The Stacks project

Proof. The schemes $X$, $X'$, $E$, $Z$ are smooth and projective over $k$ and hence we have $K'_0(X) = K_0(X) = K_0(\textit{Vect}(X)) = K_0(D^ b_{\textit{Coh}}(X)))$ and similarly for the other $3$. See Derived Categories of Schemes, Lemmas 36.38.1, 36.38.4, and 36.38.5. We will switch between these versions at will in this proof. Consider the short exact sequence

of finite locally free $\mathcal{O}_ E$-modules defining $\mathcal{F}$. Observe that $\mathcal{C}_{E/X'} = \mathcal{O}_{X'}(-E)|_ E$ is the restriction of the invertible $\mathcal{O}_ X$-module $\mathcal{O}_{X'}(-E)$. Let $\alpha \in K_0(X)$ be an element such that $i^*\alpha = [\mathcal{C}_{Z/X}]$ in $K_0(Z)$. Let $\alpha ' = b^*\alpha - [\mathcal{O}_{X'}(-E)]$. Then $j^*\alpha ' = [\mathcal{F}]$. We deduce that $j^*\lambda ^ i(\alpha ') = [\wedge ^ i(\mathcal{F})]$ by Weil Cohomology Theories, Lemma 45.13.1. This means that $[\mathcal{O}_ E] \cdot \alpha ' = [\wedge ^ i\mathcal{F}]$ in $K_0(X)$, see Derived Categories of Schemes, Lemma 36.38.8. Let $r$ be the maximum codimension of an irreducible component of $Z$ in $X$. A computation which we omit shows that $H^{-i}(Lb^*\mathcal{O}_ Z) = \wedge ^ i\mathcal{F}$ for $i \geq 0, 1, \ldots , r - 1$ and zero in other degrees. It follows that in $K_0(X)$ we have

This proves the lemma with $\alpha '' = \sum _{i = 0, \ldots , r - 1} (-1)^ i \lambda ^ i(\alpha ')$. $\square$