A question about Poincare duality
$begingroup$
Let $k$ be a field. Let $C$ be a small category and assume that for every $igeq 0$ we have a functor $H^i:Crightarrow FinDimVect_k$. Assume that there is a function $dim:Obj(C)rightarrow mathbb{Z}_{geq 0}$ such that for every $cin Obj(C)$ and every $0leq i leq 2 dim(c)$ we have a functorial perfect pairing $H^i(c)times H^{2dim(c)-i}(c)rightarrow H^{2dim(c)}(c)$.
Now assume we have two objects $c_1$ and $c_2$ such that $dim(c_1)=dim(c_2)=n$ and such that for some $igeq 0$ the spaces $H^i(c_1)$ and $H^i(c_2)$ are non-isomorphic. Is it true that there can not simultaneously exist morphisms $f:c_1rightarrow c_2$ and $g:c_2rightarrow c_1$ inducing isomorphisms in $H^{2n}$?
at.algebraic-topology ct.category-theory cohomology
edited Feb 16 at 0:04
YCor
29.1k487141
asked Feb 15 at 19:17
paulpaul
1604
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field. Let $C$ be a small category and assume that for every $igeq 0$ we have a functor $H^i:Crightarrow FinDimVect_k$. Assume that there is a function $dim:Obj(C)rightarrow mathbb{Z}_{geq 0}$ such that for every $cin Obj(C)$ and every $0leq i leq 2 dim(c)$ we have a functorial perfect pairing $H^i(c)times H^{2dim(c)-i}(c)rightarrow H^{2dim(c)}(c)$.
Now assume we have two objects $c_1$ and $c_2$ such that $dim(c_1)=dim(c_2)=n$ and such that for some $igeq 0$ the spaces $H^i(c_1)$ and $H^i(c_2)$ are non-isomorphic. Is it true that there can not simultaneously exist morphisms $f:c_1rightarrow c_2$ and $g:c_2rightarrow c_1$ inducing isomorphisms in $H^{2n}$?
at.algebraic-topology ct.category-theory cohomology
edited Feb 16 at 0:04
YCor
29.1k487141
asked Feb 15 at 19:17
paulpaul
1604
$endgroup$
$begingroup$
Do you assume that $H^i(c) = 0$ for $i ge dim c$? Otherwise there are obvious counterexamples.
$endgroup$
– Najib Idrissi
Feb 16 at 15:15
add a comment |
$begingroup$
Let $k$ be a field. Let $C$ be a small category and assume that for every $igeq 0$ we have a functor $H^i:Crightarrow FinDimVect_k$. Assume that there is a function $dim:Obj(C)rightarrow mathbb{Z}_{geq 0}$ such that for every $cin Obj(C)$ and every $0leq i leq 2 dim(c)$ we have a functorial perfect pairing $H^i(c)times H^{2dim(c)-i}(c)rightarrow H^{2dim(c)}(c)$.
Now assume we have two objects $c_1$ and $c_2$ such that $dim(c_1)=dim(c_2)=n$ and such that for some $igeq 0$ the spaces $H^i(c_1)$ and $H^i(c_2)$ are non-isomorphic. Is it true that there can not simultaneously exist morphisms $f:c_1rightarrow c_2$ and $g:c_2rightarrow c_1$ inducing isomorphisms in $H^{2n}$?
at.algebraic-topology ct.category-theory cohomology
edited Feb 16 at 0:04
YCor
29.1k487141
asked Feb 15 at 19:17
paulpaul
1604
$endgroup$
Let $k$ be a field. Let $C$ be a small category and assume that for every $igeq 0$ we have a functor $H^i:Crightarrow FinDimVect_k$. Assume that there is a function $dim:Obj(C)rightarrow mathbb{Z}_{geq 0}$ such that for every $cin Obj(C)$ and every $0leq i leq 2 dim(c)$ we have a functorial perfect pairing $H^i(c)times H^{2dim(c)-i}(c)rightarrow H^{2dim(c)}(c)$.
