The Stacks project

Proof. We first write the arrow as a composition of three arrows. Namely, set

and denote $F_{X/S_0} : X \to X^{(1)}$ the relative Frobenius morphism. Denote $\mathcal{E}^{(1)}$ the base change of $\mathcal{E}$ by $\mathop{\mathrm{Spec}}(\sigma )$, in other words the pullback of $\mathcal{E}$ to $\text{Cris}(X^{(1)}/S)$ by the morphism of crystalline topoi associated to the commutative diagram

Then we have the base change map

see Remark 60.24.8. Note that the composition of $F_{X/S_0} : X \to X^{(1)}$ with the projection $X^{(1)} \to X$ is the absolute Frobenius morphism $F_ X$. Hence we see that $F_{X/S_0}^*\mathcal{E}^{(1)} = (F_ X)_{\text{cris}}^*\mathcal{E}$. Thus pullback by $F_{X/S_0}$ is a map

Finally we can use $F_\mathcal {E}$ to get a map

The map of the theorem is the composition of the three maps (60.26.4.1), (60.26.4.2), and (60.26.4.3) above. The first is a quasi-isomorphism modulo all powers of $p$ by Remark 60.24.9. Hence it is a quasi-isomorphism since the complexes involved are perfect in $D(A)$ see Remark 60.24.13. The third map is a quasi-isomorphism after inverting $p$ simply because $F_\mathcal {E}$ has an inverse up to a power of $p$, see Remark 60.26.3. Finally, the second is an isomorphism after inverting $p$ by Lemma 60.25.6. $\square$