Lemma 23.6.11. Let $R$ be a commutative ring. Suppose that $(A, \text{d}, \gamma )$ and $(B, \text{d}, \gamma )$ are as in Definition 23.6.5. Let $\overline{\varphi } : H_0(A) \to H_0(B)$ be an $R$-algebra map. Assume

$A$ is a graded divided power polynomial algebra over $R$.

$H_ k(B) = 0$ for $k > 0$.

Then there exists a map $\varphi : A \to B$ of differential graded $R$-algebras compatible with divided powers that lifts $\overline{\varphi }$.

**Proof.**
The assumption means that $A$ is obtained from $R$ by successively adjoining some set of polynomial generators in degree zero, exterior generators in positive odd degrees, and divided power generators in positive even degrees. So we have a filtration $R \subset A(0) \subset A(1) \subset \ldots $ of $A$ such that $A(m + 1)$ is obtained from $A(m)$ by adjoining generators of the appropriate type (which we simply call “divided power generators”) in degree $m + 1$. In particular, $A(0) \to H_0(A)$ is a surjection from a (usual) polynomial algebra over $R$ onto $H_0(A)$. Thus we can lift $\overline{\varphi }$ to an $R$-algebra map $\varphi (0) : A(0) \to B_0$.

Write $A(1) = A(0)\langle T_ j:j\in J\rangle $ for some set $J$ of divided power variables $T_ j$ of degree $1$. Let $f_ j \in B_0$ be $f_ j = \varphi (0)(\text{d}(T_ j))$. Observe that $f_ j$ maps to zero in $H_0(B)$ as $\text{d}T_ j$ maps to zero in $H_0(A)$. Thus we can find $b_ j \in B_1$ with $\text{d}(b_ j) = f_ j$. By the universal property of divided power polynomial algebras from Lemma 23.5.1, we find a lift $\varphi (1) : A(1) \to B$ of $\varphi (0)$ mapping $T_ j$ to $f_ j$.

Having constructed $\varphi (m)$ for some $m \geq 1$ we can construct $\varphi (m + 1) : A(m + 1) \to B$ in exactly the same manner. We omit the details.
$\square$

## Comments (0)

There are also: