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Let $k$ be a field. Let $C$ be a small category and assume that for every $i\geq 0$ we have a functor $H^i:C\rightarrow FinDimVect_k$. Assume that there is a function $dim:Obj(C)\rightarrow \mathbb{Z}_{\geq 0}$ such that for every $c\in Obj(C)$ and every $0\leq i \leq 2 dim(c)$ 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 $i\geq 0$ the spaces $H^i(c_1)$ and $H^i(c_2)$ are non-isomorphic. Is it true that there can not simultaneously exist morphisms $f:c_1\rightarrow c_2$ and $g:c_2\rightarrow c_1$ 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
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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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