The Grothendieck-Verdier duality:
$$
Rf\_\*\big(R\mathcal{H}\textit{om}\_X^\bullet(\mathcal{E}^\bullet,f^!\mathcal{F}^\bullet)\big) \cong R\mathcal{H}\textit{om}^\bullet\_Y(Rf\_\*\mathcal{E}^\bullet,\mathcal{G}^\bullet)
$$
is known to hold for $f:X\to Y$ being a proper map of noetherian schemes. 


Is there a way to get rid of this requirement on schemes to be noetherian, possibly by putting some extra conditions on the morphism $f$? 


For example: let $X$ be smooth (projective in necessary) and consider the projection to the second factor $f:X\times U\to U$, where $U$ is some affine scheme. Does one have the duality in this setting?


**Edit:** After reading about the topic in Neeman's and Lipman's work, I have not managed to find an explicit construction of the right adjoint $f^\times$. In what cases is the explicit construction of $f^\times$ known? What would it be in the above example?



Grothendieck-Verdier duality without the noetherian condition

> 
> Does one have the duality in this setting?
> 
> 
> 


Yes, we do have duality in a very general setting. Your question is equivalent to asking for the existence of a right adjoint to the derived pushforward functor $\mathbf{R}f\_\*\colon \mathbf{D}\_{qc}(X) \to \mathbf{D}\_{qc}(Y)$. It turns out that it exists for any morphism $f\colon X \to Y$ of qcqs schemes. Look at [tag 0A9D](https://stacks.math.columbia.edu/tag/0A9D) for more details.


**P. S.** It is not a good idea to call this adjoint functor by $f^!$ unless $f$ is proper. Usually, *another* functor is denoted by $f^!$ that actually differs from the right adjoint $f^\times$ (SP denotes this functor by $a(f)$). 


**UPD:** If $Y$ is qcqs and $f\colon X \to Y$ is proper and smooth of (pure) relative dimension $d$, then $f^!(-)\cong f^{\times}(-)\cong \Omega^d\_{X/Y}[d]\otimes\_{\mathcal O\_X} f^\*(-)$. Look at [tag 0BRT](https://stacks.math.columbia.edu/tag/0BRT) and [tag 0A9U](https://stacks.math.columbia.edu/tag/0A9U) for a proof in the noetherian case. The proof in the non-noetherian setup is essentially the same. Just note that noetherianness of $Y$ is used only to show that $\mathbf{R}f\_\*$ preserves perfect objects. However, it is true without this assumption as it is briefly explained in Example 2.2 of the paper [Quasi-perfect scheme-maps and boundedness of the twisted inverse image functor](https://www.math.purdue.edu/~lipman/papers/quasiperfect.pdf).


