Lemma 45.14.16. Assume given data (D0), (D1), and (D2') satisfying axioms (A1) – (A8). Then the cycle class map $\gamma $ commutes with pushforward.
Proof. Let $i : Z \to X$ be as in Lemma 45.14.13. Consider the diagram
\[ \xymatrix{ E \ar[r]_ j \ar[d]_\pi & X' \ar[d]^ b \\ Z \ar[r]^ i & X } \]
Let $\theta \in \mathop{\mathrm{CH}}\nolimits ^{r - 1}(X')$ be as in Lemma 45.14.3. Then $\pi _*j^!\theta = [Z]$ in $\mathop{\mathrm{CH}}\nolimits _*(Z)$ implies that $\pi _*\gamma (j^!\theta ) = 1$ by Lemma 45.14.9 because $\pi $ is a projective space bundle. Hence we see that
\[ i_*(1) = i_*(\pi _*(\gamma (j^!\theta ))) = b_*j_*(j^*\gamma (\theta )) = b_*(j_*(1) \cup \gamma (\theta )) \]
We have $j_*(1) = \gamma ([E])$ by (A9). Thus this is equal to
\[ b_*(\gamma ([E]) \cup \gamma (\theta )) = b_*(\gamma ([E] \cdot \theta )) = b_*(\gamma (b^*[Z])) = b_*b^*\gamma ([Z]) = b_*(1) \cup \gamma ([Z]) \]
Since $b_*(1) = 1$ by Lemma 45.14.15 the proof is complete. $\square$
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