The Stacks project

Lemma 48.16.5. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. There are canonical maps

\[ \mu _{f, K} : Lf^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y \longrightarrow f^!K \]

functorial in $K$ in $D^+_\mathit{QCoh}(\mathcal{O}_ Y)$. If $g : Y \to Z$ is another morphism of $\textit{FTS}_ S$, then the diagram

\[ \xymatrix{ Lf^*(Lg^*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} g^!\mathcal{O}_ Z) \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y \ar@{=}[d] \ar[r]_-{\mu _ f} & f^!(Lg^*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} g^!\mathcal{O}_ Z) \ar[r]_-{f^!\mu _ g} & f^!g^!K \ar@{=}[d] \\ Lf^*Lg^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^* g^!\mathcal{O}_ Z \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y \ar[r]^-{\mu _ f} & Lf^*Lg^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!g^!\mathcal{O}_ Z \ar[r]^-{\mu _{g \circ f}} & f^!g^!K } \]

commutes for all $K \in D^+_\mathit{QCoh}(\mathcal{O}_ Z)$.

Proof. If $f$ is proper, then $f^! = a$ and we can use ( and if $g$ is also proper, then Lemma 48.8.4 proves the commutativity of the diagram (in greater generality).

Let us define the map $\mu _{f, K}$. Choose a compactification $j : X \to \overline{X}$ of $X$ over $Y$. Since $f^!$ is defined as $j^* \circ \overline{a}$ we obtain $\mu _{f, K}$ as the restriction of the map (

\[ L\overline{f}^*K \otimes _{\mathcal{O}_{\overline{X}}}^\mathbf {L} \overline{a}(\mathcal{O}_ Y) \longrightarrow \overline{a}(K) \]

to $X$. To see this is independent of the choice of the compactification we argue as in the proof of Lemma 48.16.2. We urge the reader to read the proof of that lemma first.

Assume given a morphism $g : \overline{X}_1 \to \overline{X}_2$ between compactifications $j_ i : X \to \overline{X}_ i$ over $Y$ such that $g^{-1}(j_2(X)) = j_1(X)$. Denote $\overline{c}$ the right adjoint for pushforward of Lemma 48.3.1 for the morphism $g$. The maps

\[ L\overline{f}_1^*K \otimes _{\mathcal{O}_{\overline{X}}}^\mathbf {L} \overline{a}_1(\mathcal{O}_ Y) \longrightarrow \overline{a}_1(K) \quad \text{and}\quad L\overline{f}_2^*K \otimes _{\mathcal{O}_{\overline{X}}}^\mathbf {L} \overline{a}_2(\mathcal{O}_ Y) \longrightarrow \overline{a}_2(K) \]

fit into the commutative diagram

\[ \xymatrix{ Lg^*(L\overline{f}_2^*K \otimes ^\mathbf {L} \overline{a}_2(\mathcal{O}_ Y)) \otimes ^\mathbf {L} \overline{c}(\mathcal{O}_{\overline{X}_2}) \ar@{=}[d] \ar[r]_-\sigma & \overline{c}(L\overline{f}_2^*K \otimes ^\mathbf {L} \overline{a}_2(\mathcal{O}_ Y)) \ar[r] & \overline{c}(\overline{a}_2(K)) \ar@{=}[d] \\ L\overline{f}_1^*K \otimes ^\mathbf {L} Lg^*\overline{a}_2(\mathcal{O}_ Y) \otimes ^\mathbf {L} \overline{c}(\mathcal{O}_{\overline{X}_2}) \ar[r]^-{1 \otimes \tau } & L\overline{f}_1^*K \otimes ^\mathbf {L} \overline{a}_1(\mathcal{O}_ Y) \ar[r] & \overline{a}_1(K) } \]

by Lemma 48.8.4. By Lemma 48.8.3 the maps $\sigma $ and $\tau $ restrict to an isomorphism over $X$. In fact, we can say more. Recall that in the proof of Lemma 48.16.2 we used the map ( $\gamma : j_1^* \circ \overline{c} \to j_2^*$ to construct our isomorphism $\alpha _ g : j_1^* \circ \overline{a}_1 \to j_2^* \circ \overline{a}_2$. Pulling back to map $\sigma $ by $j_1$ we obtain the identity map on $j_2^*\left(L\overline{f}_2^*K \otimes ^\mathbf {L} \overline{a}_2(\mathcal{O}_ Y)\right)$ if we identify $j_1^*\overline{c}(\mathcal{O}_{\overline{X}_2})$ with $\mathcal{O}_ X$ via $j_1^* \circ \overline{c} \to j_2^*$, see Lemma 48.8.2. Similarly, the map $\tau : Lg^*\overline{a}_2(\mathcal{O}_ Y) \otimes ^\mathbf {L} \overline{c}(\mathcal{O}_{\overline{X}_2}) \to \overline{a}_1(\mathcal{O}_ Y) = \overline{c}(\overline{a}_2(\mathcal{O}_ Y))$ pulls back to the identity map on $j_2^*\overline{a}_2(\mathcal{O}_ Y)$. We conclude that pulling back by $j_1$ and applying $\gamma $ wherever we can we obtain a commutative diagram

\[ \xymatrix{ j_2^*\left(L\overline{f}_2^*K \otimes ^\mathbf {L} \overline{a}_2(\mathcal{O}_ Y)\right) \ar[r] \ar[d] & j_2^*\overline{a}_2(K) \\ j_1^*L\overline{f}_1^*K \otimes ^\mathbf {L} j_2^*\overline{a}_2(\mathcal{O}_ Y) & j_1^*(L\overline{f}_1^*K \otimes ^\mathbf {L} \overline{a}_1(\mathcal{O}_ Y)) \ar[r] \ar[l]_{1 \otimes \alpha _ g} & j_1^* \overline{a}_1(K) \ar[lu]_{\alpha _ g} } \]

The commutativity of this diagram exactly tells us that the map $\mu _{f, K}$ constructed using the compactification $\overline{X}_1$ is the same as the map $\mu _{f, K}$ constructed using the compactification $\overline{X}_2$ via the identification $\alpha _ g$ used in the proof of Lemma 48.16.2. Some categorical arguments exactly as in the proof of Lemma 48.16.2 now show that $\mu _{f, K}$ is well defined (small detail omitted).

Having said this, the commutativity of the diagram in the statement of our lemma follows from the construction of the isomorphism $(g \circ f)^! \to f^! \circ g^!$ (first part of the proof of Lemma 48.16.3 using $\overline{X} \to \overline{Y} \to Z$) and the result of Lemma 48.8.4 for $\overline{X} \to \overline{Y} \to Z$. $\square$

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 0B6T. Beware of the difference between the letter 'O' and the digit '0'.