Given a oriented manifold $M$ of dimension $n$ . Let $a_1$, $a_2$ , $a_3$ be three cohomology classes such that $\sum \text{deg}(a_i) = n$. Also suppose that the poincare duals of $a_1$, $a_2$ and $a_3$ are represented by oriented submanifolds $X_1$, $X_2$ and $X_3$ respectively.
Then why is it true that the number of signed triple intersections of $X_1$, $X_2$ and $X_3$ is given by $$ \int_M a_1 \smile a_2 \smile a_3 $$.
I understand the proof for when there are only two homology classes say $b_1$ and $b_2$ (such that $\sum\text{deg}(b_i) = n$) and submanifolds $Y_1$ and $Y_2$ representing $b_i$ respectively. Then we can count the signed intersections by the intersection product of their fundamental classes which by definition is just the integral of their poicare duals.
How does one make this work for triple intersections?
If you understand the case for 2 then the case for three follows since we can call $a_1\smile a_2=b_1$ and $a_3=b_2$, these are now two cohomology classes the first of which is dual to $X_1\cap X_2$ and the second is dual to $X_3$.