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What is the mistake in the following cohomology computation?
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Consider the cell complex $X$ which is the circle with two discs attached along the attaching maps of degree $3$ and $5$. I would like to compute cohomology with integral coefficients using cellular cohomology and the universal coefficient theorem.
First, the cellular chain complex is as follows (from degree $2$ to degree $0$):
$$ 0 tomathbb Zoplusmathbb Z to mathbb Z to mathbb Z to 0$$
where the first (non trivial, degree $2$) boundary map sends $(a,b) to (3a+5b)$ while the degree $1$ map is simply $0$.
Computing the homology, we get $H_0(X) cong H_2(X) cong mathbb Z$ and $H_1(X) cong mathbb Z/3timesmathbb Z/5$.
Dualizing the complex to compute cohomology, we have (from degree $0$ to degree $2$):
$$0 to mathbb Z to mathbb Z to mathbb Zoplus mathbb Z to 0$$
where the final non trivial map is now $1 to (3,5)$ and the other maps are trivial. Computing the cohomology, we now have:
$$H^2(X) cong H^0(X) cong mathbb Z$$
since $(3,5),(1,2)$ forms a basis for $mathbb Zoplusmathbb Z$ while $H^1(X)$ is $0$.
Now, let me compute the cohomology rings using UCT:
We have the exact sequence:
$$0 to Ext^1(H_0(X),mathbb Z)to H^1(X) to Hom(H_1(X),mathbb Z)$$
and since both the $Ext$ and $Hom$ terms vanish, $H^1(X) cong 0$ as expected.
On the other hand, we also have:
$$0 to Ext^1(H_1(X),mathbb Z) to H^2(X) to Hom(H_2(X),mathbb Z)to 0.$$
Here however, the $Hom$ term is $mathbb Z$ while the $Ext$ term is $mathbb Z/3mathbb Z oplus mathbb Z/5mathbb Z$ so that $H^2 cong mathbb Zoplus mathbb Z/3oplusmathbb Z/5$ which does not match my earlier commputation.
Where did I go wrong?
algebraic-topology homology-cohomology
asked Aug 3 at 10:30
ArithmeticGeometer
3,09521329
add a comment |Â
up vote
5
down vote
favorite
Consider the cell complex $X$ which is the circle with two discs attached along the attaching maps of degree $3$ and $5$. I would like to compute cohomology with integral coefficients using cellular cohomology and the universal coefficient theorem.
First, the cellular chain complex is as follows (from degree $2$ to degree $0$):
$$ 0 tomathbb Zoplusmathbb Z to mathbb Z to mathbb Z to 0$$
where the first (non trivial, degree $2$) boundary map sends $(a,b) to (3a+5b)$ while the degree $1$ map is simply $0$.
Computing the homology, we get $H_0(X) cong H_2(X) cong mathbb Z$ and $H_1(X) cong mathbb Z/3timesmathbb Z/5$.
Dualizing the complex to compute cohomology, we have (from degree $0$ to degree $2$):
$$0 to mathbb Z to mathbb Z to mathbb Zoplus mathbb Z to 0$$
where the final non trivial map is now $1 to (3,5)$ and the other maps are trivial. Computing the cohomology, we now have:
$$H^2(X) cong H^0(X) cong mathbb Z$$
since $(3,5),(1,2)$ forms a basis for $mathbb Zoplusmathbb Z$ while $H^1(X)$ is $0$.
Now, let me compute the cohomology rings using UCT:
We have the exact sequence:
$$0 to Ext^1(H_0(X),mathbb Z)to H^1(X) to Hom(H_1(X),mathbb Z)$$
and since both the $Ext$ and $Hom$ terms vanish, $H^1(X) cong 0$ as expected.
On the other hand, we also have:
$$0 to Ext^1(H_1(X),mathbb Z) to H^2(X) to Hom(H_2(X),mathbb Z)to 0.$$
Here however, the $Hom$ term is $mathbb Z$ while the $Ext$ term is $mathbb Z/3mathbb Z oplus mathbb Z/5mathbb Z$ so that $H^2 cong mathbb Zoplus mathbb Z/3oplusmathbb Z/5$ which does not match my earlier commputation.
Where did I go wrong?
algebraic-topology homology-cohomology
asked Aug 3 at 10:30
ArithmeticGeometer
3,09521329
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Consider the cell complex $X$ which is the circle with two discs attached along the attaching maps of degree $3$ and $5$. I would like to compute cohomology with integral coefficients using cellular cohomology and the universal coefficient theorem.
First, the cellular chain complex is as follows (from degree $2$ to degree $0$):
$$ 0 tomathbb Zoplusmathbb Z to mathbb Z to mathbb Z to 0$$
where the first (non trivial, degree $2$) boundary map sends $(a,b) to (3a+5b)$ while the degree $1$ map is simply $0$.
Computing the homology, we get $H_0(X) cong H_2(X) cong mathbb Z$ and $H_1(X) cong mathbb Z/3timesmathbb Z/5$.
Dualizing the complex to compute cohomology, we have (from degree $0$ to degree $2$):
$$0 to mathbb Z to mathbb Z to mathbb Zoplus mathbb Z to 0$$
where the final non trivial map is now $1 to (3,5)$ and the other maps are trivial. Computing the cohomology, we now have:
$$H^2(X) cong H^0(X) cong mathbb Z$$
since $(3,5),(1,2)$ forms a basis for $mathbb Zoplusmathbb Z$ while $H^1(X)$ is $0$.
