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The Stacks project

Weak form of [Theorem 17.1, F]

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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