Remark 91.7.5. We sketch an alternative, perhaps simpler, proof of the existence of the fundamental triangle. Let $A \to B \to C$ be ring maps and assume that $B \to C$ is injective. Let $P_\bullet \to B$ be the standard resolution of $B$ over $A$ and let $Q_\bullet \to C$ be the standard resolution of $C$ over $B$. Picture

$\xymatrix{ P_\bullet : & A[A[A[B]]] \ar[d] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A[A[B]] \ar[d] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A[B] \ar[d] \ar@<0ex>[l] \ar[r] & B \\ Q_\bullet : & A[A[A[C]]] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A[A[C]] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A[C] \ar@<0ex>[l] \ar[r] & C }$

Observe that since $B \to C$ is injective, the ring $Q_ n$ is a polynomial algebra over $P_ n$ for all $n$. Hence we obtain a cosimplicial object in $\mathcal{C}_{C/B/A}$ (beware reversal arrows). Now set $\overline{Q}_\bullet = Q_\bullet \otimes _{P_\bullet } B$. The key to the proof of Proposition 91.7.4 is to show that $\overline{Q}_\bullet$ is a resolution of $C$ over $B$. This follows from Cohomology on Sites, Lemma 21.39.12 applied to $\mathcal{C} = \Delta$, $\mathcal{O} = P_\bullet$, $\mathcal{O}' = B$, and $\mathcal{F} = Q_\bullet$ (this uses that $Q_ n$ is flat over $P_ n$; see Cohomology on Sites, Remark 21.39.11 to relate simplicial modules to sheaves). The key fact implies that the distinguished triangle of Proposition 91.7.4 is the distinguished triangle associated to the short exact sequence of simplicial $C$-modules

$0 \to \Omega _{P_\bullet /A} \otimes _{P_\bullet } C \to \Omega _{Q_\bullet /A} \otimes _{Q_\bullet } C \to \Omega _{\overline{Q}_\bullet /B} \otimes _{\overline{Q}_\bullet } C \to 0$

which is deduced from the short exact sequences $0 \to \Omega _{P_ n/A} \otimes _{P_ n} Q_ n \to \Omega _{Q_ n/A} \to \Omega _{Q_ n/P_ n} \to 0$ of Algebra, Lemma 10.138.9. Namely, by Remark 91.5.5 and the key fact the complex on the right hand side represents $L_{C/B}$ in $D(C)$.

If $B \to C$ is not injective, then we can use the above to get a fundamental triangle for $A \to B \to B \times C$. Since $L_{B \times C/B} \to L_{B/B} \oplus L_{C/B}$ and $L_{B \times C/A} \to L_{B/A} \oplus L_{C/A}$ are quasi-isomorphism in $D(B \times C)$ (Lemma 91.6.4) this induces the desired distinguished triangle in $D(C)$ by tensoring with the flat ring map $B \times C \to C$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).