The Stacks project

Proof. The assumptions on $E$ imply that there is an open $U \subset X$ containing $X_1$ such that $E|_ U$ is zero, resp. isomorphic to a finite locally free $\mathcal{O}_ U$-module of rank $< p$. See More on Algebra, Lemma 15.75.7. Let $Z \subset X$ be the complement of $U$ in $X$ endowed with the reduced induced closed subscheme structure. Then $P_ p(X_2 \to X, E) = (Z \to X_2)_* \circ P_ p(Z \to X, E)$, resp. $c_ p(X_2 \to X, E) = (Z \to X_2)_* \circ c_ p(Z \to X, E)$ by Lemma 42.50.8. Now we can prove that $P_ p(X_2 \to X, E)$, resp. $c_ p(X_2 \to X, E)$ satisfies the characterization of $P'_ p(E_2)$, resp. $c'_ p(E_2)$ given in Lemma 42.47.1. Namely, by the relation $P_ p(X_2 \to X, E) = (Z \to X_2)_* \circ P_ p(Z \to X, E)$, resp. $c_ p(X_2 \to X, E) = (Z \to X_2)_* \circ c_ p(Z \to X, E)$ just proven and the fact that $X_1 \cap Z = \emptyset$, the composition $P_ p(X_2 \to X, E) \circ i_{1, *}$, resp. $c_ p(X_2 \to X, E) \circ i_{1, *}$ is zero by Lemma 42.50.7. On the other hand, $P_ p(X_2 \to X, E) \circ i_{2, *} = P_ p(E_2)$, resp. $c_ p(X_2 \to X, E) \circ i_{2, *} = c_ p(E_2)$ by Lemma 42.50.6. $\square$