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 $Y$.


Any suggestions/comments are welcome!



Intersection pairing and birational morphisms
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 $C \to f(C)$.




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