Now, let me compute the cohomology rings using UCT:
We have the exact sequence:
$$0 to Ext^1(H_0(X),mathbb Z)to H^1(X) to Hom(H_1(X),mathbb Z)$$
and since both the $Ext$ and $Hom$ terms vanish, $H^1(X) cong 0$ as expected.
On the other hand, we also have:
$$0 to Ext^1(H_1(X),mathbb Z) to H^2(X) to Hom(H_2(X),mathbb Z)to 0.$$
Here however, the $Hom$ term is $mathbb Z$ while the $Ext$ term is $mathbb Z/3mathbb Z oplus mathbb Z/5mathbb Z$ so that $H^2 cong mathbb Zoplus mathbb Z/3oplusmathbb Z/5$ which does not match my earlier commputation.
Where did I go wrong?
algebraic-topology homology-cohomology
asked Aug 3 at 10:30
ArithmeticGeometer
3,09521329
Consider the cell complex $X$ which is the circle with two discs attached along the attaching maps of degree $3$ and $5$. I would like to compute cohomology with integral coefficients using cellular cohomology and the universal coefficient theorem.
First, the cellular chain complex is as follows (from degree $2$ to degree $0$):
$$ 0 tomathbb Zoplusmathbb Z to mathbb Z to mathbb Z to 0$$
where the first (non trivial, degree $2$) boundary map sends $(a,b) to (3a+5b)$ while the degree $1$ map is simply $0$.
Computing the homology, we get $H_0(X) cong H_2(X) cong mathbb Z$ and $H_1(X) cong mathbb Z/3timesmathbb Z/5$.
Dualizing the complex to compute cohomology, we have (from degree $0$ to degree $2$):
$$0 to mathbb Z to mathbb Z to mathbb Zoplus mathbb Z to 0$$
where the final non trivial map is now $1 to (3,5)$ and the other maps are trivial. Computing the cohomology, we now have:
$$H^2(X) cong H^0(X) cong mathbb Z$$
since $(3,5),(1,2)$ forms a basis for $mathbb Zoplusmathbb Z$ while $H^1(X)$ is $0$.
Now, let me compute the cohomology rings using UCT:
We have the exact sequence:
$$0 to Ext^1(H_0(X),mathbb Z)to H^1(X) to Hom(H_1(X),mathbb Z)$$
and since both the $Ext$ and $Hom$ terms vanish, $H^1(X) cong 0$ as expected.
On the other hand, we also have:
$$0 to Ext^1(H_1(X),mathbb Z) to H^2(X) to Hom(H_2(X),mathbb Z)to 0.$$
Here however, the $Hom$ term is $mathbb Z$ while the $Ext$ term is $mathbb Z/3mathbb Z oplus mathbb Z/5mathbb Z$ so that $H^2 cong mathbb Zoplus mathbb Z/3oplusmathbb Z/5$ which does not match my earlier commputation.
Where did I go wrong?
algebraic-topology homology-cohomology
asked Aug 3 at 10:30
ArithmeticGeometer
3,09521329
asked Aug 3 at 10:30
ArithmeticGeometer
3,09521329
asked Aug 3 at 10:30
ArithmeticGeometer
3,09521329
asked Aug 3 at 10:30
ArithmeticGeometer
3,09521329
3,09521329
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
The image of $(a,b)mapsto 3a+5b$ as a map from $Bbb ZoplusBbb Z$ to $Bbb Z$ is all of $Bbb Z$. So $H_1(X)=0$. There's now no problem
with UCT: everything is free, so there's no Ext terms.
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
Stupid mistake :(
– ArithmeticGeometer
Aug 3 at 10:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The image of $(a,b)mapsto 3a+5b$ as a map from $Bbb ZoplusBbb Z$ to $Bbb Z$ is all of $Bbb Z$. So $H_1(X)=0$. There's now no problem
with UCT: everything is free, so there's no Ext terms.
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
Stupid mistake :(
– ArithmeticGeometer
Aug 3 at 10:44
add a comment |Â
up vote
4
down vote
accepted
The image of $(a,b)mapsto 3a+5b$ as a map from $Bbb ZoplusBbb Z$ to $Bbb Z$ is all of $Bbb Z$. So $H_1(X)=0$. There's now no problem
with UCT: everything is free, so there's no Ext terms.
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
Stupid mistake :(
– ArithmeticGeometer
Aug 3 at 10:44
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The image of $(a,b)mapsto 3a+5b$ as a map from $Bbb ZoplusBbb Z$ to $Bbb Z$ is all of $Bbb Z$. So $H_1(X)=0$. There's now no problem
with UCT: everything is free, so there's no Ext terms.
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
The image of $(a,b)mapsto 3a+5b$ as a map from $Bbb ZoplusBbb Z$ to $Bbb Z$ is all of $Bbb Z$. So $H_1(X)=0$. There's now no problem
with UCT: everything is free, so there's no Ext terms.
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 10:44
Lord Shark the Unknown
84.1k950111
84.1k950111
Stupid mistake :(
– ArithmeticGeometer
Aug 3 at 10:44
add a comment |Â
Stupid mistake :(
– ArithmeticGeometer
Aug 3 at 10:44
Stupid mistake :(
– ArithmeticGeometer
Aug 3 at 10:44
Stupid mistake :(
– ArithmeticGeometer
Aug 3 at 10:44
add a comment |Â
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