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 $H^{2n}$?






at.algebraic-topology ct.category-theory cohomology







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    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 $H^{2n}$?






    at.algebraic-topology ct.category-theory cohomology







    share|cite|improve this question









    New contributor




    paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







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      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 $H^{2n}$?






      at.algebraic-topology ct.category-theory cohomology







      share|cite|improve this question









      New contributor




      paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      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






      share|cite|improve this question









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      paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question









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      edited 4 hours ago









      YCor

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      asked 9 hours ago









      paulpaul

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          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.






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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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                7





                $begingroup$

                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 7 hours ago









                John RognesJohn Rognes

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