The Stacks project

Proof. Since formation of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits $ commutes with restrictions to opens we may as well assume $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$. Observe that relatively perfect objects of $D(\mathcal{O}_ X)$ are pseudo-coherent and hence are in $D_\mathit{QCoh}(\mathcal{O}_ X)$ (Derived Categories of Schemes, Lemma 36.10.1). Thus the statement makes sense. Observe that taking $\Delta _*$, $L\text{pr}_1^*$, and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits $ is compatible with what happens on the algebraic side by Derived Categories of Schemes, Lemmas 36.3.7, 36.3.8, 36.10.8. For the last one we observe that $L\text{pr}_1^*K$ is $S$-perfect (hence bounded below) and that $\Delta _*\mathcal{O}_ X$ is a pseudo-coherent object of $D(\mathcal{O}_ W)$; translated into algebra this means that $A$ is pseudo-coherent as an $A \otimes _ R A$-module which follows from More on Algebra, Lemma 15.82.8 applied to $R \to A \otimes _ R A \to A$. Thus we recover exactly the conditions in Dualizing Complexes, Definition 47.27.1. $\square$

Proof. Looking affine locally this reduces using Lemma 48.28.2 to the algebraic case which is Dualizing Complexes, Lemma 47.27.5. $\square$

Proof. Let $U \subset X$ be an affine open mapping into an affine open of $S$. Then there is an isomorphism $K|_ U \to L|_ U$ by Lemma 48.28.2 and Dualizing Complexes, Lemma 47.27.2. The reader can reuse the argument of that lemma in the schemes case to obtain a proof in this case. We will instead use a glueing argument.

Suppose we have an isomorphism $\alpha : K \to L$. Then $\alpha (\xi ) = u \eta $ for some invertible section $u \in H^0(W, \Delta _*\mathcal{O}_ X) = H^0(X, \mathcal{O}_ X)$. (Because both $\eta $ and $\alpha (\xi )$ are generators of an invertible $\Delta _*\mathcal{O}_ X$-module by assumption.) Hence after replacing $\alpha $ by $u^{-1}\alpha $ we see that $\alpha (\xi ) = \eta $. Since the automorphism group of $K$ is $H^0(X, \mathcal{O}_ X^*)$ by Lemma 48.28.3 there is at most one such $\alpha $.

Let $\mathcal{B}$ be the collection of affine opens of $X$ which map into an affine open of $S$. For each $U \in \mathcal{B}$ we have a unique isomorphism $\alpha _ U : K|_ U \to L|_ U$ mapping $\xi $ to $\eta $ by the discussion in the previous two paragraphs. Observe that $\text{Ext}^ i(K|_ U, K|_ U) = 0$ for $i < 0$ and any open $U$ of $X$ by Lemma 48.28.3. By Cohomology, Lemma 20.45.2 applied to $\text{id} : X \to X$ we get a unique morphism $\alpha : K \to L$ agreeing with $\alpha _ U$ for all $U \in \mathcal{B}$. Then $\alpha $ sends $\xi $ to $\eta $ as this is true locally. $\square$

Proof. Let $\mathcal{B}$ be the collection of affine opens of $X$ which map into an affine open of $S$. For each $U$ we have a relative dualizing complex $(K_ U, \xi _ U)$ for $U$ over $S$. Namely, choose an affine open $V \subset S$ such that $U \to X \to S$ factors through $V$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(R)$. By Dualizing Complexes, Lemma 47.27.4 there exists a relative dualizing complex $K_ A \in D(A)$ for $R \to A$. Arguing backwards through the proof of Lemma 48.28.2 this determines an $V$-perfect object $K_ U \in D(\mathcal{O}_ U)$ and a map

in $D(\mathcal{O}_{U \times _ V U})$. Since being $V$-perfect is the same as being $S$-perfect and since $U \times _ V U = U \times _ S U$ we find that $(K_ U, \xi _ U)$ is as desired.

If $U' \subset U \subset X$ with $U', U \in \mathcal{B}$, then we have a unique isomorphism $\rho _{U'}^ U : K_ U|_{U'} \to K_{U'}$ in $D(\mathcal{O}_{U'})$ sending $\xi _ U|_{U' \times _ S U'}$ to $\xi _{U'}$ by Lemma 48.28.4 (note that trivially the restriction of a relative dualizing complex to an open is a relative dualizing complex). The uniqueness guarantees that $\rho ^ U_{U''} = \rho ^ V_{U''} \circ \rho ^ U_{U'}|_{U''}$ for $U'' \subset U' \subset U$ in $\mathcal{B}$. Observe that $\text{Ext}^ i(K_ U, K_ U) = 0$ for $i < 0$ for $U \in \mathcal{B}$ by Lemma 48.28.3 applied to $U/S$ and $K_ U$. Thus the BBD glueing lemma (Cohomology, Theorem 20.45.8) tells us there is a unique solution, namely, an object $K \in D(\mathcal{O}_ X)$ and isomorphisms $\rho _ U : K|_ U \to K_ U$ such that we have $\rho ^ U_{U'} \circ \rho _ U|_{U'} = \rho _{U'}$ for all $U' \subset U$, $U, U' \in \mathcal{B}$.

