The Stacks project

Proof. We may replace $X$ by a connected component of $X$ (some details omitted). Thus we may assume $X$ is connected and hence irreducible. Set $k' = \Gamma (X, \mathcal{O}_ X) = \Gamma (X', \mathcal{O}_{X'})$; we omit the proof of the equality. Choose a closed point $x' \in X'$ which isn't contained in the exceptional divisor and whose residue field $k''$ is separable over $k$; this is possible by Varieties, Lemma 33.25.6. Denote $x \in X$ the image (whose residue field is equal to $k''$ as well of course). Consider the diagram

The class of the diagonal $\Delta = \Delta _ X$ pulls back to the class of the “diagonal point” $\delta _ x : x \to x \times X$ and similarly for the class of the diagonal $\Delta '$. On the other hand, the diagonal point $\delta _ x$ pulls back to the diagonal point $\delta _{x'}$ by the left vertical arrow. Write $\gamma ([\Delta ]) = \sum \eta _ i$ with $\eta _ i \in H^ i(X) \otimes H^{2d - i}(X)(d)$ and $\gamma ([\Delta ']) = \sum \eta '_ i$ with $\eta '_ i \in H^ i(X') \otimes H^{2d - i}(X')(d)$. The arguments above show that $\eta _0$ and $\eta '_0$ map to the same class in

We have $H^0(\mathop{\mathrm{Spec}}(k')) = H^0(X) = H^0(X')$ by axiom (A8). By Lemma 45.14.14 this common value maps injectively into $H^0(x')$. We conclude that $\eta _0$ maps to $\eta '_0$ by the map

This means that $\int _ X$ is equal to $\int _{X'}$ composed with the pullback map. This proves the lemma. $\square$