A “natural homomorphism” $H^n(B; Bbb Z) rightarrow H^n(B; Bbb Z_2)$.
$begingroup$
This is Proposition 4.12, pg33 The claim of the statement is:
The natural homomorphism $Gamma:H^n(B;Bbb Z) rightarrow H^n(B ; Bbb Z_2)$ sends the Euler class to the top Stiefel Whitney class.
My confusion, is, what exactly is this natural homomorphism?
Is this given by taking a cochain $varphi:C_n(B) rightarrow Bbb Z$ and post compose by quotient map - if so, is it clear this well defined?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
asked Jan 12 at 11:17
CL.CL.
2,2012825
$endgroup$
add a comment |
$begingroup$
This is Proposition 4.12, pg33 The claim of the statement is:
The natural homomorphism $Gamma:H^n(B;Bbb Z) rightarrow H^n(B ; Bbb Z_2)$ sends the Euler class to the top Stiefel Whitney class.
My confusion, is, what exactly is this natural homomorphism?
Is this given by taking a cochain $varphi:C_n(B) rightarrow Bbb Z$ and post compose by quotient map - if so, is it clear this well defined?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
asked Jan 12 at 11:17
CL.CL.
2,2012825
$endgroup$
$begingroup$
You can always compose maps, can't you ?
$endgroup$
– Max
Jan 12 at 11:39
add a comment |
$begingroup$
This is Proposition 4.12, pg33 The claim of the statement is:
The natural homomorphism $Gamma:H^n(B;Bbb Z) rightarrow H^n(B ; Bbb Z_2)$ sends the Euler class to the top Stiefel Whitney class.
My confusion, is, what exactly is this natural homomorphism?
Is this given by taking a cochain $varphi:C_n(B) rightarrow Bbb Z$ and post compose by quotient map - if so, is it clear this well defined?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
asked Jan 12 at 11:17
CL.CL.
2,2012825
$endgroup$
This is Proposition 4.12, pg33 The claim of the statement is:
The natural homomorphism $Gamma:H^n(B;Bbb Z) rightarrow H^n(B ; Bbb Z_2)$ sends the Euler class to the top Stiefel Whitney class.
My confusion, is, what exactly is this natural homomorphism?
Is this given by taking a cochain $varphi:C_n(B) rightarrow Bbb Z$ and post compose by quotient map - if so, is it clear this well defined?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
algebraic-topology definition differential-topology vector-bundles characteristic-classes
asked Jan 12 at 11:17
CL.CL.
2,2012825
asked Jan 12 at 11:17
CL.CL.
2,2012825
asked Jan 12 at 11:17
CL.CL.
2,2012825
asked Jan 12 at 11:17
CL.CL.
2,2012825
asked Jan 12 at 11:17
CL.CL.
2,2012825
2,2012825
$begingroup$
You can always compose maps, can't you ?
$endgroup$
– Max
Jan 12 at 11:39
add a comment |
$begingroup$
You can always compose maps, can't you ?
$endgroup$
– Max
Jan 12 at 11:39
$begingroup$
You can always compose maps, can't you ?
$endgroup$
– Max
Jan 12 at 11:39
$begingroup$
You can always compose maps, can't you ?
$endgroup$
– Max
Jan 12 at 11:39
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
We have a natural transformation $Hom(-,mathbb Z) to Hom(-,mathbb Z/2)$ of endofunctors of abelian groups induced by the quotient map. Applying this to the chain complex $C_n(B;mathbb Z)$ determines the morphism of cochain complexes $$C^n(B; mathbb Z) to C^n(B; mathbb Z/2)$$
morphisms of complexes always induce morphisms on cohomology.
answered Jan 12 at 12:20
BenBen
3,423616
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
We have a natural transformation $Hom(-,mathbb Z) to Hom(-,mathbb Z/2)$ of endofunctors of abelian groups induced by the quotient map. Applying this to the chain complex $C_n(B;mathbb Z)$ determines the morphism of cochain complexes $$C^n(B; mathbb Z) to C^n(B; mathbb Z/2)$$
morphisms of complexes always induce morphisms on cohomology.
answered Jan 12 at 12:20
BenBen
3,423616
$endgroup$
add a comment |
$begingroup$
We have a natural transformation $Hom(-,mathbb Z) to Hom(-,mathbb Z/2)$ of endofunctors of abelian groups induced by the quotient map. Applying this to the chain complex $C_n(B;mathbb Z)$ determines the morphism of cochain complexes $$C^n(B; mathbb Z) to C^n(B; mathbb Z/2)$$
morphisms of complexes always induce morphisms on cohomology.
answered Jan 12 at 12:20
BenBen
3,423616
$endgroup$
add a comment |
$begingroup$
We have a natural transformation $Hom(-,mathbb Z) to Hom(-,mathbb Z/2)$ of endofunctors of abelian groups induced by the quotient map. Applying this to the chain complex $C_n(B;mathbb Z)$ determines the morphism of cochain complexes $$C^n(B; mathbb Z) to C^n(B; mathbb Z/2)$$
morphisms of complexes always induce morphisms on cohomology.
answered Jan 12 at 12:20
BenBen
3,423616
$endgroup$
We have a natural transformation $Hom(-,mathbb Z) to Hom(-,mathbb Z/2)$ of endofunctors of abelian groups induced by the quotient map. Applying this to the chain complex $C_n(B;mathbb Z)$ determines the morphism of cochain complexes $$C^n(B; mathbb Z) to C^n(B; mathbb Z/2)$$
morphisms of complexes always induce morphisms on cohomology.
answered Jan 12 at 12:20
BenBen
3,423616
answered Jan 12 at 12:20
BenBen
3,423616
answered Jan 12 at 12:20
BenBen
3,423616
answered Jan 12 at 12:20
BenBen
3,423616
3,423616
add a comment |
add a comment |
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$begingroup$
You can always compose maps, can't you ?
$endgroup$
– Max
Jan 12 at 11:39