Lemma 42.43.3. In Situation 42.7.1 let X be locally of finite type over S. Let \mathcal{E} be a finite locally free \mathcal{O}_ X-module with dual \mathcal{E}^\vee . Then
c_ i(\mathcal{E}^\vee ) = (-1)^ i c_ i(\mathcal{E})
in A^ i(X).
Proof. Choose a morphism \pi : P \to X as in Lemma 42.43.1. By the injectivity of \pi ^* (after any base change) it suffices to prove the relation between the Chern classes of \mathcal{E} and \mathcal{E}^\vee after pulling back to P. Thus we may assume there exist invertible \mathcal{O}_ X-modules {\mathcal L}_ i, i = 1, \ldots , r and a filtration
0 = \mathcal{E}_0 \subset \mathcal{E}_1 \subset \mathcal{E}_2 \subset \ldots \subset \mathcal{E}_ r = \mathcal{E}
such that \mathcal{E}_ i/\mathcal{E}_{i - 1} \cong \mathcal{L}_ i. Then we obtain the dual filtration
0 = \mathcal{E}_ r^\perp \subset \mathcal{E}_1^\perp \subset \mathcal{E}_2^\perp \subset \ldots \subset \mathcal{E}_0^\perp = \mathcal{E}^\vee
such that \mathcal{E}_{i - 1}^\perp /\mathcal{E}_ i^\perp \cong \mathcal{L}_ i^{\otimes -1}. Set x_ i = c_1(\mathcal{L}_ i). Then c_1(\mathcal{L}_ i^{\otimes -1}) = - x_ i by Lemma 42.25.2. By Lemma 42.40.4 we have
c(\mathcal{E}) = \prod \nolimits _{i = 1}^ r (1 + x_ i) \quad \text{and}\quad c(\mathcal{E}^\vee ) = \prod \nolimits _{i = 1}^ r (1 - x_ i)
in A^*(X). The result follows from a formal computation which we omit. \square
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