Lemma 50.18.1. Let $R$ be a ring and consider a commutative diagram
\[ \xymatrix{ 0 \ar[r] & K^0 \ar[r] & L^0 \ar[r] & M^0 \ar[r] & 0 \\ & & L^{-1} \ar[u]_\partial \ar@{=}[r] & M^{-1} \ar[u] } \]
of $R$-modules with exact top row and $M^0$ and $M^{-1}$ finite free of the same rank. For $p \geq 0$ there are canonical maps
\[ c^ p : \wedge ^ p(H^0(L^\bullet )) \longrightarrow \wedge ^ p(K^0) \otimes _ R \det (M^\bullet ) \]
with the following properties:
$c^0(1) = \delta (M^\bullet )$, and
$c^{q + p}(\omega \wedge \eta ) = \omega \wedge c^ p(\eta )$ for $\omega \in \wedge ^ q(K^0)$ and $\eta \in \wedge ^ p(H^0(L^\bullet ))$.
In particular, the composition of $c^ p$ with the map $\wedge ^ p(K^0) \to \wedge ^ p(H^0(L^\bullet ))$ is multiplication with $\delta (M^\bullet )$.
Proof.
Say $M^0$ and $M^{-1}$ are free of rank $n$. For every $p \geq 0$ there is a unique surjection
\[ \pi _ p : \wedge ^{p + n}(L^0) \longrightarrow \wedge ^ p(K^0) \otimes \wedge ^ n(M^0) \]
with the following two properties; (i) we have $ \pi _ p(k_1 \wedge \ldots \wedge k_ p \wedge l_1 \wedge \ldots \wedge l_ n) = k_1 \wedge \ldots \wedge k_ p \otimes m_1 \wedge \ldots \wedge m_ n $ if $k_1, \ldots , k_ p \in K^0$ and $l_1, \ldots , l_ n \in L^0$ map to $m_1, \ldots , m_ n \in M^0$ and (ii) the kernel of $\pi _ p$ is the submodule generated by wedges $l_1 \wedge \ldots \wedge l_{p + n}$ such that $> p$ of the $l_ j$ are in $K^0$. Pick a basis $e_1, \ldots , e_ n$ for $L^{-1} = M^{-1}$. Then we define our map by the rule
\[ \eta \mapsto \pi _ p(\tilde\eta \wedge \partial (e_1) \wedge \ldots \wedge \partial (e_ n)) \otimes (e_1 \wedge \ldots \wedge e_ n)^{\otimes -1} \]
where $\eta \in \wedge ^ p(H^0(L^\bullet ))$ and $\tilde\eta \in \wedge ^ p(L^0)$ is a lift of $\eta $. The expression on the right hand side makes sense because $\det (M^\bullet ) = \wedge ^ n(M^0) \otimes \wedge ^ n(M^{-1})^{\otimes -1}$. The expression on the right is independent of the choice of $\tilde\eta $ since $\partial (\xi ) \wedge \partial (e_1) \wedge \ldots \wedge \partial (e_ n) = 0$ for any $\xi \in L^{-1}$. A straightforward calculation shows that the map is independent of the choice of the basis of $L^{-1} = M^{-1}$. It is immediate from the definition of $\delta (M^\bullet )$ in More on Algebra, Section 15.123 that $c^0(1) = \delta (M^\bullet )$. Finally, the rule in statement (2) can be directly verified from the definition.
$\square$
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