One form of poincare duality imply the other? I saw in my differential geometry class that Poincare duality states:
$H_{DR}^k(X,\mathbb{R}) \times H_{DR}^{n - k}(X,\mathbb{R})  \rightarrow \mathbb{R}$
given by $([v],[r]) \mapsto \int_{X} n \wedge r$ is non-singular 
Prof mentioned that If X is compact oriented manifold of dimension n, then this is really the same as saying:
$H_{DR}^i(X,\mathbb{R}) \cong H_{n - i}(X,\mathbb{R})$ where the right side of the equation we have singular homology. Can someone explain why this is the case ? 
 A: The first statement gives an isomorphism $H_{dR}^k(X,\mathbb{R})\to H_{dR}^{n-k}(X,\mathbb{R})^\vee$ where $^\vee$ denotes the dual vector space.  By the de Rham theorem, $H^{n-k}_{dR}(X,\mathbb{R})\cong H^{n-k}(X,\mathbb{R})$ (where the right-hand side is singular cohomology), and by the universal coefficient theorem $H^{n-k}(X,\mathbb{R})\cong H_{n-k}(X,\mathbb{R})^\vee$.  So putting these isomorphisms together, you get $H^k_{dR}(X,\mathbb{R})\cong (H_{n-k}(X,\mathbb{R})^\vee)^\vee\cong H_{n-k}(X,\mathbb{R})$.
Conversely, if you start with $H^k_{dR}(X,\mathbb{R})\cong H_{n-k}(X,\mathbb{R})$ you can reverse these steps and obtain an isomorphism $H^k_{dR}(X,\mathbb{R})\cong H_{dR}^{n-k}(X,\mathbb{R})^\vee$.  However, this isn't quite as strong as your first statement, because it doesn't tell you exactly what the isomorphism in question is (namely, the isomorphism given by the perfect pairing you wrote).
A: There's no geometry here. It's an algebra thing. See Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott Section 5.

