Lemma 42.35.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$. Let $p \in \mathbf{Z}$. Suppose given a rule which assigns to every locally of finite type morphism $Y' \to Y$ and every $k$ a map

\[ c \cap - : \mathop{\mathrm{CH}}\nolimits _ k(Y') \longrightarrow \mathop{\mathrm{CH}}\nolimits _{k - p}(X') \]

where $Y' = X' \times _ X Y$, satisfying conditions (1), (2) of Definition 42.33.1 and condition (3) whenever $\mathcal{L}'|_{D'} \cong \mathcal{O}_{D'}$. Then $c \cap -$ is a bivariant class.

**Proof.**
Let $Y' \to Y$ be a morphism of schemes which is locally of finite type. Let $(\mathcal{L}', s', i' : D' \to Y')$ be as in Definition 42.29.1 with pullback $(\mathcal{N}', t', j' : E' \to X')$ to $X'$. We have to show that $c \cap (i')^*\alpha ' = (j')^*(c \cap \alpha ')$ for all $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ k(Y')$.

Denote $g : Y'' \to Y'$ the smooth morphism of relative dimension $1$ with $i'' : D'' \to Y''$ and $p : D'' \to D'$ constructed in Lemma 42.32.7. (Warning: $D''$ isn't the full inverse image of $D'$.) Denote $f : X'' \to X'$ and $E'' \subset X''$ their base changes by $X' \to Y'$. Picture

\[ \xymatrix{ & X'' \ar[rr] \ar '[d][dd]_ h & & Y'' \ar[dd]^ g \\ E'' \ar[rr] \ar[dd]_ q \ar[ru]^{j''} & & D'' \ar[dd]^ p \ar[ru]^{i''} & \\ & X' \ar '[r][rr] & & Y' \\ E' \ar[rr] \ar[ru]^{j'} & & D' \ar[ru]^{i'} } \]

By the properties given in the lemma we know that $\beta ' = (i')^*\alpha '$ is the unique element of $\mathop{\mathrm{CH}}\nolimits _{k - 1}(D')$ such that $p^*\beta ' = (i'')^*g^*\alpha '$. Similarly, we know that $\gamma ' = (j')^*(c \cap \alpha ')$ is the unique element of $\mathop{\mathrm{CH}}\nolimits _{k - 1 - p}(E')$ such that $q^*\gamma ' = (j'')^*h^*(c \cap \alpha ')$. Since we know that

\[ (j'')^*h^*(c \cap \alpha ') = (j'')^*(c \cap g^*\alpha ') = c \cap (i'')^*g^*\alpha ' \]

by our assumptions on $c$; note that the modified version of (3) assumed in the statement of the lemma applies to $i''$ and its base change $j''$. We similarly know that

\[ q^*(c \cap \beta ') = c \cap p^*\beta ' \]

We conclude that $\gamma ' = c \cap \beta '$ by the uniqueness pointed out above.
$\square$

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