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Let $M^n$ be a connected, closed manifold. It has Poincaré duality with $\mathbb{Z}/2$ coefficients $H^k(M;\mathbb{Z}/2)\cong H_{n-k}(M;\mathbb{Z}/2)$, induced by cap product with the fundamental class $[M]_2\in H_n(M;\mathbb{Z}/2)$. It also has Poincaré duality with twisted integral coefficients, induced by cap product with a twisted fundamental class. It has Poincaré duality with $\mathbb{Z}$ coefficients if and only if it's orientable.

Let us say that $M$ satisfies "almost Poincaré duality in dimension $k$" if for every $x\in H^k(M;\mathbb{Z})$ there exists $a\in H_{n-k}(N;\mathbb{Z})$ such that $$ \rho(x)\cap[M]_2 = \rho(a), $$ where $\rho$ denotes reduction mod $2$ in both cohomology and homology. That is, the Poincaré dual of the reduction of an integral cohomology class admits an integral lift.

Examples:

  • An orientable manifold satisfies almost Poincaré duality in all dimensions (because reduction commutes with the duality isomorphisms).
  • A non-orientable manifold cannot satisfy almost Poincaré duality in dimension $0$, since $[M]_2$ is not the reduction of an integral class.
  • $\mathbb{R}P^{2m}$ satisfies almost Poincaré duality dimension $k$ if and only if $k$ is odd or $k=2m$.

Question: Are there conditions we can place on the (co)homology of $M$ which imply that it satisfies almost Poincaré duality in a certain dimension? I am particularly interested in closed $4$-manifolds which satisfy almost Poincaré duality in dimension $2$.

More generally, I would be interested to know if this notion has come up in the literature anywhere before.

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    $\begingroup$ Just an observation: "almost duality in dimension $k$ is the same as "every element of $H^k(M; \mathbb Z/2)$ that is in the image of integral cohomology is also in the image of twisted integral cohomology". $\endgroup$ – Tom Goodwillie May 19 '17 at 17:47
  • $\begingroup$ @TomGoodwillie: Yes! That's actually why I got interested in this property. I thought this formulation might arouse more interest, but now that I think about it, maybe it obscures the fact that any such condition would have to involve not only the (co)homology groups but also $w_1(M)$. $\endgroup$ – Mark Grant May 19 '17 at 18:58

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