The Stacks project

Lemma 92.26.3. 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 $L_{X/Y} = \epsilon ^*L_{X_0/Y_0}$ in $D(\mathcal{O}_ X)$ where $\epsilon $ is as in Derived Categories of Spaces, Section 75.4 and $L_{X_0/Y_0}$ is as in Definition 92.24.1.

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

\[ \epsilon ^{-1}L_{X_0/Y_0} = \epsilon ^{-1}L_{\mathcal{O}_{X_0}/f_0^{-1}\mathcal{O}_{Y_0}} = L_{\epsilon ^{-1}\mathcal{O}_{X_0}/\epsilon ^{-1}f_0^{-1}\mathcal{O}_{Y_0}} \longrightarrow L_{\mathcal{O}_ X/f^{-1}_{small}\mathcal{O}_ Y} = L_{X/Y} \]

by the functoriality discussed in Section 92.18 and Lemma 92.18.3. To see that the induced map $\epsilon ^*L_{X_0/Y_0} \to L_{X/Y}$ is an isomorphism we may check on stalks at geometric points (Properties of Spaces, Theorem 66.19.12). We will use Lemma 92.18.9 to compute the stalks. 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

\[ L_{X/Y, \overline{x}} = L_{\mathcal{O}_{X, \overline{x}}/\mathcal{O}_{Y, \overline{y}}} \]


\[ (\epsilon ^*L_{X_0/Y_0})_{\overline{x}} = L_{X_0/Y_0, x} \otimes _{\mathcal{O}_{X_0, x}} \mathcal{O}_{X, \overline{x}} = L_{\mathcal{O}_{X_0, x}/\mathcal{O}_{Y_0, y}} \otimes _{\mathcal{O}_{X_0, x}} \mathcal{O}_{X, \overline{x}} \]

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 Lemma 92.8.7. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08VF. Beware of the difference between the letter 'O' and the digit '0'.