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Suppose $X,Y$ are varieties over $\mathbb{C}$, $Y$ is smooth and $X$ is Gorenstein ($X$ is not smooth in my case). Let $f: X \to Y$ be an affine morphism, and each fibre of $f$ has the same dimension $n$. Moreover, $f$ can be assumed to be flat.

For $F \in D^b(X)$ and $G \in D^b(Y)$, I expect to have a Grothendieck-Verdier duality: $${\rm R}f_* {\rm R}\mathcal{H}om(F, f^!(G)) \cong {\rm R}\mathcal{H}om({\rm R}f_*F, G).$$

However, I don't know how to define the functor $f^!$ here.

I have checked the book "Residue and Duality" by Hartshorne, it seems that he requires $f$ at least to be proper in order to define the duality. Besides, in the book "Fourier-Mukai Transforms in Algebraic Geometry" by Huybrechts (c.f. Theorem 3.34 page 86), he did not require $f$ to be proper, but both $X$ and $Y$ are required to be smooth. So I don't know how to establish the duality in my affine setting.

Any suggestion is welcome!!

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    $\begingroup$ I humbly think that Huybrechts forgot a properness assumption, either on X and Y or f. To add to dracula's comment below: the right adjoint to $Rf_*$ exists pretty much always, purely for formal reasons. However, people often denote it by $f^\times$, as $f^!$ is reserved for the more sophisticated "twisted inverse image". The two $f^\times, f^!$ coincide when $f$ is proper. $\endgroup$
    – pro
    Commented Jun 16, 2015 at 17:57
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    $\begingroup$ All varieties in the book of Huybrechts are taken to be (smooth) projective, and all morphisms between these are projective themselves. So he didn't have to put in the properness, it is there to start with. $\endgroup$
    – pbelmans
    Commented Jun 16, 2015 at 19:56
  • $\begingroup$ Okay, I see.... $\endgroup$
    – Li Yutong
    Commented Jun 16, 2015 at 19:57
  • $\begingroup$ I just looked at page 86, and there he doesn't explicitly say that his varieties / schemes are projective. But later on in the context of Fourier--Mukai transforms everything is projective, so it turns out fine where he applies things. I guess he didn't want to be too elaborate on $f^\times$ and $f^!$ in this preliminary section... $\endgroup$
    – pbelmans
    Commented Jun 16, 2015 at 20:08

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You can get some kind right adjoint for $Rf_*$ by work of Lipman and Neeman. See the references quoted in the first few sentences here: Section Tag 0A9D. But this does not mean necessarily that one can "compute" the right adjoint. In the generality you are asking about the statement will not sheafify well (so there will not be a formula of the type you are asking for). Anyway, I suggest working out what happens when $Y$ is the spectrum of a field to see the trouble you get into.

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  • $\begingroup$ Thank you so much for your answer! I curious why they can not sheafify? In "Foundations of Grothendieck Duality for Diagrams of Schemes", they can work with "quasi-proper" morphism instead of proper morphism (See Cor 4.3.6 Page 174). It seems that the only property they use is that for such morphism, $Rf_*$ sends coherent complexes to coherent complexes. Is this some obstacle prevents the sheafification? $\endgroup$
    – Li Yutong
    Commented Jun 16, 2015 at 19:54

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