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How is following map isomorphism by Poincare duality?

H^{p-1}(M)--------> H^{p}(M) by cup product whit w_1 in H^{1}(M). please help me!

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It seems reasonable to suppose that w_1 means the first Stiefel-Whitney class and therefore that the coefficients of these cohomology groups are supposed to be $\mathbb Z/2$. Some hypotheses on M and p will be needed, because the map surely isn't an isomorphism in general. Offhand, I can't think of reasonable hypotheses, but presumably the source of this question contained some hypotheses. – Andreas Blass Nov 9 2011 at 22:35
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 There's no reason to think that cup product with w1 will be an isomorphism; after all, $w_1$ could be 0, and even if it is non-0, it will generally fail to be a zero-divisor on $H∗(M)$. Please rethink your question! – Charles Rezk Nov 9 2011 at 22:37
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 It seems like you don't learn how to appropriately ask questions, after you've asked problems in the past with the same outcome of comments! To me, this is spam. – Chris Gerig Nov 9 2011 at 22:41

closed as not a real question by Charles Rezk, Will Jagy, Yemon Choi, George Lowther, Andres Caicedo Nov 9 2011 at 22:57

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