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Triangle NLs when the last arrow is lci

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aisejohan committed Nov 27, 2019
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@@ -3838,6 +3838,33 @@ \section{The naive cotangent complex}
Modules, Lemma \ref{modules-lemma-exact-sequence-NL-ringed-topoi}.
\end{proof}

\begin{lemma}
\label{lemma-get-triangle-NL}
Let $f : X \to Y$ and $Y \to Z$ be morphisms of schemes. Assume
$X \to Y$ is a complete intersection morphism. Then there is
a canonical distinguished triangle
$$
f^*\NL_{Y/Z} \to \NL_{X/Z} \to \NL_{X/Y} \to f^*\NL_{Y/Z}[1]
$$
in $D(\mathcal{O}_X)$ which recovers the $6$-term exact sequence of
Lemma \ref{lemma-exact-sequence-NL}.
\end{lemma}

\begin{proof}
It suffices to show the canonical map
$$
f^*\NL_{Y/Z} \to \text{Cone}(\NL_{X/Y} \to \NL_{X/Z})[-1]
$$
of Modules, Lemma \ref{modules-lemma-exact-sequence-NL-ringed-topoi}
is an isomorphism in $D(\mathcal{O}_X)$. In order to show this, it
suffices to show that the $6$-term sequence has
a zero on the left, i.e., that $H^{-1}(f^*\NL_{Y/Z}) \to H^{-1}(\NL_{X/Z})$
is injective. Affine locally this follows from the corresponding
algebra result in More on Algebra, Lemma
\ref{more-algebra-lemma-transitive-lci-at-end}.
To translate into algebra use Lemma \ref{lemma-NL-affine}.
\end{proof}

\begin{lemma}
\label{lemma-get-NL}
Let $f : X \to Y$ be a morphism of schemes which factors

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