Lemma 37.13.2. Let $f : X \to Y$ be a morphism of schemes. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset S$ be affine opens with $f(U) \subset V$. There is a canonical map

\[ \widetilde{\mathop{N\! L}\nolimits _{A/R}} \longrightarrow \mathop{N\! L}\nolimits _{X/Y}|_ U \]

of complexes which is an isomorphism in $D(\mathcal{O}_ U)$.

**Proof.**
From the construction of $\mathop{N\! L}\nolimits _{X/Y}$ in Modules, Section 17.30 we see there is a canonical map of complexes $\mathop{N\! L}\nolimits _{\mathcal{O}_ X(U)/f^{-1}\mathcal{O}_ Y(U)} \to \mathop{N\! L}\nolimits _{X/Y}(U)$ of $A = \mathcal{O}_ X(U)$-modules, which is compatible with further restrictions. Using the canonical map $R \to f^{-1}\mathcal{O}_ Y(U)$ we obtain a canonical map $\mathop{N\! L}\nolimits _{A/R} \to \mathop{N\! L}\nolimits _{\mathcal{O}_ X(U)/f^{-1}\mathcal{O}_ Y(U)}$ of complexes of $A$-modules. Using the universal property of the $\widetilde{\ }$ functor (see Schemes, Lemma 26.7.1) we obtain a map as in the statement of the lemma. We may check this map is an isomorphism on cohomology sheaves by checking it induces isomorphisms on stalks. This follows from Algebra, Lemma 10.134.11 and 10.134.13 and Modules, Lemma 17.30.4 (and the description of the stalks of $\mathcal{O}_ X$ and $f^{-1}\mathcal{O}_ Y$ at a point $\mathfrak p \in \mathop{\mathrm{Spec}}(A)$ as $A_\mathfrak p$ and $R_\mathfrak q$ where $\mathfrak q = R \cap \mathfrak p$; references used are Schemes, Lemma 26.5.4 and Sheaves, Lemma 6.21.5).
$\square$

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