Is there an example of compact nonorientable $n$-manifold s.t. $H^i(M)\cong H_{n-i}(M)$ fails ?
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Presumably you mean nonorientable, not unoriented. – anomaly Nov 08 '16 at 05:35
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@anomaly thanks, I have fixed – 6666 Nov 08 '16 at 05:37
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Take $M = \mathbb{RP}^2$. It's compact, being a quotient of $S^2$, but it's easy to show that $H_1(M) = \mathbb{Z}_2$ and $H^1(M) = 0$.
anomaly
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Every connected compact nonorientable manifold is a counterexample. Indeed, $H_n(M)=0$ and $H^0(M)=\mathbb{Z}$ for any such $M$, giving a counterexample with $i=0$.
Eric Wofsey
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This is a nontrivial statement that is part of the theory of Poincare duality. See, for instance, Theorem 3.26(b) in Hatcher. – Eric Wofsey Nov 08 '16 at 07:00