Compare with discussion in [Rizzardo].

Lemma 56.9.8. Let $S$ be a Noetherian scheme. Let $Y \to S$ be a flat proper Gorenstein morphism and let $X \to S$ be a finite type morphism. Denote $\omega ^\bullet _{Y/S}$ the relative dualizing complex of $Y$ over $S$. Let $\Phi : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ be a Fourier-Mukai functor with perfect kernel $K \in D_\mathit{QCoh}(\mathcal{O}_{X \times _ S Y})$. Denote

$K' = (Y \times _ S X \to X \times _ S Y)^*(K^\vee \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} L\text{pr}_2^*\omega ^\bullet _{Y/S}) \in D_\mathit{QCoh}(\mathcal{O}_{Y \times _ S X})$

and denote $\Phi ' : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ the corresponding Fourier-Mukai transform. There is a canonical isomorphism

$\mathop{\mathrm{Hom}}\nolimits _ Y(N, \Phi (M)) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ X(\Phi '(N), M)$

functorial in $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and $N$ in $D_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. By Lemma 56.9.2 we obtain a functor $\Phi$ as in the statement.

Observe that formation of the relative dualizing complex commutes with base change in our setting, see Duality for Schemes, Remark 48.12.5. Thus $L\text{pr}_2^*\omega ^\bullet _{Y/S} = \omega ^\bullet _{X \times _ S Y/X}$. Moreover, we observe that $\omega ^\bullet _{Y/S}$ is an invertible object of the derived category, see Duality for Schemes, Lemma 48.25.10, and a fortiori perfect.

To actually prove the lemma we're going to cheat. Namely, we will show that if we replace the roles of $X$ and $Y$ and $K$ and $K'$ then these are as in Lemma 56.9.7 and we get the result. It is clear that $K'$ is perfect as a tensor product of perfect objects so that the discussion in Lemma 56.9.7 applies to it. To show that the procedure of Lemma 56.9.7 applied to $K'$ on $Y \times _ S X$ produces a complex isomorphic to $K$ it suffices (details omitted) to show that

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \omega ^\bullet _{X \times _ S Y/X}), \omega ^\bullet _{X \times _ S Y/X}) = K$

This is clear because $K$ is perfect and $\omega ^\bullet _{X \times _ S Y/X}$ is invertible; details omitted. Thus Lemma 56.9.7 produces a map

$\mathop{\mathrm{Hom}}\nolimits _ Y(N, \Phi (M)) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ X(\Phi '(N), M)$

functorial in $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and $N$ in $D_\mathit{QCoh}(\mathcal{O}_ Y)$ which is an isomorphism because $K'$ is perfect. This finishes the proof. $\square$

## Comments (4)

Comment #5341 by Bogdan on

Don't we want to assume that $Y \to S$ is flat and proper as opposed to $X \to S$?

This would guarantee that $pr_1\colon X\times_S Y \to X$ is flat and proper, so $a(\mathcal O_X)$ is in $D^b_{coh}(X\times_S Y)$. Otherwise, it seems false unless I miss something.

If we take $X=S=\mathrm{Spec} k$ and $Y=\mathbf{A}^1_k$, we get that $a(\mathcal O_X)$ is not coherent. And if we take $X=S=\mathrm{Spec} k[x]/x^2$ and $Y=\mathrm{Spec} k$, we get that $a(\mathcal O_X)$ is not bounded.

Comment #5342 by Bogdan on

Also, it seems that the argument only constructs a map $\mathrm{Hom}_Y(N,\Phi(M))\to \mathrm{Hom}_X(\Phi′(N),M)$. Tag 0FYW guarantees that it is an isomorphism if $K'$ is perfect. If $K=\mathcal O$, it boils down to perfectness of $a(\mathcal O_X)$. And if we further assume that $X=S=\mathrm{Spec} \ k$ and $Y$ is proper over $S$, the complex $a(\mathcal O_X)$ is perfect if and only if $Y$ is Gorenstein.

Comment #5343 by on

Yes, I think you are right. I will carefully look at this and fix this tomorrow. Luckily the assumptions hold in the only place we use it. Thanks very much!

Comment #5345 by on

Again thanks. I have now fixed this here in exactly the manner you suggested.

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