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Can anyone explain to me what it is said in the following article : toperkin.mysite.syr.edu/talks/intersections.pdf , page : $3$, by Mr. Tony perkins :

The corollary says :

The pairing :

$$ H_{ \mathrm{DR} }^{k} (M) \otimes H_{ \mathrm{DR} }^{n-k} (M) \to \mathbb{R} $$ given by : $$ ( [ \phi ] , [ \psi ]) \to \int_M \phi \wedge \psi $$ is non degenerate, or that for any closed $k$ - form $ \phi $ on $ M $ there exists an $ (n-k) $ - cycle $ A $, unique up to homology, such that for any closed $ (n-k) $ - form $ \psi $, $$ \int_M \phi \wedge \psi = \int_A \psi $$ So, what i'm not able to understand, is, why is, for any closed $k$ - form $ \phi $ on $ M $ there exists an $ (n-k) $ - cycle $ A $, unique up to homology, such that for any closed $ (n-k) $ - form $ \psi $, $$ \int_M \phi \wedge \psi = \int_A \psi $$ In other words, why is : $ H_{n-k} ( M ) \to H^{k} (M ) $ surjective ? Is this the analogue of the Hodge conjecture for the real case ?

Thanks in advance for your help.

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Think about singular cohomology,$\psi\rightarrow \int_M \phi\wedge \psi$ is a linear for defined on $H^k_{DR}(M)=H^k_{Sing}(M)$, and the dual space (for real coefficients) of $H^k_{Sing}(M)$ is $H_k(M)$ via the universal coefficient theorem. So $H^{n-k}(M,R)=H_{n-k}(M,R)^*$ so $H_{n-k}(M,R)^{**}=H^{n-k}(M,R)^*$ with the bidual identification, so for every linear form $f$ in $H^{n-k}(M,R)$ there exists a class $A\in H^{n-k}(M,R)$ such that $f(\phi)=\phi(A)$.

Is homology with coefficients in a field isomorphic to cohomology?

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  • $\begingroup$ Thank you, $ H_{n-k} (M) \to H^k (M) = \mathrm{Hom} ( H_k (M) , \mathbb{R} ) $ is surjective means that, $ H_{n-k} (M) \times H_k (M) \to \mathbb{R} $ is non degenerate, right ? but, what is the expression of $ H_{n-k} (M) \times H_k (M) \to \mathbb{R} $ ? is it : $ (A , B) \to \int_A \psi $, but where is $ B $ ? $\endgroup$ – Lina45 Aug 9 '16 at 20:33
  • $\begingroup$ In $(A,B)\rightarrow \int_A\psi$ what is $\psi$ ? $\endgroup$ – Tsemo Aristide Aug 9 '16 at 20:36
  • $\begingroup$ $ \psi \in H^{n-k} ( M ) $. So : $ \psi \to \int_A \psi \in H_{n-k} (M) $, right ? so, $ B $ is $ \psi \to \int_A \psi $, no ? Thank you. :-) $\endgroup$ – Lina45 Aug 9 '16 at 20:38
  • $\begingroup$ Yes it is something like that $\endgroup$ – Tsemo Aristide Aug 9 '16 at 20:39

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