# Intersection pairing and birational morphisms

Let $$f:X\to Y$$ be a birational morphism of smooth projective variety. We assume that $$f(V)\simeq U$$ isomorphism induced by $$f$$, where $$V\subset X$$ and $$U\subset Y$$ are two Zariski open sets. Let $$x\in V$$, $$C$$ be a curve passing through $$x$$ in $$X$$ and $$L$$ be a line bundle over $$Y$$. Then is the following true?

$$f^{*}L\cdot C=L\cdot\overline{f(C\cap V)}$$

where $$f^{*}L$$ denotes the pullback of the line bundle $$L$$ and $$\overline{f(C\cap V)}$$ is the Zariski closure of $$f(C\cap V)$$ in $$Y$$.

First note $$\overline{f(C \cap V)} = f(C)$$, since $$f$$ is closed and $$C$$ (and hence also $$f(C)$$) is irreducible. Also $$f$$ induces a birational map $$C \to f(C)$$, so $$f_* [C] = [f(C)]$$ where $$[\cdot]$$ denotes rational equivalence classes.

Then by the projection formula¹ $$f_* (f^* L \cdot [C]) = L \cdot f_* [C] = L \cdot [f(C)].$$

In a more general situation you will get $$f_*(f^*L \cdot [C]) = \deg(f|_C) L \cdot [f(C)]$$, if $$f$$ is not a birational map $$C \to f(C)$$.

¹ see e.g. Fulton's Intersection Theory, Proposition 2.3.

• Thanks!! @red_trumpet for your answer. I missed the point that birational morphism of curves is proper.
– tota
Sep 23 at 14:06
• @tota Not sure if I understand you correctly, but properness just comes from the projectivity of $X$ and $Y$, and has nothing to do with birationality. Sep 23 at 16:28