains.
$endgroup$
– Justin Young
Dec 13 '18 at 17:14
$begingroup$
You could try GW Whitehead's paper 'Generalized Homology Theories' for a stable homotopy theoretic approach to the whole business, including cup, cap, slant, etc.. products.
$endgroup$
– Tyrone
Dec 13 '18 at 10:09
$begingroup$
You could try GW Whitehead's paper 'Generalized Homology Theories' for a stable homotopy theoretic approach to the whole business, including cup, cap, slant, etc.. products.
$endgroup$
– Tyrone
Dec 13 '18 at 10:09
$begingroup$
The cup product can be defined on the chain level by the Alexander-Whitney map and the diagonal. These maps are given by explicit formulas. If you want to go deeper, there are $E_infty$ operads acting on cochains, with a huge number of explicit operations spelling out the higher homotopy commutativity of cochains.
$endgroup$
– Justin Young
Dec 13 '18 at 17:14
$begingroup$
The cup product can be defined on the chain level by the Alexander-Whitney map and the diagonal. These maps are given by explicit formulas. If you want to go deeper, there are $E_infty$ operads acting on cochains, with a huge number of explicit operations spelling out the higher homotopy commutativity of cochains.
$endgroup$
– Justin Young
Dec 13 '18 at 17:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The cup product is graded commutative in homology, but not on the chain level. Maps witnessing higher non-commutativity in a coherent way are known as $i$-cup products, and were introduced by N. Steenrod in this paper. Computations there are very explicit. The fundamental result for (usual) cup products is that if $a$ and $b$ are cochains in degree $p$ and $q$, there is a cochain $asmile_1 b$, called the 1-cup product of $a$ with $b$, so that
$$d(asmile_1 b) -dasmile_1 b-(-1)^p asmile_1 b= (-1)^{p+q+1}[a,b]$$
where the right hand side is the graded commutator. This MO post contains more information on these operations.
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
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add a comment |
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$begingroup$
The cup product is graded commutative in homology, but not on the chain level. Maps witnessing higher non-commutativity in a coherent way are known as $i$-cup products, and were introduced by N. Steenrod in this paper. Computations there are very explicit. The fundamental result for (usual) cup products is that if $a$ and $b$ are cochains in degree $p$ and $q$, there is a cochain $asmile_1 b$, called the 1-cup product of $a$ with $b$, so that
$$d(asmile_1 b) -dasmile_1 b-(-1)^p asmile_1 b= (-1)^{p+q+1}[a,b]$$
where the right hand side is the graded commutator. This MO post contains more information on these operations.
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
$endgroup$
add a comment |
$begingroup$
The cup product is graded commutative in homology, but not on the chain level. Maps witnessing higher non-commutativity in a coherent way are known as $i$-cup products, and were introduced by N. Steenrod in this paper. Computations there are very explicit. The fundamental result for (usual) cup products is that if $a$ and $b$ are cochains in degree $p$ and $q$, there is a cochain $asmile_1 b$, called the 1-cup product of $a$ with $b$, so that
$$d(asmile_1 b) -dasmile_1 b-(-1)^p asmile_1 b= (-1)^{p+q+1}[a,b]$$
where the right hand side is the graded commutator. This MO post contains more information on these operations.
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
$endgroup$
add a comment |
$begingroup$
The cup product is graded commutative in homology, but not on the chain level. Maps witnessing higher non-commutativity in a coherent way are known as $i$-cup products, and were introduced by N. Steenrod in this paper. Computations there are very explicit. The fundamental result for (usual) cup products is that if $a$ and $b$ are cochains in degree $p$ and $q$, there is a cochain $asmile_1 b$, called the 1-cup product of $a$ with $b$, so that
$$d(asmile_1 b) -dasmile_1 b-(-1)^p asmile_1 b= (-1)^{p+q+1}[a,b]$$
where the right hand side is the graded commutator. This MO post contains more information on these operations.
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
$endgroup$
The cup product is graded commutative in homology, but not on the chain level. Maps witnessing higher non-commutativity in a coherent way are known as $i$-cup products, and were introduced by N. Steenrod in this paper. Computations there are very explicit. The fundamental result for (usual) cup products is that if $a$ and $b$ are cochains in degree $p$ and $q$, there is a cochain $asmile_1 b$, called the 1-cup product of $a$ with $b$, so that
$$d(asmile_1 b) -dasmile_1 b-(-1)^p asmile_1 b= (-1)^{p+q+1}[a,b]$$
where the right hand side is the graded commutator. This MO post contains more information on these operations.
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
edited Dec 13 '18 at 12:45
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
answered Dec 13 '18 at 12:39
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
96.9k10153297
add a comment |
add a comment |
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$begingroup$
You could try GW Whitehead's paper 'Generalized Homology Theories' for a stable homotopy theoretic approach to the whole business, including cup, cap, slant, etc.. products.
$endgroup$
– Tyrone
Dec 13 '18 at 10:09
$begingroup$
The cup product can be defined on the chain level by the Alexander-Whitney map and the diagonal. These maps are given by explicit formulas. If you want to go deeper, there are $E_infty$ operads acting on cochains, with a huge number of explicit operations spelling out the higher homotopy commutativity of cochains.
$endgroup$
– Justin Young
Dec 13 '18 at 17:14