# Intersection form and poincaré duality

Let $M$ be a $2n$-dimensional compact connected oriented smooth manifold and let $A$, $B$ be two $n$-dimensional submanifolds that intersect transversally. Denote by $A \cdot B$ the sum of the algebraic intersection numbers defined by the orientations of $A$ and $B$.

The intersection pairing is defined on $H^n(M;\mathbb{Z})$ by $\alpha \cdot \beta =<\alpha \cup \beta , [M] >$ where $[M]$ is the orientation class of $M$. Denote by $D:H_n(M) \rightarrow H^n(M)$ the inverse of the Poincaré duality isomorphism. The intersection pairing can be defined on $H_n(M;\mathbb{Z})$ via $a \cdot b = D(a) \cdot D(b)$.

One desires to show that $a \cdot b = A \cdot B$ where $A$ and $B$ are two submanifolds representing $a$ and $b$ in the sense that $a=i_*[A]=:[A]_M$ and $b=i_*[B]=:[B]_M$ where the $i$'s are the inclusions maps.

Suppose that one already knows the following equation $$[A \cap B]^*=[A]^* \cup [B]^*$$ (where $[A]^*=D([A]_M$.)

Is the following reasonning correct?

As $M$ is connected, we have \begin{align*} [A \cap B]_M &= (A \cdot B) [pt] \end{align*} where $[pt]$ is the generator of $H_0(M)$. Passing to the Poincaré dual gives, \begin{align*} [A \cap B]^* &= (A \cdot B) [pt]^* \end{align*}

Evaluating the known expression at $[M]$ gives \begin{align*} ([A]^* \cup [B]^*)[M]&= [A \cap B]^*([M]) \\ &= (A \cdot B) [pt]^* ([M]) \\ &= A \cdot B \end{align*} because the orientation class is dual to the poincaré dual of generator of $H_0(M)$

Finally using the above notations, we have \begin{align*} a \cdot b &= D{a} \cdot D{b} \\ &=D([A]_M) \cdot D([B]_M) \\ &=[A]^* \cdot [B]^* \\ &=([A]^* \cup [B]^*)[M] \\ &= A \cdot B \end{align*}

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