Lemma 42.34.5. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then

1. $c_1(\mathcal{L}) \in A^1(X)$ is in the center of $A^*(X)$ and

2. if $f : X' \to X$ is locally of finite type and $c \in A^*(X' \to X)$, then $c \circ c_1(\mathcal{L}) = c_1(f^*\mathcal{L}) \circ c$.

Proof. Of course (2) implies (1). Let $p : L \to X$ be as in Lemma 42.32.2 and let $o : X \to L$ be the zero section. Denote $p' : L' \to X'$ and $o' : X' \to L'$ their base changes. By Lemma 42.32.4 we have

$p^*(c_1(\mathcal{L}) \cap \alpha ) = - o_* \alpha \quad \text{and}\quad (p')^*(c_1(f^*\mathcal{L}) \cap \alpha ') = - o'_* \alpha '$

Since $c$ is a bivariant class we have

\begin{align*} (p')^*(c \cap c_1(\mathcal{L}) \cap \alpha ) & = c \cap p^*(c_1(\mathcal{L}) \cap \alpha ) \\ & = - c \cap o_* \alpha \\ & = - o'_*(c \cap \alpha ) \\ & = (p')^*(c_1(f^*\mathcal{L}) \cap c \cap \alpha ) \end{align*}

Since $(p')^*$ is injective by one of the lemmas cited above we obtain $c \cap c_1(\mathcal{L}) \cap \alpha = c_1(f^*\mathcal{L}) \cap c \cap \alpha$. The same is true after any base change by $Y \to X$ locally of finite type and hence we have the equality of bivariant classes stated in (2). $\square$

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