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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!

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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.

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  • $\begingroup$ Thanks!! @red_trumpet for your answer. I missed the point that birational morphism of curves is proper. $\endgroup$
    – tota
    Commented Sep 23, 2022 at 14:06
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    $\begingroup$ @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. $\endgroup$ Commented Sep 23, 2022 at 16:28

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