To finish the proof we have to construct the map

in $D(\mathcal{O}_ W)$ inducing an isomorphism from $\Delta _*\mathcal{O}_ X$ to $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W)$. Since we may change $W$, we choose $W = \bigcup _{U \in \mathcal{B}} U \times _ S U$. We can use $\rho _ U$ to get isomorphisms

As $W$ is covered by the opens $U \times _ S U$ we conclude that the cohomology sheaves of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W)$ are zero except in degree $0$. Moreover, we obtain isomorphisms

Let $\tau _ U$ in the LHS be an element mapping to $\xi _ U$ under this map. The compatibilities between $\rho ^ U_{U'}$, $\xi _ U$, $\xi _{U'}$, $\rho _ U$, and $\rho _{U'}$ for $U' \subset U \subset X$ open $U', U \in \mathcal{B}$ imply that $\tau _ U|_{U' \times _ S U'} = \tau _{U'}$. Thus we get a global section $\tau $ of the $0$th cohomology sheaf $H^0(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W))$. Since the other cohomology sheaves of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W)$ are zero, this global section $\tau $ determines a morphism $\xi $ as desired. Since the restriction of $\xi $ to $U \times _ S U$ gives $\xi _ U$, we see that it satisfies the final condition of Definition 48.28.1. $\square$

Proof. Consider the cartesian square

Choose $W \subset X \times _ S X$ open such that $\Delta _{X/S}$ factors through a closed immersion $\Delta : X \to W$. Choose $W' \subset X' \times _{S'} X'$ open such that $\Delta _{X'/S'}$ factors through a closed immersion $\Delta ' : X \to W'$ and such that $(g' \times g')(W') \subset W$. Let us still denote $g' \times g' : W' \to W$ the induced morphism. We have

The first equality holds because $X$ and $X' \times _{S'} X'$ are tor independent over $X \times _ S X$ (see for example More on Morphisms, Lemma 37.69.1). The second holds by transitivity of derived pullback (Cohomology, Lemma 20.27.2). Thus $\xi ' = L(g' \times g')^*\xi $ can be viewed as a map

Having said this the proof of the lemma is straightforward. First, $K'$ is $S'$-perfect by Derived Categories of Schemes, Lemma 36.35.6. To check that $\xi '$ induces an isomorphism of $\Delta '_*\mathcal{O}_{X'}$ to $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{W'}}( \Delta '_*\mathcal{O}_{X'}, L\text{pr}_1^*K'|_{W'})$ we may work affine locally. By Lemma 48.28.2 we reduce to the corresponding statement in algebra which is proven in Dualizing Complexes, Lemma 47.27.4. $\square$

Proof. In Lemma 48.12.7 we proved that $\omega _{X/S}^\bullet $ is $S$-perfect. Let $c$ be the right adjoint of Lemma 48.3.1 for the diagonal $\Delta : X \to X \times _ S X$. Then we can apply $\Delta _*$ to (48.12.8.1) to get an isomorphism

The equality holds by Lemmas 48.9.7 and 48.9.3. This finishes the proof. $\square$

Proof. By Lemma 48.17.10 we have that $f^!\mathcal{O}_ Y$ is $Y$-perfect. As $f$ is separated the diagonal $\Delta : X \to X \times _ Y X$ is a closed immersion and $\Delta _*\Delta ^!(-) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X \times _ Y X}}(\mathcal{O}_ X, -)$, see Lemmas 48.9.7 and 48.9.3. Hence to finish the proof it suffices to show $\Delta ^!(L\text{pr}_1^*f^!(\mathcal{O}_ Y)) \cong \mathcal{O}_ X$ where $\text{pr}_1 : X \times _ Y X \to X$ is the first projection. We have

where $\text{pr}_2 : X \times _ Y X \to X$ is the second projection and where we have used the base change isomorphism $\text{pr}_1^! \circ L\text{pr}_2^* = L\text{pr}_1^* \circ f^!$ of Lemma 48.18.1. $\square$

Proof. Using Lemma 48.28.2 and the algebraic version of this lemma (Dualizing Complexes, Lemma 47.27.6) we see that $E$ is affine locally the first component of a relative dualizing complex. In particular we see that $E$ is $S$-perfect since this may be checked affine locally, see Derived Categories of Schemes, Lemma 36.35.3.

Let us first prove the existence of $\zeta $ in case the morphisms $X \to S$ and $Y \to X$ are separated so that $\Delta _{X/S}$, $\Delta _{Y/X}$, and $\Delta _{Y/S}$ are closed immersions. Consider the following diagram

where $p$, $q$, $r$ are the first projections. By Lemma 48.9.4 we have

By Lemma 48.10.3 we have

By Lemma 48.10.2 we have

The last expression is isomorphic (via $\xi $) to $Lm^*\mathcal{O}_ X = \mathcal{O}_{Y \times _ X Y}$. Hence the expression preceding is isomorphic to $Lq^*M$. Hence

The material inside the parentheses is isomorphic to $\Delta _{Y/X, *}*\mathcal{O}_ X$ via $\eta $. This finishes the proof in the separated case.

In the general case we choose an open $W \subset X \times _ S X$ such that $\Delta _{X/S}$ factors through a closed immersion $\Delta : X \to W$ and we choose an open $V \subset Y \times _ X Y$ such that $\Delta _{Y/X}$ factors through a closed immersion $\Delta ' : Y \to V$. Finally, choose an open $W' \subset Y \times _ S Y$ whose intersection with $Y \times _ X Y$ gives $V$ and which maps into $W$. Then we consider the diagram

and we use exactly the same argument as before. $\square$