The Stacks project

Lemma 42.47.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $i_ j : X_ j \to X$, $j = 1, 2$ be closed immersions such that $X = X_1 \cup X_2$ set theoretically. Let $E_2 \in D(\mathcal{O}_{X_2})$ be a perfect object. Assume

  1. Chern classes of $E_2$ are defined,

  2. the restriction $E_2|_{X_1 \cap X_2}$ is zero, resp. isomorphic to a finite locally free $\mathcal{O}_{X_1 \cap X_2}$-module of rank $< p$ sitting in cohomological degree $0$.

Then there is a canonical bivariant class

\[ P'_ p(E_2),\text{ resp. }c'_ p(E_2) \in A^ p(X_2 \to X) \]

characterized by the property

\[ P'_ p(E_2) \cap i_{2, *} \alpha _2 = P_ p(E_2) \cap \alpha _2 \quad \text{and}\quad P'_ p(E_2) \cap i_{1, *} \alpha _1 = 0, \]

respectively

\[ c'_ p(E_2) \cap i_{2, *} \alpha _2 = c_ p(E_2) \cap \alpha _2 \quad \text{and}\quad c'_ p(E_2) \cap i_{1, *} \alpha _1 = 0 \]

for $\alpha _ i \in \mathop{\mathrm{CH}}\nolimits _ k(X_ i)$ and similarly after any base change $X' \to X$ locally of finite type.

Proof. We are going to use the material of Section 42.46 without further mention.

Assume $E_2|_{X_1 \cap X_2}$ is zero. Consider a morphism of schemes $X' \to X$ which is locally of finite type and denote $i'_ j : X'_ j \to X'$ the base change of $i_ j$. By Lemma 42.19.4 we can write any element $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ k(X')$ as $i'_{1, *}\alpha '_1 + i'_{2, *}\alpha '_2$ where $\alpha '_2 \in \mathop{\mathrm{CH}}\nolimits _ k(X'_2)$ is well defined up to an element in the image of pushforward by $X'_1 \cap X'_2 \to X'_2$. Then we can set $P'_ p(E_2) \cap \alpha ' = P_ p(E_2) \cap \alpha '_2 \in \mathop{\mathrm{CH}}\nolimits _{k - p}(X'_2)$. This is well defined by our assumption that $E_2$ restricts to zero on $X_1 \cap X_2$.

If $E_2|_{X_1 \cap X_2}$ is isomorphic to a finite locally free $\mathcal{O}_{X_1 \cap X_2}$-module of rank $< p$ sitting in cohomological degree $0$, then $c_ p(E_2|_{X_1 \cap X_2}) = 0$ by rank considerations and we can argue in exactly the same manner. $\square$

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