The Stacks project

Proof. Choose $a : D \to D'$ and $b : D' \to D$ as in Lemma 60.17.5. Note that the base change $M = M' \otimes _{D', b} D$ with its connection $\nabla $ corresponds to $\mathcal{F}$. Hence we know that $M \otimes ^\wedge _ D \Omega _ D^*$ computes the crystalline cohomology of $\mathcal{F}$, see Proposition 60.21.3. Hence it suffices to show that the base change maps (induced by $a$ and $b$)

are quasi-isomorphisms. Since $a \circ b = \text{id}_{D'}$ we see that the composition one way around is the identity on the complex $M' \otimes ^\wedge _{D'} \Omega ^*_{D'}$. Hence it suffices to show that the map

induced by $b \circ a : D \to D$ is a quasi-isomorphism. (Note that we have the same complex on both sides as $M = M' \otimes ^\wedge _{D', b} D$, hence $M \otimes ^\wedge _{D, b \circ a} D = M' \otimes ^\wedge _{D', b \circ a \circ b} D = M' \otimes ^\wedge _{D', b} D = M$.) In fact, we claim that for any divided power $A$-algebra homomorphism $\rho : D \to D$ compatible with the augmentation to $C$ the induced map $M \otimes ^\wedge _ D \Omega ^*_ D \to M \otimes ^\wedge _{D, \rho } \Omega ^*_ D$ is a quasi-isomorphism.

Write $\rho (x_ i) = x_ i + z_ i$. The elements $z_ i$ are in the divided power ideal of $D$ because $\rho $ is compatible with the augmentation to $C$. Hence we can factor the map $\rho $ as a composition

where the first map is given by $x_ i \mapsto x_ i + \xi _ i$ and the second map is the divided power $D$-algebra map which maps $\xi _ i$ to $z_ i$. (This uses the universal properties of polynomial algebra, divided power polynomial algebras, divided power envelopes, and $p$-adic completion.) Note that there exists an automorphism $\alpha $ of $D\langle \xi _ i \rangle ^\wedge $ with $\alpha (x_ i) = x_ i - \xi _ i$ and $\alpha (\xi _ i) = \xi _ i$. Applying Lemma 60.20.2 to $\alpha \circ \sigma $ (which maps $x_ i$ to $x_ i$) and using that $\alpha $ is an isomorphism we conclude that $\sigma $ induces a quasi-isomorphism of $M \otimes ^\wedge _ D \Omega ^*_ D$ with $M \otimes ^\wedge _{D, \sigma } \Omega ^*_{D\langle x_ i \rangle ^\wedge }$. On the other hand the map $\tau $ has as a left inverse the map $D \to D\langle x_ i \rangle ^\wedge $, $x_ i \mapsto x_ i$ and we conclude (using Lemma 60.20.2 once more) that $\tau $ induces a quasi-isomorphism of $M \otimes ^\wedge _{D, \sigma } \Omega ^*_{D\langle x_ i \rangle ^\wedge }$ with $M \otimes ^\wedge _{D, \tau \circ \sigma } \Omega ^*_ D$. Composing these two quasi-isomorphisms we obtain that $\rho $ induces a quasi-isomorphism $M \otimes ^\wedge _ D \Omega ^*_ D \to M \otimes ^\wedge _{D, \rho } \Omega ^*_ D$ as desired. $\square$