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Let $f:X\rightarrow Y$ be a surjective finite morphism of varieties, with $X$ normal and $Y$ smooth. Let $D\subset X$ be a divisor and $C\subset Y$ a curve. Does the equality $$C\cdot f_{*}D = f^{*}C\cdot D$$ always hold under these hypothesis?

Thank you very much.

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  • $\begingroup$ Is $D$ a Cartier divisor? $\endgroup$ – Mohan Nov 17 '20 at 17:47
  • $\begingroup$ Yes, $D$ is a Cartier divisor on $X$. $\endgroup$ – SmY Nov 17 '20 at 17:57
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Let $W = C \times_Y X$. I imagine $f^*C$ is suitably interpreted as the chow class on $W$ given by $C \cdot X$. Write $i : C \subseteq Y$, $f': W \to C$, $i' : W \to X$. (Suppose $i$ is l.c.i. so $i^!$ makes sense, or use obstruction theories. This is automatic if $C, Y$ are smooth). Then Gysin pullback and pushforward commute, so $i^! f_* [D] = f'_* i'^![D] \in A_*(C)$. This is one interpretation of your formula, but beware that $i'^!$ and $i^!$ may differ by the "excess intersection formula" if $f$ isn't flat.

If you're just interested in verifying that the intersection numbers agree, those are given by pushing forward to a point.

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  • $\begingroup$ Thank you for the answer. Just a question. Where did you use that $f:X\rightarrow Y$ is finite in your argument? $\endgroup$ – SmY Nov 19 '20 at 8:50
  • $\begingroup$ I suppose the same argument works for f proper, but be careful because in that case the pushforward could just send D to zero. $\endgroup$ – Leo Herr Nov 25 '20 at 3:13

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