Lemma 60.25.1 (07Q7)—The Stacks project

Lemma 60.25.1. In the situation above there exists a map of complexes

\[ e_ M^\bullet : M \otimes _ B (\Omega ')^\bullet \longrightarrow M \otimes _ B \Omega ^\bullet \]

such that $c_ M^\bullet \circ e_ M^\bullet $ and $e_ M^\bullet \circ c_ M^\bullet $ are homotopic to multiplication by $a$.

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

\[ M \otimes (\Omega ')^ i = (B' \otimes M \otimes \Omega ^ i) \oplus (B' \text{d}z \otimes M \otimes \Omega ^{i - 1}) \]

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

\[ \epsilon (f) = af - \theta (\partial _ z(f)) \quad \text{and}\quad \epsilon '(f) = (\theta \otimes 1)(\text{d}_1(f)) - \text{d}_1(\theta (f)) \]

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

\[ h^ i : M \otimes _ B (\Omega ')^ i \longrightarrow M \otimes _ B (\Omega ')^{i - 1} \]

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

\[ \text{d} \circ h + h \circ \text{d} = a - c_ M^\bullet \circ e_ M^\bullet \]

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

\begin{align*} (\text{d} \circ h + h \circ \text{d})(f \omega ) & = h\left(\text{d}(f) \wedge \omega + f \nabla (\omega )\right) \\ & = \theta (\partial _ z(f)) \omega \\ & = a f\omega - \epsilon (f)\omega \\ & = a f \omega - c^ i_ M(e^ i_ M(f\omega )) \end{align*}

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

\begin{align*} (\text{d} \circ h + h \circ \text{d})(f \text{d}z \wedge \eta ) & = \text{d}(\theta (f) \eta ) + h\left(\text{d}(f) \wedge \text{d}z \wedge \eta - f \text{d}z \wedge \nabla (\eta )\right) \\ & = \text{d}(\theta (f)) \wedge \eta + \theta (f) \nabla (\eta ) - (\theta \otimes 1)(\text{d}_1(f)) \wedge \eta - \theta (f) \nabla (\eta ) \\ & = \text{d}_1(\theta (f)) \wedge \eta + \partial _ z(\theta (f)) \text{d}z \wedge \eta - (\theta \otimes 1)(\text{d}_1(f)) \wedge \eta \\ & = a f \text{d}z \wedge \eta - \epsilon '(f) \wedge \eta \\ & = a f \text{d}z \wedge \eta - c^ i_ M(e^ i_ M(f \text{d}z \wedge \eta )) \end{align*}

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. $\square$

The Stacks project

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