Remark 45.9.3. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C)(a). Let $X$ be a smooth projective scheme over $k$. We obtain maps

where the first arrow is as in axiom (B) and $\Delta ^*$ is pullback along the diagonal morphism $\Delta : X \to X \times X$. The composition is the cup product as pullback is an algebra homomorphism and $\text{pr}_ i \circ \Delta = \text{id}$. On the other hand, given cycles $\alpha , \beta $ on $X$ the intersection product is defined by the formula

In other words, $\alpha \cdot \beta $ is the pullback of the exterior product $\alpha \times \beta $ on $X \times X$ by the diagonal. Note also that $\alpha \times \beta = \text{pr}_1^*\alpha \cdot \text{pr}_2^*\beta $ in $\mathop{\mathrm{CH}}\nolimits ^*(X \times X)$ (we omit the proof). Hence, given axiom (C)(a), axiom (C)(c) is equivalent to the statement that $\gamma $ is compatible with exterior product in the sense that $\gamma (\alpha \times \beta )$ is equal to $\text{pr}_1^*\gamma (\alpha ) \cup \text{pr}_2^*\gamma (\beta )$.

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