# Poincaré Duality and Intersection Pairing

Suppose $M$ is an oriented compact $n$-dimensional manifold, then Poincaré duality is an isomorphism \begin{equation} \alpha \in H^i(M,\mathbb{Z}) \mapsto [M] \cap \alpha \in H_{n-i}(M,\mathbb{Z}),~ \alpha \end{equation} where $[M]$ is the homology class associated to the manifold $M$. For details, see e.g. Algebraic Topology by Allen Hatcher.

In Griffiths and Harris' book Principles of Algebraic Geometry, the intersection of two homology classes $A$ and $B$ is defined, and if we let $A^\vee$ be the Poincaré dual of $A$, etc, then there is \begin{equation} A^\vee \cup B^\vee =(A\cap B)^{\vee} \end{equation}

I am wondering whether there are interesting explanations of this equality or interesting examples which could show the essence of it?

Here is an example. Consider the torus, $T = S^1 \times S^1$. You have two obvious generators for the homology, $a$ and $b$, which go exactly once around either the first $S^1$ or the second $S^1$. These two homology classes have Poincaré dual cohomology classes $\alpha$ and $\beta$.
Now how do you compute the cup products? For $\alpha \smile \beta$, your formula tells you that you can consider the intersection $a \frown b$. This is just a point. The Poincaré dual of a point is the top cohomology class $\omega \in H^2(T)$, hence $\alpha \smile \beta = \omega$. That was easy!
And let's say you also want to compute $\alpha \smile \alpha$ (nevermind that we know it's zero for algebraic reasons: $\alpha$ has odd degree...). Your formula tells you that you need to consider the intersection $a \frown a$. Of course you can't take the intersection randomly, it has to be transverse intersection. So you move $a$ a little bit to the right, it becomes $a'$, the same homology class, but now the intersection is empty (hence transverse): $a \frown a' = \varnothing$, the zero homology class. So $\alpha \smile \alpha = 0$, because the Poincaré dual of $0$ is $0$.