Now assume we have two objects $c_1$ and $c_2$ such that $dim(c_1)=dim(c_2)=n$ and such that for some $igeq 0$ the spaces $H^i(c_1)$ and $H^i(c_2)$ are non-isomorphic. Is it true that there can not simultaneously exist morphisms $f:c_1rightarrow c_2$ and $g:c_2rightarrow c_1$ inducing isomorphisms in $H^{2n}$?
at.algebraic-topology ct.category-theory cohomology
at.algebraic-topology ct.category-theory cohomology
edited Feb 16 at 0:04
YCor
29.1k487141
asked Feb 15 at 19:17
paulpaul
1604
edited Feb 16 at 0:04
YCor
29.1k487141
asked Feb 15 at 19:17
paulpaul
1604
edited Feb 16 at 0:04
YCor
29.1k487141
edited Feb 16 at 0:04
YCor
29.1k487141
edited Feb 16 at 0:04
YCor
29.1k487141
29.1k487141
asked Feb 15 at 19:17
paulpaul
1604
asked Feb 15 at 19:17
paulpaul
1604
asked Feb 15 at 19:17
paulpaul
1604
1604
$begingroup$
Do you assume that $H^i(c) = 0$ for $i ge dim c$? Otherwise there are obvious counterexamples.
$endgroup$
– Najib Idrissi
Feb 16 at 15:15
add a comment |
$begingroup$
Do you assume that $H^i(c) = 0$ for $i ge dim c$? Otherwise there are obvious counterexamples.
$endgroup$
– Najib Idrissi
Feb 16 at 15:15
$begingroup$
Do you assume that $H^i(c) = 0$ for $i ge dim c$? Otherwise there are obvious counterexamples.
$endgroup$
– Najib Idrissi
Feb 16 at 15:15
$begingroup$
Do you assume that $H^i(c) = 0$ for $i ge dim c$? Otherwise there are obvious counterexamples.
$endgroup$
– Najib Idrissi
Feb 16 at 15:15
add a comment |
1 Answer
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votes
$begingroup$
It is true. For ordinary cohomology this is a standard result about degree one maps; I think it is discussed at the beginning of Browder's book "Surgery on simply-connected manifolds". Let me treat each $H^i$ as a contravariant functor, and write $cup$ for the functorial perfect pairing. Suppose that $f : c_1 to c_2$ is such that $H^{2n}(f)$ is an isomorphism. Then $H^i(f) : H^i(c_2) to H^i(c_1)$ is injective, since if $H^i(f)(x) = 0$ then $H^{2n}(f)(x cup y) = H^i(f)(x) cup H^{2n-i}(f)(y) = 0$ for all $y in H^{2n-i}(c_2)$, which implies $x cup y = 0$, since $H^{2n}(f)$ is an isomorphism. This implies $x = 0$, since $y in H^{2n-i}(c_2)$ was arbitrary and $cup$ is perfect. Hence $dim H^i(c_2) le dim H^i(c_1)$. The same argument for $g$ implies the opposite inequality, so $H^i(c_1)$ and $H^i(c_2)$ are $k$-vector spaces of the same finite dimension, hence are isomorphic.
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
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add a comment |
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$begingroup$
It is true. For ordinary cohomology this is a standard result about degree one maps; I think it is discussed at the beginning of Browder's book "Surgery on simply-connected manifolds". Let me treat each $H^i$ as a contravariant functor, and write $cup$ for the functorial perfect pairing. Suppose that $f : c_1 to c_2$ is such that $H^{2n}(f)$ is an isomorphism. Then $H^i(f) : H^i(c_2) to H^i(c_1)$ is injective, since if $H^i(f)(x) = 0$ then $H^{2n}(f)(x cup y) = H^i(f)(x) cup H^{2n-i}(f)(y) = 0$ for all $y in H^{2n-i}(c_2)$, which implies $x cup y = 0$, since $H^{2n}(f)$ is an isomorphism. This implies $x = 0$, since $y in H^{2n-i}(c_2)$ was arbitrary and $cup$ is perfect. Hence $dim H^i(c_2) le dim H^i(c_1)$. The same argument for $g$ implies the opposite inequality, so $H^i(c_1)$ and $H^i(c_2)$ are $k$-vector spaces of the same finite dimension, hence are isomorphic.
