Lemma 90.11.3. Let $A \to B$ be a ring map. Then $\tau _{\geq -1}L_{B/A}$ is canonically quasi-isomorphic to the naive cotangent complex.

**Proof.**
Consider $P = A[B] \to B$ with kernel $I$. The naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}$ of $B$ over $A$ is the complex $I/I^2 \to \Omega _{P/A} \otimes _ P B$, see Algebra, Definition 10.133.1. Observe that in (90.11.1.2) we have already constructed a canonical map

Consider the distinguished triangle (90.7.0.1)

associated to the ring maps $A \to A[B] \to B$. We know that $L_{P/A} = \Omega _{P/A}[0] = \mathop{N\! L}\nolimits _{P/A}$ in $D(P)$ (Lemma 90.4.7 and Algebra, Lemma 10.133.3) and that $\tau _{\geq -1}L_{B/P} = I/I^2[1] = \mathop{N\! L}\nolimits _{B/P}$ in $D(B)$ (Lemma 90.11.2 and Algebra, Lemma 10.133.6). To show $c$ is a quasi-isomorphism it suffices by Algebra, Lemma 10.133.4 and the long exact cohomology sequence associated to the distinguished triangle to show that the maps $L_{P/A} \to L_{B/A} \to L_{B/P}$ are compatible on cohomology groups with the corresponding maps $\mathop{N\! L}\nolimits _{P/A} \to \mathop{N\! L}\nolimits _{B/A} \to \mathop{N\! L}\nolimits _{B/P}$ of the naive cotangent complex. We omit the verification. $\square$

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