Lemma 42.49.7. In Lemma 42.49.1 assume $Q|_ T$ is isomorphic to a finite locally free $\mathcal{O}_ T$-module of rank $< p$. Assume we have another perfect object $Q' \in D(\mathcal{O}_ W)$ whose Chern classes are defined with $Q'|_ T$ isomorphic to a finite locally free $\mathcal{O}_ T$-module of rank $< p'$ placed in cohomological degree $0$. With notation as in Remark 42.34.7 set

$c^{(p)}(Q) = 1 + c_1(Q|_{X \times \{ 0\} }) + \ldots + c_{p - 1}(Q|_{X \times \{ 0\} }) + c'_{p}(Q) + c'_{p + 1}(Q) + \ldots$

in $A^{(p)}(Z \to X)$ with $c'_ i(Q)$ for $i \geq p$ as in Lemma 42.49.1. Similarly for $c^{(p')}(Q')$ and $c^{(p + p')}(Q \oplus Q')$. Then $c^{(p + p')}(Q \oplus Q') = c^{(p)}(Q)c^{(p')}(Q')$ in $A^{(p + p')}(Z \to X)$.

Proof. Recall that the image of $c'_ i(Q)$ in $A^ p(X)$ is equal to $c_ i(Q|_{X \times \{ 0\} })$ for $i \geq p$ and similarly for $Q'$ and $Q \oplus Q'$, see Lemma 42.49.1. Hence the equality in degrees $< p + p'$ follows from the additivity of Lemma 42.46.7.

Let's take $n \geq p + p'$. As in the proof of Lemma 42.49.1 let $E \subset W_\infty$ denote the inverse image of $Z$. Observe that we have the equality

$c^{(p + p')}(Q|_ E \oplus Q'|_ E) = c^{(p)}(Q|_ E)c^{(p')}(Q'|_ E)$

in $A^{(p + p')}(E \to W_\infty )$ by Lemma 42.47.8. Since by construction

$c'_ p(Q \oplus Q') = (E \to Z)_* \circ c'_ p(Q|_ E \oplus Q'|_ E) \circ C$

we conclude that suffices to show for all $i + j = n$ we have

$(E \to Z)_* \circ c^{(p)}_ i(Q|_ E)c^{(p')}_ j(Q'|_ E) \circ C = c^{(p)}_ i(Q)c^{(p')}_ j(Q')$

in $A^ n(Z \to X)$ where the multiplication is the one from Remark 42.34.7 on both sides. There are three cases, depending on whether $i \geq p$, $j \geq p'$, or both.

Assume $i \geq p$ and $j \geq p'$. In this case the products are defined by inserting $(E \to W_\infty )_*$, resp. $(Z \to X)_*$ in between the two factors and taking compositions as bivariant classes, see Remark 42.34.8. In other words, we have to show

$(E \to Z)_* \circ c'_ i(Q|_ E) \circ (E \to W_\infty )_* \circ c'_ j(Q'|_ E) \circ C = c'_ i(Q) \circ (Z \to X)_* \circ c'_ j(Q')$

By Lemma 42.47.1 the left hand side is equal to

$(E \to Z)_* \circ c'_ i(Q|_ E) \circ c_ j(Q'|_{W_\infty }) \circ C$

Since $c'_ i(Q) = (E \to Z)_* \circ c'_ i(Q|_ E) \circ C$ the right hand side is equal to

$(E \to Z)_* \circ c'_ i(Q|_ E) \circ C \circ (Z \to X)_* \circ c'_ j(Q')$

which is immediately seen to be equal to the above by Lemma 42.49.4.

Assume $i \geq p$ and $j < p$. Unwinding the products in this case we have to show

$(E \to Z)_* \circ c'_ i(Q|_ E) \circ c_ j(Q'|_{W_\infty }) \circ C = c'_ i(Q) \circ c_ j(Q'|_{X \times \{ 0\} })$

Again using that $c'_ i(Q) = (E \to Z)_* \circ c'_ i(Q|_ E) \circ C$ we see that it suffices to show $c_ j(Q'|_{W_\infty }) \circ C = C \circ c_ j(Q'|_{X \times \{ 0\} })$ which is part of Lemma 42.49.4.

Assume $i < p$ and $j \geq p'$. Unwinding the products in this case we have to show

$(E \to Z)_* \circ c_ i(Q|_ E) \circ c'_ j(Q'|_ E) \circ C = c_ i(Q|_{Z \times \{ 0\} }) \circ c'_ j(Q')$

However, since $c'_ j(Q|_ E)$ and $c'_ j(Q')$ are bivariant classes, they commute with capping with Chern classes (Lemma 42.38.9). Hence it suffices to prove

$(E \to Z)_* \circ c'_ j(Q'|_ E) \circ c_ i(Q|_{W_\infty }) \circ C = c'_ j(Q') \circ c_ i(Q|_{X \times \{ 0\} })$

which we reduces us to the case discussed in the preceding paragraph. $\square$

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