The Stacks project

Proof. Denote $p : \mathbf{P}^1_ X \to X$ and $p' : \mathbf{P}^1_{X'} \to X'$ the structure morphisms. Recall that $b : W \to \mathbf{P}^1_ X$ and $b' : W' \to \mathbf{P}^1_{X'}$ are the morphism constructed from the triples $(\mathbf{P}^1_ X, (\mathbf{P}^1_ X)\infty , p^*E)$ and $(\mathbf{P}^1_{X'}, (\mathbf{P}^1_{X'})\infty , (p')^*E')$ in More on Flatness, Lemma 38.43.6. Furthermore $Q = L\eta _{\mathcal{I}_\infty }p^*E$ and $Q = L\eta _{\mathcal{I}'_\infty }(p')^*E'$ where $\mathcal{I}_\infty \subset \mathcal{O}_ W$ is the ideal sheaf of $W_\infty$ and $\mathcal{I}'_\infty \subset \mathcal{O}_{W'}$ is the ideal sheaf of $W'_\infty$. Next, $h : \mathbf{P}^1_{X'} \to \mathbf{P}^1_ X$ is a morphism of schemes such that the pullback of the effective Cartier divisor $(\mathbf{P}^1_ X)_\infty$ is the effective Cartier divisor $(\mathbf{P}^1_{X'})_\infty$ and such that $h^*p^*E = (p')^*E'$. By More on Flatness, Lemma 38.43.8 we obtain a commutative diagram

such that $W'$ is the “strict transform” of $\mathbf{P}^1_{X'}$ with respect to $b$ and such that $Q' = (q \circ g)^*Q$. Now recall that $P_ p(Z \to X, E) = P'_ p(Q)$, resp. $c_ p(Z \to X, E) = c'_ p(Q)$ where $P'_ p(Q)$, resp. $c'_ p(Q)$ are constructed in Lemma 42.49.1 using $b, Q, T'$ where $T'$ is a closed subscheme $T' \subset W_\infty$ with the following two properties: (a) $T'$ contains all points of $W_\infty$ lying over $X \setminus Z$, and (b) $Q|_{T'}$ is zero, resp. isomorphic to a finite locally free module of rank $< p$ placed in degree $0$. In the construction of Lemma 42.49.1 we chose a particular closed subscheme $T'$ with properties (a) and (b) but the precise choice of $T'$ is immaterial, see Lemma 42.49.3.

Next, by Lemma 42.49.2 the restriction of the bivariant class $P_ p(Z \to X, E) = P'_ p(Q)$, resp. $c_ p(Z \to X, E) = c_ p(Q')$ to $X'$ corresponds to the class $P'_ p(q^*Q)$, resp. $c'_ p(q^*Q)$ constructed as in Lemma 42.49.1 using $r : \mathbf{P}^1_{X'} \times _{\mathbf{P}^1_ X} W \to \mathbf{P}^1_{X'}$, the complex $q^*Q$, and the inverse image $q^{-1}(T')$.

Now by the second statement of Lemma 42.49.3 we have $P'_ p(Q') = P'_ p(q^*Q)$, resp. $c'_ p(q^*Q) = c'_ p(Q')$. Since $P_ p(Z' \to X', E') = P'_ p(Q')$, resp. $c_ p(Z' \to X', E') = c'_ p(Q')$ we conclude that the lemma is true. $\square$