Lemma 92.18.8. Let $\mathcal{D}$ be a site. Let $\mathcal{A} \to \mathcal{B} \to \mathcal{C}$ be homomorphisms of sheaves of rings on $\mathcal{D}$. There is a canonical distinguished triangle
in $D(\mathcal{C})$.
Lemma 92.18.8. Let $\mathcal{D}$ be a site. Let $\mathcal{A} \to \mathcal{B} \to \mathcal{C}$ be homomorphisms of sheaves of rings on $\mathcal{D}$. There is a canonical distinguished triangle
in $D(\mathcal{C})$.
Proof. We will use the method described in Remarks 92.7.5 and 92.7.6 to construct the triangle; we will freely use the results mentioned there. As in those remarks we first construct the triangle in case $\mathcal{B} \to \mathcal{C}$ is an injective map of sheaves of rings. In this case we set
$\mathcal{P}_\bullet $ is the standard resolution of $\mathcal{B}$ over $\mathcal{A}$,
$\mathcal{Q}_\bullet $ is the standard resolution of $\mathcal{C}$ over $\mathcal{A}$,
$\mathcal{R}_\bullet $ is the standard resolution of $\mathcal{C}$ over $\mathcal{B}$,
$\mathcal{S}_\bullet $ is the standard resolution of $\mathcal{B}$ over $\mathcal{B}$,
$\overline{\mathcal{Q}}_\bullet = \mathcal{Q}_\bullet \otimes _{\mathcal{P}_\bullet } \mathcal{B}$, and
$\overline{\mathcal{R}}_\bullet = \mathcal{R}_\bullet \otimes _{\mathcal{S}_\bullet } \mathcal{B}$.
The distinguished triangle is the distinguished triangle associated to the short exact sequence of simplicial $\mathcal{C}$-modules
The first two terms are equal to the first two terms of the triangle of the statement of the lemma. The identification of the last term with $L_{\mathcal{C}/\mathcal{B}}$ uses the quasi-isomorphisms of complexes
All the constructions used above can first be done on the level of presheaves and then sheafified. Hence to prove sequences are exact, or that map are quasi-isomorphisms it suffices to prove the corresponding statement for the ring maps $\mathcal{A}(U) \to \mathcal{B}(U) \to \mathcal{C}(U)$ which are known. This finishes the proof in the case that $\mathcal{B} \to \mathcal{C}$ is injective.
In general, we reduce to the case where $\mathcal{B} \to \mathcal{C}$ is injective by replacing $\mathcal{C}$ by $\mathcal{B} \times \mathcal{C}$ if necessary. This is possible by the argument given in Remark 92.7.5 by Lemma 92.18.7. $\square$
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