Let $f: X \to Y$ be a birational morphism between proper integral surfaces, $\mathcal{L}, \mathcal{N}$ two invertible sheaves on $X$, and $\mathcal{L}'= f^*\mathcal{L}, \mathcal{N}= f^*\mathcal{N}$ their pull backs.
My question is how to see that for the intersection holds
$$(\mathcal{L},\mathcal{N})= (\mathcal{L}',\mathcal{N}')$$
My ideas: I want to apply the formula $(\mathcal{L},\mathcal{N}) = deg(\mathcal{L} \vert _{D(\mathcal{N})})$ where $D(\mathcal{N})$ is the corresponding Cartier divisor to $\mathcal{N}$.
But to apply this formula I have to get Cartier divisors $D(\mathcal{L}),D(\mathcal{N})$ which hasn't common irreducible component. How can I force it?
I know futhermore that every invertible sheaf can be expressed as "difference" of tho ample sheaves, so by bilinearity of intersection I can find candidates for the Cartier divisors. But how to make them don't having common irreducible components and pass the equality to the pull back?
The equality follows from the more general projection formula, which states that if $\pi: Y\rightarrow X$ is a surjective morphism between proper varieties, and $D_1,\ldots,D_r$ are Cartier divisors on $X$ with $r\ge \dim(Y)$, then $$\pi^*D_1 \cdot \ldots \pi^* D_r = \deg(\pi) (D_1\cdot \ldots \cdot D_r)$$. The proof of this can be found in Proposition 1.10 in Debarre's book on higher dimensional algebraic geometry.
Using this your claim is immediate, since $\deg(f) = 1$ as $f$ is birational.
Alternatively, there is another way of proving the projection formula in higher generality following intersection theory in Fulton. This is covered in Chapter 2 of Fulton's Intersection Theory.