Lemma 45.14.7. Assume given data (D0), (D1), and (D2') satisfying axioms (A1) – (A7). Then axiom (B) of Section 45.9 holds.
Proof. Axiom (B)(a) is immediate from axiom (A5). Let $X$ and $Y$ be nonempty smooth projective schemes over $k$ equidimensional of dimensions $d$ and $e$. To see that axiom (B)(b) holds, observe that the diagonal $\Delta _{X \times Y}$ of $X \times Y$ is the intersection product of the pullbacks of the diagonals $\Delta _ X$ of $X$ and $\Delta _ Y$ of $Y$ by the projections $p : X \times Y \times X \times Y \to X \times X$ and $q : X \times Y \times X \times Y \to Y \times Y$. Compatibility of $\gamma $ with intersection products then gives that
\[ \gamma ([\Delta _{X \times Y}]) \in H^{2d + 2e}(X \times Y \times X \times Y)(d + e) \]
is the cup product of the pullbacks of $\gamma ([\Delta _ X])$ and $\gamma ([\Delta _ Y])$ by $p$ and $q$. Write
\[ \gamma ([\Delta _{X \times Y}]) = \sum \eta _{X \times Y, i} \text{ with } \eta _{X \times Y, i} \in H^ i(X \times Y) \otimes H^{2d + 2e - i}(X \times Y)(d + e) \]
and similarly $\gamma ([\Delta _ X]) = \sum \eta _{X, i}$ and $\gamma ([\Delta _ Y]) = \sum \eta _{Y, i}$. The observation above implies we have
\[ \eta _{X \times Y, 0} = \sum \nolimits _{i \in \mathbf{Z}} p^*\eta _{X, i} \cup q^*\eta _{Y, -i} \]
(If our cohomology theory vanishes in negative degrees, which will be true in almost all cases, then only the term for $i = 0$ contributes and $\eta _{X \times Y, 0}$ lies in $H^0(X) \otimes H^0(Y) \otimes H^{2d}(X)(d) \otimes H^{2e}(Y)(e)$ as expected, but we don't need this.) Since $\lambda _ X : H^{2d}(X)(d) \to F$ and $\lambda _ Y : H^{2e}(Y)(e) \to F$ send $\eta _{X, 0}$ and $\eta _{Y, 0}$ to $1$ in $H^*(X)$ and $H^*(Y)$, we see that $\lambda _ X \otimes \lambda _ Y$ sends $\eta _{X \times Y, 0}$ to $1$ in $H^*(X) \otimes H^*(Y) = H^*(X \times Y)$ and the proof is complete. $\square$
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