The Stacks project

Lemma 41.53.4. In the situation described just above assume $\dim _\delta (Y) = n$ and that $\mathcal{C}_{Y \times _ X Z/Z}$ has constant rank $r$. Then

\[ c(Z \to X, \mathcal{N}) \cap [Y]_ n = c_{top}(\mathcal{E}) \cap [Z \times _ X Y]_{n - r} \]

in $\mathop{\mathrm{CH}}\nolimits _*(Z \times _ X Y)$.

Proof. The bivariant class $c_{top}(\mathcal{E}) \in A^*(Z \times _ X Y)$ was defined in Remark 41.37.11. By Lemma 41.53.2 we may replace $X$ by $Y$. Thus we may assume $Z \to X$ is a regular closed immersion of codimension $r$, we have $\dim _\delta (X) = n$, and we have to show that $c(Z \to X, \mathcal{N}) \cap [X]_ n = c_{top}(\mathcal{E}) \cap [Z]_{n - r}$ in $\mathop{\mathrm{CH}}\nolimits _*(Z)$. By Lemma 41.53.3 we may even assume $\mathcal{N}^\vee \to \mathcal{C}_{Z/X}$ is an isomorphism. In other words, we have to show $c(Z \to X, \mathcal{C}_{Z/X}^\vee ) \cap [X]_ n = [Z]_{n - r}$ in $\mathop{\mathrm{CH}}\nolimits _*(Z)$.

Let us trace through the steps in the definition of $c(Z \to X, \mathcal{C}_{Z/X}^\vee ) \cap [X]_ n$. Let $b : W \to \mathbf{P}^1_ X$ be the blowing up of $\infty (Z)$. We first have to compute $C \cap [X]_ n$ where $C \in A^0(W_\infty \to X)$ is the class of Lemma 41.47.1. To do this, note that $[W]_{n + 1}$ is a cycle on $W$ whose restriction to $\mathbf{A}^1_ X$ is equal to the flat pullback of $[X]_ n$. Hence $C \cap [X]_ n$ is equal to $i_\infty ^*[W]_{n + 1}$. Since $W_\infty $ is an effective Cartier divisor on $W$ we have $i_\infty ^*[W]_{n + 1} = [W_\infty ]_ n$, see Lemma 41.28.5. The restriction of this class to the open $C_ ZX \subset W_\infty $ is of course just $[C_ ZX]_ n$. Because $Z \subset X$ is regularly embedded we have

\[ \mathcal{C}_{Z/X, *} = \text{Sym}(\mathcal{C}_{Z/X}) \]

as graded $\mathcal{O}_ Z$-algebras, see Divisors, Lemma 30.21.5. Hence $p : N = C_ ZX \to Z$ is the structure morphism of the vector bundle associated to the finite locally free module $\mathcal{C}_{Z/X}$ of rank $r$. Then it is clear that $p^*[Z]_{n - r} = [C_ ZX]_ n$ and the proof is complete. $\square$


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