Lemma 76.21.3. Let S be a scheme. Let f : X \to Y be a morphism of algebraic spaces over S. Assume X and Y representable by schemes X_0 and Y_0. Then there is a canonical identification \mathop{N\! L}\nolimits _{X/Y} = \epsilon ^*\mathop{N\! L}\nolimits _{X_0/Y_0} in D(\mathcal{O}_ X) where \epsilon is as in Derived Categories of Spaces, Section 75.4 and \mathop{N\! L}\nolimits _{X_0/Y_0} is as in More on Morphisms, Definition 37.13.1.
Proof. Let f_0 : X_0 \to Y_0 be the morphism of schemes corresponding to f. There is a canonical map \epsilon ^{-1}f_0^{-1}\mathcal{O}_{Y_0} \to f_{small}^{-1}\mathcal{O}_ Y compatible with \epsilon ^\sharp : \epsilon ^{-1}\mathcal{O}_{X_0} \to \mathcal{O}_ X because there is a commutative diagram
\xymatrix{ X_{0, Zar} \ar[d]_{f_0} & X_{\acute{e}tale}\ar[l]^\epsilon \ar[d]^ f \\ Y_{0, Zar} & Y_{\acute{e}tale}\ar[l]_\epsilon }
see Derived Categories of Spaces, Remark 75.6.3. Thus we obtain a canonical map
\epsilon ^{-1}\mathop{N\! L}\nolimits _{X_0/Y_0} = \epsilon ^{-1}\mathop{N\! L}\nolimits _{\mathcal{O}_{X_0}/f_0^{-1}\mathcal{O}_{Y_0}} = \mathop{N\! L}\nolimits _{\epsilon ^{-1}\mathcal{O}_{X_0}/\epsilon ^{-1}f_0^{-1}\mathcal{O}_{Y_0}} \to \mathop{N\! L}\nolimits _{\mathcal{O}_ X/f^{-1}_{small}\mathcal{O}_ Y} = \mathop{N\! L}\nolimits _{X/Y}
by functoriality of the naive cotangent complex. To see that the induced map \epsilon ^*\mathop{N\! L}\nolimits _{X_0/Y_0} \to \mathop{N\! L}\nolimits _{X/Y} is an isomorphism in D(\mathcal{O}_ X) we may check on stalks at geometric points (Properties of Spaces, Theorem 66.19.12). Let \overline{x} : \mathop{\mathrm{Spec}}(k) \to X_0 be a geometric point lying over x \in X_0, with \overline{y} = f \circ \overline{x} lying over y \in Y_0. Then
\mathop{N\! L}\nolimits _{X/Y, \overline{x}} = \mathop{N\! L}\nolimits _{\mathcal{O}_{X, \overline{x}}/\mathcal{O}_{Y, \overline{y}}}
This is true because taking stalks at \overline{x} is the same as taking inverse image via \overline{x} : \mathop{\mathrm{Spec}}(k) \to X and we may apply Modules on Sites, Lemma 18.35.3. On the other hand we have
(\epsilon ^*\mathop{N\! L}\nolimits _{X_0/Y_0})_{\overline{x}} = \mathop{N\! L}\nolimits _{X_0/Y_0, x} \otimes _{\mathcal{O}_{X_0, x}} \mathcal{O}_{X, \overline{x}} = \mathop{N\! L}\nolimits _{\mathcal{O}_{X_0, x}/\mathcal{O}_{Y_0, y}} \otimes _{\mathcal{O}_{X_0, x}} \mathcal{O}_{X, \overline{x}}
Some details omitted (hint: use that the stalk of a pullback is the stalk at the image point, see Sites, Lemma 7.34.2, as well as the corresponding result for modules, see Modules on Sites, Lemma 18.36.4). Observe that \mathcal{O}_{X, \overline{x}} is the strict henselization of \mathcal{O}_{X_0, x} and similarly for \mathcal{O}_{Y, \overline{y}} (Properties of Spaces, Lemma 66.22.1). Thus the result follows from More on Algebra, Lemma 15.33.8. \square
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