Mixing
$F:Mto N$ is surjective if $int_M F^* eta ne 0$ for some $eta in Omega^n(N)$
Let $M$ and $N$ be compact orientable and connected smooth $n$-manifolds and $F:M to N$ a smooth map. Suppose $$int_M F^* eta ne 0$$ for some $eta in Omega^n(N)$. Then $F$ is surjective. Give an example that shows the converse is not true.
A non-surjective map has degree $0$ so the first part is clear. I could not think of an example for the converse, however. I want to find two compact oriented connected manifolds such that $F$ is surjective but $int_M F^* eta = 0$ for all $eta in Omega^n(N)$.
differential-geometry smooth-manifolds de-rham-cohomology
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
add a comment |
Let $M$ and $N$ be compact orientable and connected smooth $n$-manifolds and $F:M to N$ a smooth map. Suppose $$int_M F^* eta ne 0$$ for some $eta in Omega^n(N)$. Then $F$ is surjective. Give an example that shows the converse is not true.
A non-surjective map has degree $0$ so the first part is clear. I could not think of an example for the converse, however. I want to find two compact oriented connected manifolds such that $F$ is surjective but $int_M F^* eta = 0$ for all $eta in Omega^n(N)$.
differential-geometry smooth-manifolds de-rham-cohomology
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
1
@MoisheCohen Yes I'm looking for a counterexample
– mysatellite
Dec 31 '18 at 2:01
1
Choose $F$ surjective but null-homotopic.
– Mike Miller
Dec 31 '18 at 2:20
I don't know if you read the deleted answer and all the comments therein before it was deleted. Are you still interested in an answer?
– Amitai Yuval
Dec 31 '18 at 16:24
@AmitaiYuval yes
– mysatellite
Dec 31 '18 at 17:11
add a comment |
Let $M$ and $N$ be compact orientable and connected smooth $n$-manifolds and $F:M to N$ a smooth map. Suppose $$int_M F^* eta ne 0$$ for some $eta in Omega^n(N)$. Then $F$ is surjective. Give an example that shows the converse is not true.
A non-surjective map has degree $0$ so the first part is clear. I could not think of an example for the converse, however. I want to find two compact oriented connected manifolds such that $F$ is surjective but $int_M F^* eta = 0$ for all $eta in Omega^n(N)$.
differential-geometry smooth-manifolds de-rham-cohomology
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
Let $M$ and $N$ be compact orientable and connected smooth $n$-manifolds and $F:M to N$ a smooth map. Suppose $$int_M F^* eta ne 0$$ for some $eta in Omega^n(N)$. Then $F$ is surjective. Give an example that shows the converse is not true.
A non-surjective map has degree $0$ so the first part is clear. I could not think of an example for the converse, however. I want to find two compact oriented connected manifolds such that $F$ is surjective but $int_M F^* eta = 0$ for all $eta in Omega^n(N)$.
differential-geometry smooth-manifolds de-rham-cohomology
differential-geometry smooth-manifolds de-rham-cohomology
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
asked Dec 31 '18 at 1:44
mysatellitemysatellite
2,14221130
2,14221130
1
@MoisheCohen Yes I'm looking for a counterexample
– mysatellite
Dec 31 '18 at 2:01
1
Choose $F$ surjective but null-homotopic.
– Mike Miller
Dec 31 '18 at 2:20
I don't know if you read the deleted answer and all the comments therein before it was deleted. Are you still interested in an answer?
– Amitai Yuval
Dec 31 '18 at 16:24
@AmitaiYuval yes
– mysatellite
Dec 31 '18 at 17:11
add a comment |
1
@MoisheCohen Yes I'm looking for a counterexample
– mysatellite
Dec 31 '18 at 2:01
1
Choose $F$ surjective but null-homotopic.
– Mike Miller
Dec 31 '18 at 2:20
I don't know if you read the deleted answer and all the comments therein before it was deleted. Are you still interested in an answer?
– Amitai Yuval
Dec 31 '18 at 16:24
@AmitaiYuval yes
– mysatellite
Dec 31 '18 at 17:11
1
1
@MoisheCohen Yes I'm looking for a counterexample
– mysatellite
Dec 31 '18 at 2:01
@MoisheCohen Yes I'm looking for a counterexample
– mysatellite
Dec 31 '18 at 2:01
1
1
Choose $F$ surjective but null-homotopic.
– Mike Miller
Dec 31 '18 at 2:20
Choose $F$ surjective but null-homotopic.
– Mike Miller
Dec 31 '18 at 2:20
I don't know if you read the deleted answer and all the comments therein before it was deleted. Are you still interested in an answer?
– Amitai Yuval
Dec 31 '18 at 16:24
I don't know if you read the deleted answer and all the comments therein before it was deleted. Are you still interested in an answer?
– Amitai Yuval
Dec 31 '18 at 16:24
@AmitaiYuval yes
– mysatellite
Dec 31 '18 at 17:11
@AmitaiYuval yes
– mysatellite
Dec 31 '18 at 17:11
add a comment |
1 Answer
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Here is a concrete realization of Mike Miller's comment. Think of $S^1$ as sitting in $mathbb{C}$ and consider the map
begin{align*}
varphi: S^1 & to S^1 \
x+iy & mapsto e^{2pi i x}.
end{align*}
Then $varphi$ is both surjective and null-homotopic and thus serves as a counterexample.
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
add a comment |
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1 Answer
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1 Answer
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Here is a concrete realization of Mike Miller's comment. Think of $S^1$ as sitting in $mathbb{C}$ and consider the map
begin{align*}
varphi: S^1 & to S^1 \
x+iy & mapsto e^{2pi i x}.
end{align*}
Then $varphi$ is both surjective and null-homotopic and thus serves as a counterexample.
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
add a comment |
Here is a concrete realization of Mike Miller's comment. Think of $S^1$ as sitting in $mathbb{C}$ and consider the map
begin{align*}
varphi: S^1 & to S^1 \
x+iy & mapsto e^{2pi i x}.
end{align*}
Then $varphi$ is both surjective and null-homotopic and thus serves as a counterexample.
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
add a comment |
Here is a concrete realization of Mike Miller's comment. Think of $S^1$ as sitting in $mathbb{C}$ and consider the map
begin{align*}
varphi: S^1 & to S^1 \
x+iy & mapsto e^{2pi i x}.
end{align*}
Then $varphi$ is both surjective and null-homotopic and thus serves as a counterexample.
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
Here is a concrete realization of Mike Miller's comment. Think of $S^1$ as sitting in $mathbb{C}$ and consider the map
begin{align*}
varphi: S^1 & to S^1 \
x+iy & mapsto e^{2pi i x}.
end{align*}
Then $varphi$ is both surjective and null-homotopic and thus serves as a counterexample.
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
answered Dec 31 '18 at 21:08
Or EisenbergOr Eisenberg
1346
1346
add a comment |
add a comment |
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1
@MoisheCohen Yes I'm looking for a counterexample
– mysatellite
Dec 31 '18 at 2:01
1
Choose $F$ surjective but null-homotopic.
– Mike Miller
Dec 31 '18 at 2:20
I don't know if you read the deleted answer and all the comments therein before it was deleted. Are you still interested in an answer?
– Amitai Yuval
Dec 31 '18 at 16:24
@AmitaiYuval yes
– mysatellite
Dec 31 '18 at 17:11