Let $N^{4k+1}$ be a compact oriented manifold with boundary $i:M^{4k} \hookrightarrow N$. Suppose $c \in H^{4k}(N,A)$ for some abelian group $A$. I have to prove that $ \langle i^*(c), [M] \rangle =0 $. In order to do this I'd like to prove that $$ \langle i^*(c),[M] \rangle = \langle c, i_*[M] \rangle .$$ But why this identity is true? I denote with $[M] \in H_n(M)$ the fundalmental class and with $c$ an element in $H^n(M;A)$. Then how can I conclude my thesis?
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the identity $\langle f^c,r\rangle=\langle c,f_r$ is true for any $f:X\to Y$ and any $c\in H^k(Y,A)$, $r\in H_k(X)$; it's true already for singular chains/cochains (to see that $i_*[M]=0$ use a triangulation of $N$) – user8268 Sep 08 '13 at 18:45
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@user8268 But why is it true? – user93772 Sep 08 '13 at 20:53
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because it's the definition of $f^*$ – user8268 Sep 08 '13 at 20:56
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I think you just need to write everything out explicitly here;
$$<i^*(c),y> = i^*(c)(y) = c(i(y)) = c(i_*(y)) = <c,i_*(y)> $$
Be careful to understand what is going on at each step.
EHH
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