Lemma 42.48.3. In Lemma 42.48.1 let $g : W' \to W$ be a proper morphism which is an isomorphism over $\mathbf{A}^1_ X$. Let $C' \in A^0(W'_\infty \to X)$ and $C \in A^0(W_\infty \to X)$ be the classes constructed in Lemma 42.48.1. Then $g_{\infty , *} \circ C' = C$ in $A^0(W_\infty \to X)$.
Proof. Set $b' = b \circ g : W' \to \mathbf{P}^1_ X$. Denote $i'_\infty : W'_\infty \to W'$ the inclusion morphism. Denote $g_\infty : W'_\infty \to W_\infty $ the restriction of $g$. Given $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$ choose $\beta ' \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W')$ restricting to the flat pullback of $\alpha $ on $(b')^{-1}\mathbf{A}^1_ X$. Then $\beta = g_*\beta ' \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W)$ restricts to the flat pullback of $\alpha $ on $b^{-1}\mathbf{A}^1_ X$. Then $i_\infty ^*\beta = g_{\infty , *}(i'_\infty )^*\beta '$ by Lemma 42.29.8. This and the corresponding fact after base change by morphisms $X' \to X$ locally of finite type, corresponds to the assertion made in the lemma. $\square$
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