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. Then there are canonical maps

$\wedge ^ i(H^0(L^\bullet )) \longrightarrow \wedge ^ i(K^0) \otimes _ R \det (M^\bullet )$

whose composition with $\wedge ^ i(K^0) \to \wedge ^ i(H^0(L^\bullet ))$ is equal to multiplication with $\delta (M^\bullet )$.

Proof. Say $M^0$ and $M^{-1}$ are free of rank $n$. For every $i \geq 0$ there is a canonical surjection

$\pi _ i : \wedge ^{n + i}(L^0) \longrightarrow \wedge ^ i(K^0) \otimes \wedge ^ n(M^0)$

whose kernel is the submodule generated by wedges $l_1 \wedge \ldots \wedge l_{n + i}$ such that $> i$ of the $l_ j$ are in $K^0$. On the other hand, the exact sequence

$L^{-1} \to L^0 \to H^0(L^\bullet ) \to 0$

similarly produces canonical maps

$\wedge ^ i(H^0(L^\bullet )) \otimes \wedge ^ n(L^{-1}) \longrightarrow \wedge ^{n + i}(L^0)$

by sending $\eta \otimes \theta$ to $\tilde\eta \wedge \partial (\theta )$ where $\tilde\eta \in \wedge ^ i(L^0)$ is a lift of $\eta$. The composition of these two maps, combined with the identification $\wedge ^ n(L^{-1}) = \wedge ^ n(M^{-1})$ gives a map

$\wedge ^ i(H^0(L^\bullet )) \otimes \wedge ^ n(M^{-1}) \longrightarrow \wedge ^ i(K^0) \otimes \wedge ^ n(M^0)$

Since $\det (M^\bullet ) = \wedge ^ n(M^0) \otimes (\wedge ^ n(M^{-1}))^{\otimes -1}$ this produces a map as in the statement of the lemma. If $\eta$ is the image of $\omega \in \wedge ^ i(K^0)$, then we see that $\theta \otimes \eta$ is mapped to $\pi _ i(\omega \wedge \partial (\theta )) = \omega \otimes \overline{\theta }$ in $\wedge ^ i(K^0) \otimes \wedge ^ n(M^0)$ where $\overline{\theta }$ is the image of $\theta$ in $\wedge ^ n(M^0)$. Since $\delta (M^\bullet )$ is simply the determinant of the map $M^{-1} \to M^0$ this proves the last statement. $\square$

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