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
$endgroup$
add a comment |
$begingroup$
It is true. For ordinary cohomology this is a standard result about degree one maps; I think it is discussed at the beginning of Browder's book "Surgery on simply-connected manifolds". Let me treat each $H^i$ as a contravariant functor, and write $cup$ for the functorial perfect pairing. Suppose that $f : c_1 to c_2$ is such that $H^{2n}(f)$ is an isomorphism. Then $H^i(f) : H^i(c_2) to H^i(c_1)$ is injective, since if $H^i(f)(x) = 0$ then $H^{2n}(f)(x cup y) = H^i(f)(x) cup H^{2n-i}(f)(y) = 0$ for all $y in H^{2n-i}(c_2)$, which implies $x cup y = 0$, since $H^{2n}(f)$ is an isomorphism. This implies $x = 0$, since $y in H^{2n-i}(c_2)$ was arbitrary and $cup$ is perfect. Hence $dim H^i(c_2) le dim H^i(c_1)$. The same argument for $g$ implies the opposite inequality, so $H^i(c_1)$ and $H^i(c_2)$ are $k$-vector spaces of the same finite dimension, hence are isomorphic.
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
$endgroup$
add a comment |
$begingroup$
It is true. For ordinary cohomology this is a standard result about degree one maps; I think it is discussed at the beginning of Browder's book "Surgery on simply-connected manifolds". Let me treat each $H^i$ as a contravariant functor, and write $cup$ for the functorial perfect pairing. Suppose that $f : c_1 to c_2$ is such that $H^{2n}(f)$ is an isomorphism. Then $H^i(f) : H^i(c_2) to H^i(c_1)$ is injective, since if $H^i(f)(x) = 0$ then $H^{2n}(f)(x cup y) = H^i(f)(x) cup H^{2n-i}(f)(y) = 0$ for all $y in H^{2n-i}(c_2)$, which implies $x cup y = 0$, since $H^{2n}(f)$ is an isomorphism. This implies $x = 0$, since $y in H^{2n-i}(c_2)$ was arbitrary and $cup$ is perfect. Hence $dim H^i(c_2) le dim H^i(c_1)$. The same argument for $g$ implies the opposite inequality, so $H^i(c_1)$ and $H^i(c_2)$ are $k$-vector spaces of the same finite dimension, hence are isomorphic.
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
$endgroup$
It is true. For ordinary cohomology this is a standard result about degree one maps; I think it is discussed at the beginning of Browder's book "Surgery on simply-connected manifolds". Let me treat each $H^i$ as a contravariant functor, and write $cup$ for the functorial perfect pairing. Suppose that $f : c_1 to c_2$ is such that $H^{2n}(f)$ is an isomorphism. Then $H^i(f) : H^i(c_2) to H^i(c_1)$ is injective, since if $H^i(f)(x) = 0$ then $H^{2n}(f)(x cup y) = H^i(f)(x) cup H^{2n-i}(f)(y) = 0$ for all $y in H^{2n-i}(c_2)$, which implies $x cup y = 0$, since $H^{2n}(f)$ is an isomorphism. This implies $x = 0$, since $y in H^{2n-i}(c_2)$ was arbitrary and $cup$ is perfect. Hence $dim H^i(c_2) le dim H^i(c_1)$. The same argument for $g$ implies the opposite inequality, so $H^i(c_1)$ and $H^i(c_2)$ are $k$-vector spaces of the same finite dimension, hence are isomorphic.
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
answered Feb 15 at 20:43
John RognesJohn Rognes
4,6612527
4,6612527
add a comment |
add a comment |
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$begingroup$
Do you assume that $H^i(c) = 0$ for $i ge dim c$? Otherwise there are obvious counterexamples.
$endgroup$
– Najib Idrissi
Feb 16 at 15:15