# A question about Poincare duality

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 $$H^{2n}$$?

• Do you assume that $H^i(c) = 0$ for $i \ge \dim c$? Otherwise there are obvious counterexamples. – Najib Idrissi Feb 16 at 15:15

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.