Mixing
A theorem on fibre bundles, $H^k(E) cong H^{k+n}(E,E_0)$.
$begingroup$
This is a theorem about fibre bundles, stated in page 21, Theorem 3.6
Let $xi= (E,B,p)$ be an $n$-vector bundle. Let $F_0$ be the fibre $F$ without its nonzero element, and $E_0$ be the total space with out the zero section.
The group $H^i(E,E_0; Bbb Z_2)$ is zero for $i<n$ and $H^n(E,E_0; Bbb Z_2)$ contains a unique class $u$ such that for each fibre $F= pi^{-1}(b)$ the restriction,
$$ u|(F,F_0) in H^n(F,F_0; Bbb Z_2)$$
is the unique nonzero class in $H^n(F,F_0; Bbb Z_2)$. Furthermore, the correspondence $x mapsto x cup u $ defines an isomoprhism
$$ cdot cup u : H^k(E,Bbb Z_2) rightarrow H^{k+n}(E,E_0; Bbb Z_2)$$
My question is, how exactly is the last map (highlighted) defined?
My intuitive attempt: By definition an element in $H^{k+n}(E,E_0; Bbb Z_2)$ is represented by some cocycle $u:C_{n}(E,E_0) rightarrow Bbb Z_2$.
This again lifts to an element, $u':C_{n}(E) rightarrow Bbb Z_2$. Hence we can define
$$ x cup u: C_{k+n}(E) rightarrow Bbb Z_2$$
So it suffices to check this vanishes on $C_{k+n}(E_0)$, which does as $u$ vanishes on $C_n(E_0)$. So we have an induced map $x cup u:C_{k+n}(E,E_0) rightarrow Bbb Z_2$.
Now we then have a series of check: that this operation gives an element in cohomology class, and is well defined on cohomology class.
I think my argument (even if it is right) is extremely inefficient, is there a neat way to see this?
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 11 at 17:26
CL.CL.
2,2282825
$endgroup$
add a comment |
$begingroup$
This is a theorem about fibre bundles, stated in page 21, Theorem 3.6
Let $xi= (E,B,p)$ be an $n$-vector bundle. Let $F_0$ be the fibre $F$ without its nonzero element, and $E_0$ be the total space with out the zero section.
The group $H^i(E,E_0; Bbb Z_2)$ is zero for $i<n$ and $H^n(E,E_0; Bbb Z_2)$ contains a unique class $u$ such that for each fibre $F= pi^{-1}(b)$ the restriction,
$$ u|(F,F_0) in H^n(F,F_0; Bbb Z_2)$$
is the unique nonzero class in $H^n(F,F_0; Bbb Z_2)$. Furthermore, the correspondence $x mapsto x cup u $ defines an isomoprhism
$$ cdot cup u : H^k(E,Bbb Z_2) rightarrow H^{k+n}(E,E_0; Bbb Z_2)$$
My question is, how exactly is the last map (highlighted) defined?
My intuitive attempt: By definition an element in $H^{k+n}(E,E_0; Bbb Z_2)$ is represented by some cocycle $u:C_{n}(E,E_0) rightarrow Bbb Z_2$.
This again lifts to an element, $u':C_{n}(E) rightarrow Bbb Z_2$. Hence we can define
$$ x cup u: C_{k+n}(E) rightarrow Bbb Z_2$$
So it suffices to check this vanishes on $C_{k+n}(E_0)$, which does as $u$ vanishes on $C_n(E_0)$. So we have an induced map $x cup u:C_{k+n}(E,E_0) rightarrow Bbb Z_2$.
Now we then have a series of check: that this operation gives an element in cohomology class, and is well defined on cohomology class.
I think my argument (even if it is right) is extremely inefficient, is there a neat way to see this?
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 11 at 17:26
CL.CL.
2,2282825
$endgroup$
$begingroup$
An argument will depend on your definition of the cup product (in particular, it will depend on which cohomology theory you are using). For singular cohomology Hatcher gives a definition of the relative cup product on page 209 of Algebraic Toplogy. (See also, for instance, this post math.stackexchange.com/questions/1536537/…)
$endgroup$
– Tyrone
Jan 12 at 10:16
$begingroup$
Ok, thanks, yes, I am using the definition of Hatcher's, I will check out the post.
$endgroup$
– CL.
Jan 12 at 10:19
add a comment |
$begingroup$
This is a theorem about fibre bundles, stated in page 21, Theorem 3.6
Let $xi= (E,B,p)$ be an $n$-vector bundle. Let $F_0$ be the fibre $F$ without its nonzero element, and $E_0$ be the total space with out the zero section.
The group $H^i(E,E_0; Bbb Z_2)$ is zero for $i<n$ and $H^n(E,E_0; Bbb Z_2)$ contains a unique class $u$ such that for each fibre $F= pi^{-1}(b)$ the restriction,
$$ u|(F,F_0) in H^n(F,F_0; Bbb Z_2)$$
is the unique nonzero class in $H^n(F,F_0; Bbb Z_2)$. Furthermore, the correspondence $x mapsto x cup u $ defines an isomoprhism
$$ cdot cup u : H^k(E,Bbb Z_2) rightarrow H^{k+n}(E,E_0; Bbb Z_2)$$
My question is, how exactly is the last map (highlighted) defined?
My intuitive attempt: By definition an element in $H^{k+n}(E,E_0; Bbb Z_2)$ is represented by some cocycle $u:C_{n}(E,E_0) rightarrow Bbb Z_2$.
