# Why is this map about Poincaré duality surjective?

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 ?

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 $f(\phi)=\phi(A)$.
• 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 $B$ ? – Lina45 Aug 9 '16 at 20:33
• In $(A,B)\rightarrow \int_A\psi$ what is $\psi$ ? – Tsemo Aristide Aug 9 '16 at 20:36
• $\psi \in H^{n-k} ( M )$. So : $\psi \to \int_A \psi \in H_{n-k} (M)$, right ? so, $B$ is $\psi \to \int_A \psi$, no ? Thank you. :-) – Lina45 Aug 9 '16 at 20:38