The Stacks project

Proof. In this proof all tensor products are over $B$. Assumption (2) implies that

for all $i \geq 0$. A collection of additive generators for $M \otimes (\Omega ')^ i$ is formed by elements of the form $f \omega$ and elements of the form $f \text{d}z \wedge \eta$ where $f \in B'$, $\omega \in M \otimes \Omega ^ i$, and $\eta \in M \otimes \Omega ^{i - 1}$.

For $f \in B'$ we write

so that $\epsilon (f) \in B$ and $\epsilon '(f) \in \Omega$ by assumptions (6) and (7). We define $e_ M^\bullet$ by the rules $e^ i_ M(f\omega ) = \epsilon (f) \omega$ and $e^ i_ M(f \text{d}z \wedge \eta ) = \epsilon '(f) \wedge \eta$. We will see below that the collection of maps $e^ i_ M$ is a map of complexes.

We define

by the rules $h^ i(f \omega ) = 0$ and $h^ i(f \text{d}z \wedge \eta ) = \theta (f) \eta$ for elements as above. We claim that

Note that multiplication by $a$ is a map of complexes by (1). Hence, since $c_ M^\bullet$ is an injective map of complexes by assumption (5), we conclude that $e_ M^\bullet$ is a map of complexes. To prove the claim we compute

The second equality because $\text{d}z$ does not occur in $\nabla (\omega )$ and the third equality by assumption (6). Similarly, we have

The second equality because $\text{d}(f) \wedge \text{d}z \wedge \eta = - \text{d}z \wedge \text{d}_1(f) \wedge \eta$. The fourth equality by assumption (4). On the other hand it is immediate from the definitions that $e^ i_ M(c^ i_ M(\omega )) = \epsilon (1) \omega = a \omega$. This proves the lemma. $\square$