Lemma 20.55.8. In Situation 20.55.2 let $M$ be an object of $D(\mathcal{O}_ X)$. There is a canonical isomorphism

$L\eta _\mathcal {I}M \otimes ^\mathbf {L} \mathcal{O}_ X/\mathcal{I} \longrightarrow H^\bullet (M/\mathcal{I})$

in $D(\mathcal{O}_ X)$ where the right hand side is the complex (20.55.7.2).

Proof. By the construction of $L\eta _\mathcal {I}$ in Lemma 20.55.6 we may assume $M$ is represented by a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules $\mathcal{F}^\bullet$. Then $L\eta _\mathcal {I}M$ is represented by the complex $\eta _\mathcal {I}\mathcal{F}^\bullet$ which is a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules as well. Thus $L\eta _\mathcal {I}M \otimes ^\mathbf {L} \mathcal{O}_ X/\mathcal{I}$ is represented by the complex $\eta _\mathcal {I}\mathcal{F}^\bullet \otimes \mathcal{O}_ X/\mathcal{I}$. Similarly, the complex $H^\bullet (M/\mathcal{I})$ has terms $H^ i(\mathcal{F}^\bullet \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1})$.

Let $f$ be a local generator for $\mathcal{I}$. Let $s$ be a local section of $(\eta _\mathcal {I}\mathcal{F})^ i$. Then we can write $s = f^ is'$ for a local section $s'$ of $\mathcal{F}^ i$ and similarly $d^ i(s) = f^{i + 1}t$ for a local section $t$ of $\mathcal{F}^{i + 1}$. Thus $d^ i$ maps $f^ is'$ to zero in $\mathcal{F}^{i + 1} \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1}$. Hence we may map $s$ to the class of $f^ is'$ in $H^ i(\mathcal{F}^\bullet \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1})$. This rule defines a map

$(\eta _\mathcal {I}\mathcal{F})^ i \otimes \mathcal{O}_ X/\mathcal{I} \longrightarrow H^ i(\mathcal{F}^\bullet \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1})$

of $\mathcal{O}_ X$-modules. A calculation shows that these maps are compatible with differentials (essentially because $\beta$ sends the class of $f^ is'$ to the class of $f^{i + 1}t$), whence a map of complexes representing the arrow in the statement of the lemma.

To finish the proof, we observe that the construction given in the previous paragraph agrees on stalks with the maps constructed in More on Algebra, Lemma 15.95.6 hence we conclude. $\square$

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