A question about Poincare duality












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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 $H^{2n}$?










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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
















6












$begingroup$


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 $H^{2n}$?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














6












6








6


1



$begingroup$


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 $H^{2n}$?










share|cite|improve this question











$endgroup$




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






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edited Feb 16 at 0:04









YCor

29.1k487141




29.1k487141










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


















  • $begingroup$
    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










1 Answer
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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.






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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.






    share|cite|improve this answer









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      8












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







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        answered Feb 15 at 20:43









        John RognesJohn Rognes

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        4,6612527






























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