The Stacks project

Lemma 41.39.2. Let $(S, \delta )$ be as in Situation 41.7.1. Let $X$ be a scheme locally of finite type over $S$. Let

\[ 0 \to \mathcal{N}' \to \mathcal{N} \to \mathcal{E} \to 0 \]

be a short exact sequence of finite locally free $\mathcal{O}_ X$-modules. Consider the closed embedding

\[ N' = \underline{\text{Spec}}_ X(\text{Sym}((\mathcal{N}')^\vee )) \longrightarrow N = \underline{\text{Spec}}_ X(\text{Sym}(\mathcal{N}^\vee )) \]

Then we have

\[ (p')^*\alpha = p^*(c_{top}(\mathcal{E}) \cap \alpha ) \]

where $p' : N' \to X$ and $p : N \to X$ are the structure morphisms.

Proof. Here $c_{top}(\mathcal{E})$ is the bivariant class defined in Remark 41.34.10. By its very definition, in order to verify the formula, we may assume that $\mathcal{E}$ has constant rank. We may similarly assume $\mathcal{N}'$ and $\mathcal{N}$ have constant ranks, say $r'$ and $r$, so $\mathcal{E}$ has rank $r - r'$ and $c_{top}(\mathcal{E}) = c_{r - r'}(\mathcal{E})$. Observe that $p^*\mathcal{E}$ has a canonical section

\[ s \in \Gamma (N, p^*\mathcal{E}) = \Gamma (X, p_*p^*\mathcal{E}) = \Gamma (X, \mathcal{E} \otimes _{\mathcal{O}_ X} \text{Sym}(\mathcal{N}^\vee ) \supset \Gamma (X, \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{N}, \mathcal{E})) \]

corresponding to the surjection $\mathcal{N} \to \mathcal{E}$ given in the statement of the lemma. The vanishing scheme of this section is exactly $N' \subset N$. Let $Y \subset X$ be an integral closed subscheme of $\delta $-dimension $n$. Then we have

  1. $p^*[Y] = [p^{-1}(Y)]$ since $p^{-1}(Y)$ is integral of $\delta $-dimension $n + r$,

  2. $(p')^*[Y] = [(p')^{-1}(Y)]$ since $(p')^{-1}(Y)$ is integral of $\delta $-dimension $n + r'$,

  3. the restriction of $s$ to $p^{-1}Y$ has vanishing scheme $(p')^{-1}Y$ and the closed immersion $(p')^{-1}Y \to p^{-1}Y$ is a regular immersion (locally cut out by a regular sequence).

We conclude that

\[ (p')^*[Y] = c_{r - r'}(p^*\mathcal{E}) \cap p^*[Y] \quad \text{in}\quad A_*(N) \]

by Lemma 41.39.1. This proves the lemma. $\square$

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