**Proof.**
In view of the fact that formation of $\omega _{X/Y}^\bullet $ commutes with base change (see Remark 48.12.5), we may and do assume that $Y$ is affine. For a perfect object $E$ of $D(\mathcal{O}_ X)$ we have

\begin{align*} Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet ) & = Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , \omega _{X/Y}^\bullet ) \\ & = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y) \\ & = (Rf_*E^\vee )^\vee \end{align*}

For the first equality, see Cohomology, Lemma 20.50.5. For the second equality, see Lemma 48.3.6, Remark 48.3.8, and Derived Categories of Schemes, Lemma 36.30.4. The third equality is the definition of the dual. In particular these references also show that the outcome is a perfect object of $D(\mathcal{O}_ Y)$. We conclude that $\omega _{X/Y}^\bullet $ is $Y$-perfect by More on Morphisms, Lemma 37.69.6. This proves (1).

Let $M$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ Y)$. Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*\omega _{X/Y}^\bullet ) & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M, \omega _{X/Y}^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*Lf^*M, \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*\mathcal{O}_ X, \mathcal{O}_ Y) \end{align*}

The first equality holds by Cohomology, Lemma 20.28.1. The second equality by construction of $a$. The third equality by Derived Categories of Schemes, Lemma 36.22.1. Recall $Rf_*\mathcal{O}_ X$ is perfect of tor amplitude in $[0, N]$ for some $N$, see Derived Categories of Schemes, Lemma 36.30.4. Thus we can represent $Rf_*\mathcal{O}_ X$ by a complex of finite projective modules sitting in degrees $[0, N]$ (using More on Algebra, Lemma 15.74.2 and the fact that $Y$ is affine). Hence if $M = \mathcal{O}_ Y[-i]$ for some $i > 0$, then the last group is zero. Since $Y$ is affine we conclude that $H^ i(Rf_*\omega _{X/Y}^\bullet ) = 0$ for $i > 0$. This proves (2).

Let $E$ be a perfect object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. Then we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ X(E, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet ) & = \mathop{\mathrm{Hom}}\nolimits _ X(E \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet ), \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , \omega _{X/Y}^\bullet )), \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y), \mathcal{O}_ Y) \\ & = R\Gamma (Y, Rf_*E^\vee ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(E, \mathcal{O}_ X) \end{align*}

The first equality holds by Cohomology, Lemma 20.42.2. The second equality is the definition of $\omega _{X/Y}^\bullet $. The third equality comes from the construction of the dual perfect complex $E^\vee $, see Cohomology, Lemma 20.50.5. The fourth equality follows from the equality $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , \omega _{X/Y}^\bullet ) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y)$ shown in the first paragraph of the proof. The fifth equality holds by double duality for perfect complexes (Cohomology, Lemma 20.50.5) and the fact that $Rf_*E$ is perfect by Derived Categories of Schemes, Lemma 36.30.4. The last equality is Leray for $f$. This string of equalities essentially shows (3) holds by the Yoneda lemma. Namely, the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet )$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.10.8. Taking $E = \mathcal{O}_ X$ in the above we get a map $\alpha : \mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet )$ corresponding to $\text{id}_{\mathcal{O}_ X} \in \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ X, \mathcal{O}_ X)$. Since all the isomorphisms above are functorial in $E$ we see that the cone on $\alpha $ is an object $C$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$ such that $\mathop{\mathrm{Hom}}\nolimits (E, C) = 0$ for all perfect $E$. Since the perfect objects generate (Derived Categories of Schemes, Theorem 36.15.3) we conclude that $\alpha $ is an isomorphism.
$\square$

## Comments (3)

Comment #7499 by Hao Peng on

Comment #7500 by Hao Peng on

Comment #7642 by Stacks Project on