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Sorry if the question is too simple, algebraic topology is not my strong suit. Let $(M,\partial M)$ be a $2n$-dimensional manifold with boundary, with one-dimensional middle cohomology. By Poincare-Lefschetz duality and universal coefficients theorem there is a non-degenerate pairing $$\langle \cdot,\cdot\rangle\,:\,H^n(M,\partial M)\times H^n(M)\rightarrow \mathbb{R}.$$ In addition, in the long exact sequence of the pair $(M,\partial M)$, there is a homomorphism $$\alpha\,:\,H^n(M,\partial M)\rightarrow H^n(M).$$ Let $y\in H^n(M)$. Define $y^*$ by the condition $\langle y^*, y\rangle=1$. It is uniquely determined, because $$h^n(M,\partial M)=h_n(M)=h^n(M)=1.$$ Then $\alpha(y^*)=ry$ for some real number $r\in \mathbb{R}$. My question: Is it possible to determine $r$ generally?

Philip Engel
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  • What are $x$ and $y$? Why is $h^n=1$? –  Mar 26 '15 at 20:37
  • You are asking for the value of $\langle x, \alpha(x)\rangle$. To figure this out you should go through the definition of the pairing in the proof of Poincaré-Lefschetz duality. The pairing should be defined as an intersection number. – user39082 Mar 26 '15 at 20:37
  • Yes, it's possible to not only compute $r$ but to compute all the maps. Is that all you want to know? – Ryan Budney Mar 26 '15 at 20:39
  • Alex, it is one of the assumptions of the problem. Sorry, I may have phrased it unclearly, I've edited the question to clarify. –  Mar 26 '15 at 20:43
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    I remove the downvote, but it's still far from research level. It is usually possible to compute $r$ in a particular situation (and it's easier to do that in homology), but there's no universal constant. E.g., consider an $n$-disk bundle over $S^n$; then, up to sign, $r$ is the Euler class. –  Mar 26 '15 at 20:47

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