Lemma 37.13.11. 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

$f^*\mathop{N\! L}\nolimits _{Y/Z} \to \mathop{N\! L}\nolimits _{X/Z} \to \mathop{N\! L}\nolimits _{X/Y} \to f^*\mathop{N\! L}\nolimits _{Y/Z}[1]$

in $D(\mathcal{O}_ X)$ which recovers the $6$-term exact sequence of Lemma 37.13.10.

Proof. It suffices to show the canonical map

$f^*\mathop{N\! L}\nolimits _{Y/Z} \to \text{Cone}(\mathop{N\! L}\nolimits _{X/Z} \to \mathop{N\! L}\nolimits _{X/Y})[-1]$

of Modules, Lemma 17.31.7 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 $H^{-1}(f^*\mathop{N\! L}\nolimits _{Y/Z}) \to H^{-1}(\mathop{N\! L}\nolimits _{X/Z})$ is injective. Affine locally this follows from the corresponding algebra result in More on Algebra, Lemma 15.33.6. To translate into algebra use Lemma 37.13.2. $\square$

Comment #8873 by Eoin on

In the proof, in the cone of the first equation, the terms $NL_{X/Y}$ and $NL_{X/Z}$ are swapped.

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