This again lifts to an element, $u':C_{n}(E) rightarrow Bbb Z_2$. Hence we can define
$$ x cup u: C_{k+n}(E) rightarrow Bbb Z_2$$
So it suffices to check this vanishes on $C_{k+n}(E_0)$, which does as $u$ vanishes on $C_n(E_0)$. So we have an induced map $x cup u:C_{k+n}(E,E_0) rightarrow Bbb Z_2$.
Now we then have a series of check: that this operation gives an element in cohomology class, and is well defined on cohomology class.
I think my argument (even if it is right) is extremely inefficient, is there a neat way to see this?
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 11 at 17:26
CL.CL.
2,2282825
$endgroup$
This is a theorem about fibre bundles, stated in page 21, Theorem 3.6
Let $xi= (E,B,p)$ be an $n$-vector bundle. Let $F_0$ be the fibre $F$ without its nonzero element, and $E_0$ be the total space with out the zero section.
The group $H^i(E,E_0; Bbb Z_2)$ is zero for $i<n$ and $H^n(E,E_0; Bbb Z_2)$ contains a unique class $u$ such that for each fibre $F= pi^{-1}(b)$ the restriction,
$$ u|(F,F_0) in H^n(F,F_0; Bbb Z_2)$$
is the unique nonzero class in $H^n(F,F_0; Bbb Z_2)$. Furthermore, the correspondence $x mapsto x cup u $ defines an isomoprhism
$$ cdot cup u : H^k(E,Bbb Z_2) rightarrow H^{k+n}(E,E_0; Bbb Z_2)$$
My question is, how exactly is the last map (highlighted) defined?
My intuitive attempt: By definition an element in $H^{k+n}(E,E_0; Bbb Z_2)$ is represented by some cocycle $u:C_{n}(E,E_0) rightarrow Bbb Z_2$.
This again lifts to an element, $u':C_{n}(E) rightarrow Bbb Z_2$. Hence we can define
$$ x cup u: C_{k+n}(E) rightarrow Bbb Z_2$$
So it suffices to check this vanishes on $C_{k+n}(E_0)$, which does as $u$ vanishes on $C_n(E_0)$. So we have an induced map $x cup u:C_{k+n}(E,E_0) rightarrow Bbb Z_2$.
Now we then have a series of check: that this operation gives an element in cohomology class, and is well defined on cohomology class.
I think my argument (even if it is right) is extremely inefficient, is there a neat way to see this?
algebraic-topology differential-topology vector-bundles fiber-bundles
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 11 at 17:26
CL.CL.
2,2282825
asked Jan 11 at 17:26
CL.CL.
2,2282825
asked Jan 11 at 17:26
CL.CL.
2,2282825
asked Jan 11 at 17:26
CL.CL.
2,2282825
asked Jan 11 at 17:26
CL.CL.
2,2282825
2,2282825
$begingroup$
An argument will depend on your definition of the cup product (in particular, it will depend on which cohomology theory you are using). For singular cohomology Hatcher gives a definition of the relative cup product on page 209 of Algebraic Toplogy. (See also, for instance, this post math.stackexchange.com/questions/1536537/…)
$endgroup$
– Tyrone
Jan 12 at 10:16
$begingroup$
Ok, thanks, yes, I am using the definition of Hatcher's, I will check out the post.
$endgroup$
– CL.
Jan 12 at 10:19
add a comment |
$begingroup$
An argument will depend on your definition of the cup product (in particular, it will depend on which cohomology theory you are using). For singular cohomology Hatcher gives a definition of the relative cup product on page 209 of Algebraic Toplogy. (See also, for instance, this post math.stackexchange.com/questions/1536537/…)
$endgroup$
– Tyrone
Jan 12 at 10:16
$begingroup$
Ok, thanks, yes, I am using the definition of Hatcher's, I will check out the post.
$endgroup$
– CL.
Jan 12 at 10:19
$begingroup$
An argument will depend on your definition of the cup product (in particular, it will depend on which cohomology theory you are using). For singular cohomology Hatcher gives a definition of the relative cup product on page 209 of Algebraic Toplogy. (See also, for instance, this post math.stackexchange.com/questions/1536537/…)
$endgroup$
– Tyrone
Jan 12 at 10:16
$begingroup$
An argument will depend on your definition of the cup product (in particular, it will depend on which cohomology theory you are using). For singular cohomology Hatcher gives a definition of the relative cup product on page 209 of Algebraic Toplogy. (See also, for instance, this post math.stackexchange.com/questions/1536537/…)
$endgroup$
– Tyrone
Jan 12 at 10:16
$begingroup$
Ok, thanks, yes, I am using the definition of Hatcher's, I will check out the post.
$endgroup$
– CL.
Jan 12 at 10:19
$begingroup$
Ok, thanks, yes, I am using the definition of Hatcher's, I will check out the post.
$endgroup$
– CL.
Jan 12 at 10:19
add a comment |
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$begingroup$
An argument will depend on your definition of the cup product (in particular, it will depend on which cohomology theory you are using). For singular cohomology Hatcher gives a definition of the relative cup product on page 209 of Algebraic Toplogy. (See also, for instance, this post math.stackexchange.com/questions/1536537/…)
$endgroup$
– Tyrone
Jan 12 at 10:16
$begingroup$
Ok, thanks, yes, I am using the definition of Hatcher's, I will check out the post.
$endgroup$
– CL.
Jan 12 at 10:19