The Stacks project

Proof. By linearity we reduce to the case where $\alpha = [V]$ and $\beta = [W]$ for some closed subvariety $V \subset X$ and $W \subset Y$ of dimension $r$ and $s$. Then $f^{-1}(W)$ has pure dimension $s + e$. We assume the cycles $[V]$ and $f^*[W]$ intersect properly. We will use without further mention the fact that $V \cap f^{-1}(W) \to f(V) \cap W$ is surjective.

Let $a$ be the dimension of the generic fibre of $V \to f(V)$. If $a > 0$, then $f_*[V] = 0$. In particular $f_*\alpha $ and $\beta $ intersect properly. To finish this case we have to show that $f_*([V] \cdot f^*[W]) = 0$. However, since every fibre of $V \to f(V)$ has dimension $\geq a$ (see Morphisms, Lemma 29.28.4) we conclude that every irreducible component $Z$ of $V \cap f^{-1}(W)$ has fibres of dimension $\geq a$ over $f(Z)$. This certainly implies what we want.

Assume that $V \to f(V)$ is generically finite. Let $Z \subset f(V) \cap W$ be an irreducible component. Let $Z_ i \subset V \cap f^{-1}(W)$, $i = 1, \ldots , t$ be the irreducible components of $V \cap f^{-1}(W)$ dominating $Z$. By assumption each $Z_ i$ has dimension $r + s + e - \dim (X) = r + s - \dim (Y)$. Hence $\dim (Z) \leq r + s - \dim (Y)$. Thus we see that $f(V)$ and $W$ intersect properly, $\dim (Z) = r + s - \dim (Y)$, and each $Z_ i \to Z$ is generically finite. In particular, it follows that $V \to f(V)$ has finite fibre over the generic point $\xi $ of $Z$. Thus $V \to Y$ is finite in an open neighbourhood of $\xi $, see Cohomology of Schemes, Lemma 30.21.2. Using a very general projection formula for derived tensor products, we get

see Derived Categories of Schemes, Lemma 36.22.1. Since $f$ is flat, we see that $Lf^*\mathcal{O}_ W = f^*\mathcal{O}_ W$. Since $f|_ V$ is finite in an open neighbourhood of $\xi $ we have

for any coherent sheaf on $X$ whose support is contained in $V$ (see Cohomology of Schemes, Lemma 30.20.8). Thus we conclude that

for all $i$. Since $f^*[W] = [f^*\mathcal{O}_ W]_{s + e}$ by Lemma 43.7.1 we have

by Lemma 43.19.4. Applying Lemma 43.6.1 we find

Since $f_*[V] = [f_*\mathcal{O}_ V]_ r$ by Lemma 43.6.1 we have

again by Lemma 43.19.4. Comparing the formula for $f_*([V] \cdot f^*[W])$ with the formula for $f_*[V] \cdot [W]$ and looking at the coefficient of $Z$ by taking lengths of stalks at $\xi $, we see that (43.22.1.1) finishes the proof. $\square$