Lemma 42.49.6. In Lemma 42.49.1 assume $Q|_ T$ is zero. In $A^*(Z \to X)$ we have

and so on with multiplication as in Remark 42.34.7.

Lemma 42.49.6. In Lemma 42.49.1 assume $Q|_ T$ is zero. In $A^*(Z \to X)$ we have

\begin{align*} P'_1(Q) & = c'_1(Q), \\ P'_2(Q) & = c'_1(Q)^2 - 2c'_2(Q), \\ P'_3(Q) & = c'_1(Q)^3 - 3c'_1(Q)c'_2(Q) + 3c'_3(Q), \\ P'_4(Q) & = c'_1(Q)^4 - 4c'_1(Q)^2c'_2(Q) + 4c'_1(Q)c'_3(Q) + 2c'_2(Q)^2 - 4c'_4(Q), \end{align*}

and so on with multiplication as in Remark 42.34.7.

**Proof.**
The statement makes sense because the zero sheaf has rank $< 1$ and hence the classes $c'_ p(Q)$ are defined for all $p \geq 1$. In the proof of Lemma 42.49.1 we have constructed the classes $P'_ p(Q)$ and $c'_ p(Q)$ using the bivariant class $C \in A^0(W_\infty \to X)$ of Lemma 42.48.1 and the bivariant classes $P'_ p(Q|_ E)$ and $c'_ p(Q|_ E)$ of Lemma 42.47.1 for the restriction $Q|_ E$ of $Q$ to the inverse image $E$ of $Z$ in $W_\infty $. Observe that by Lemma 42.47.7 we have the desired relationship between $P'_ p(Q|_ E)$ and $c'_ p(Q|_ E)$. Recall that

\[ P'_ p(Q) = (E \to Z)_* \circ P'_ p(Q|_ E) \circ C \quad \text{and}\quad c'_ p(Q) = (E \to Z)_* \circ c'_ p(Q|_ E) \circ C \]

To finish the proof it suffices to show the multiplications defined in Remark 42.34.7 on the classes $a_ p = c'_ p(Q)$ and on the classes $b_ p = c'_ p(Q|_ E)$ agree:

\[ a_{p_1}a_{p_2} \ldots a_{p_ r} = (E \to Z)_* \circ b_{p_1}b_{p_2} \ldots b_{p_ r} \circ C \]

Some details omitted. If $r = 1$, then this is true. For $r > 1$ note that by Remark 42.34.8 the multiplication in Remark 42.34.7 proceeds by inserting $(Z \to X)_*$, resp. $(E \to W_\infty )_*$ in between the factors of the product $a_{p_1}a_{p_2} \ldots a_{p_ r}$, resp. $b_{p_1}b_{p_2} \ldots b_{p_ r}$ and taking compositions as bivariant classes. Now by Lemma 42.47.1 we have

\[ (E \to W_\infty )_* \circ b_{p_ i} = c_{p_ i}(Q|_{W_\infty }) \]

and by Lemma 42.49.4 we have

\[ C \circ (Z \to X)_* \circ a_{p_ i} = c_{p_ i}(Q|_{W_\infty }) \circ C \]

for $i = 2, \ldots , r$. A calculation shows that the left and right hand side of the desired equality both simplify to

\[ (E \to Z)_* \circ c'_{p_1}(Q|_ E) \circ c_{p_2}(Q|_{W_\infty }) \circ \ldots \circ c_{p_ r}(Q|_{W_\infty }) \circ C \]

and the proof is complete. $\square$

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