TITLE: Geometric and topological ways to define intersection number
QUESTION [14 upvotes]: Resently I'm reading Bott-Tu's differential forms in algebraic topology, and comparing it with some differential topology textbook.
While proving the Poincare-Hopf theorem, it defines the intersection number $I(M,N)$, where M, N are submanifold of another smooth manifold K with condition $\dim M+\dim N=\dim K$, to be $I(M,N)=\int_K \eta_M \wedge\eta_N.$ Here $\eta_M$ is the Poincare dual of M, so $\eta_M \wedge\eta_N$ is a top form in K, assuming the dimension condition.
My question is why this definition is coincide with the definition appeared in differential topology, which using the orientable intersection number that "sum up" $\pm 1$  locally according to the orientation.
If it's possible, I perfer an answer without using much algebraic geometry. Any links to webpage or other reference are welcomed as well. Great Thanks!

REPLY [3 votes]: Assuming readers are familiar with Bott-Tu's book, so I'll use the language in their book. Set $\dim M=m,\ \dim N=n$.
It is well known that Thom isomophism gives $\mathcal{T}:H^*(M)\cong H^{*+k}_{cv}(E)$, where E is a vector bundle over M with rank k, given by intergral along the fiber. 
Now $M$ is a submanifold of $K$, so there's a tubular neighborhood $T$, which is diffeomorphism to the normal bundle $S$ in $K$. Let$\Phi =\mathcal{T}(1) \in H^{m}(T)$ be the Thom class,$\ j:T \to M$ be the inclusion, then it's proved in Bott-Tu that the Poincare dual of M $\eta_M=j^*\Phi$.
When computing $\int_K \eta_M \wedge\eta_N$, using Poincare duality and our discussion above, we got $$\int_K \eta_M \wedge\eta_n=\int_N\eta_M|_N= \sum_{p\in M \cap N} \int_{B_{\epsilon}\ (p)\cap N}\eta_M .$$ By choosing a Riemannian metric on K, we might assume that, at every point $p\in M \cap N$, $T_p M \bot T_pN$. (The metric is clearly exist, since such $p$ is finite, so we could take some alart, choosing metric on it, then using partition of unity.) So at point $p$, pull back of Thom class $j^*\Phi_p$ lies in $T_pN$. Besides, Thom isomorphism gives by intergral along the fiber, so intergral of Thom class along fiber $S_p = T_pN$ is $\pm 1$, determinded by orientation.
Thus we only have to chech the signal of each intergral. Let $\omega_M$ be the volumn form on N, then $\omega_M \wedge \eta_M$ is a positive orientation on $T_p K =T_pS \oplus T_pM$. Therefore, if $\omega_M \wedge \omega_N$ is a positive orientation, then the orientation on $T_pS$ and $T_pN$ should be same, leading to the integral of$j^*\Phi_p \in T_pN$ shouldn't change its signal, and the negative part is exactly the same.