Lemma 59.17.5. In Situation 59.5.1. Let $A \to P' \to C$ be ring maps with $A \to P'$ smooth and $P' \to C$ surjective with kernel $J'$. Let $D'$ be the $p$-adic completion of $D_{P', \gamma }(J')$. There are homomorphisms of divided power $A$-algebras

\[ a : D \longrightarrow D',\quad b : D' \longrightarrow D \]

compatible with the maps $D \to C$ and $D' \to C$ such that $a \circ b = \text{id}_{D'}$. These maps induce an equivalence of categories of pairs $(M, \nabla )$ satisfying (1), (2), (3), and (4) over $D$ and pairs $(M', \nabla ')$ satisfying (1), (2), (3), and (4) over $D'$. In particular, the equivalence of categories of Proposition 59.17.4 also holds for the corresponding functor towards pairs over $D'$.

**Proof.**
We can pick the map $P = A[x_ i] \to C$ such that it factors through a surjection of $A$-algebras $P \to P'$ (we may have to increase the number of variables in $P$ to do this). Hence we obtain a surjective map $a : D \to D'$ by functoriality of divided power envelopes and completion. Pick $e$ large enough so that $D_ e$ is a divided power thickening of $C$ over $A$. Then $D_ e \to C$ is a surjection whose kernel is locally nilpotent, see Divided Power Algebra, Lemma 23.2.6. Setting $D'_ e = D'/p^ eD'$ we see that the kernel of $D_ e \to D'_ e$ is locally nilpotent. Hence by Algebra, Lemma 10.137.17 we can find a lift $\beta _ e : P' \to D_ e$ of the map $P' \to D'_ e$. Note that $D_{e + i + 1} \to D_{e + i} \times _{D'_{e + i}} D'_{e + i + 1}$ is surjective with square zero kernel for any $i \geq 0$ because $p^{e + i}D \to p^{e + i}D'$ is surjective. Applying the usual lifting property (Algebra, Proposition 10.137.13) successively to the diagrams

\[ \xymatrix{ P' \ar[r] & D_{e + i} \times _{D'_{e + i}} D'_{e + i + 1} \\ A \ar[u] \ar[r] & D_{e + i + 1} \ar[u] } \]

we see that we can find an $A$-algebra map $\beta : P' \to D$ whose composition with $a$ is the given map $P' \to D'$. By the universal property of the divided power envelope we obtain a map $D_{P', \gamma }(J') \to D$. As $D$ is $p$-adically complete we obtain $b : D' \to D$ such that $a \circ b = \text{id}_{D'}$.

Consider the base change functor

\[ (M, \nabla ) \longmapsto (M \otimes ^\wedge _ D D', \nabla ') \]

from pairs for $D$ to pairs for $D'$, see Remark 59.6.9. Similarly, we have the base change functor corresponding to the divided power homomorphism $D' \to D$. To finish the proof of the lemma we have to show that the base change for the compositions $b \circ a : D \to D$ and $a \circ b : D' \to D'$ are isomorphic to the identity functor. This is clear for the second as $a \circ b = \text{id}_{D'}$. To prove it for the first, we use the functorial isomorphism

\[ c_{\text{id}_ D, b \circ a} : M \otimes _{D, \text{id}_ D} D \longrightarrow M \otimes _{D, b \circ a} D \]

of the proof of Proposition 59.17.4. The only thing to prove is that these maps are horizontal, which we omit.

The last statement of the proof now follows.
$\square$

## Comments (1)

Comment #5099 by Lei Zhang on

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