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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