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Let $\mathscr{C}$ and $\mathscr{D}$ be closed symmetric monoidal categories. We fix a strong symmetric monoidal functor $f^*:\mathscr{D}\to\mathscr{C}$ with a right adjoint $f_*:\mathscr{C}\to\mathscr{D}$. In this context, there is a formal isomorphism $\hom(Y,f_*X)\cong f_*\hom(f^*Y,X)$.

Now, suppose that we're given a second adjoint pair $(f_!,f^!)$ relating $\mathscr{C}$ and $\mathscr{D}$. It would be nice if we had a internal adjunction $\hom(f_!X,Y)\cong f_*\hom(X,f^!Y)$.

While this holds in most 6 functor formalisms, this doesn't follow formally from our suppositions.

In Isomorphisms between left and right adjoints, H. Fausk, P. Hu, and J.P. May affirm that it suffices to give the existence of one of the three arrows $$f_*\hom(X,f^!Y)\to \hom(f_!X,Y),\quad \hom(f^*Y,f^!Z)\to f^!\hom(Y,Z),\quad\pi:Y\otimes f_!X\to f_!(f^*Y\otimes X)$$ for all three to exist. Moreover, if one of them is an isomorphism, so are them all.

I wonder what condition holds in practice for us to have at least the existence of these morphisms.

Perhaps the fact that we usually have a morphism $f_!\to f_*$ suffices for us to construct these morphisms? (That's how we sometimes construct the projection formula. But we usually have that this morphism is injective and this doesn't hold for $D$-modules, for example.)

Perhaps a base change theorem suffices as in Ryan Reich's answer in Ubiquity of the push-pull formula? (This answer doesn't completely solve my problem since the formula $(g\times h)_!(X\boxtimes Y)\cong g_! X\boxtimes h_! Y$ is not clear to me as well.)

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  • $\begingroup$ What a priori relationship is there between $f^*\dashv f_*$ and $f^!\dashv f_!$? What are the "6 functors"? Is this question a matter of skill with diagram-chasing or does it depend on knowing algebraic geometry or something like that? $\endgroup$ Commented Sep 20, 2021 at 12:42
  • $\begingroup$ We suppose the existence of 3 pairs of adjunct functors (our 6 functors): $f^*\dashv f_*$, $f_!\dashv f^!$ (observe that $f^*$ is the left adjoint but $f^!$ is the right adjoint), and $\otimes\dashv\hom$. A priori, there's no relation between $f^*\dashv f_*$ and $f_!\dashv f^!$, so there's no reason for the existence of the wanted morphism. What I'm asking if the existence of a morphism $f_!\to f_*$ is enough. If not, what else can we do? (The latter question depends on the knowledge of the contexts which we're modelling, but not the former.) $\endgroup$
    – Gabriel
    Commented Sep 20, 2021 at 12:50
  • $\begingroup$ I would also like to remark that $f^*\dashv f_*$ interacts with $\otimes\dashv\hom$ since $f^*$ is supposed to be monoidal. This allows us to construct a morphism $Y\otimes f_*X\to f_*(f^* Y\otimes X)$, for example. $\endgroup$
    – Gabriel
    Commented Sep 20, 2021 at 12:57
  • $\begingroup$ Do you know an example where $f^!$ is not just right adjoint to $f_*$? $\endgroup$
    – Drew Heard
    Commented Sep 21, 2021 at 19:55
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    $\begingroup$ @Gabriel In any case, if you are not aware of it (although you probably are) the paper by Balmer--Dell'Ambrogio--Sanders should be of interest (arxiv.org/pdf/1501.01999.pdf) - Equation 3.7 is what you want, although they are working under stronger conditions than what you want. But maybe you can find something useful in there anyway. $\endgroup$
    – Drew Heard
    Commented Sep 22, 2021 at 8:26

1 Answer 1

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Consider $f\colon X \to Y$ to be a smooth map between smooth complex projective varieties and $f_{*}, f_{!} \colon D(X) \rightleftarrows D(Y) \colon f^{!}$, $f^{*}$ be the associated derived functors on the level of bounded derived categories of coherent sheaves. Assuming also that $f$ is perfect, the following adjuctions hold: $f_{!}\dashv f^{*}\dashv f_{*}\dashv f^{!}$. The first (from left to right) is explained in https://arxiv.org/pdf/1004.3052.pdf, the remaining three in Huybrecht's book about Fourier Mukai kernels.

In this setup, holds $f_{*}(A\otimes B) \xrightarrow{\simeq} f_{*}(A \otimes f^{*}B)$, called projection formula. For a proof see Hartshorne's Residues and duality.

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  • $\begingroup$ Dear @Christos, I'm not supposing to be in the Grothendieck duality context. My question is purely categorical; I wonder which structure I must add in the formalism I've described in order to obtain the projection formula (or any other morphism described by Fausk-Hu-May). $\endgroup$
    – Gabriel
    Commented Sep 24, 2021 at 19:56
  • $\begingroup$ Moreover, while I know that this is somewhat usual, I would advise against denoting the left adjoint of $fˆ*$ by $f_!$. In the context of Grothendieck duality, the most natural functor to be called $f_!$ is $f_*$. $\endgroup$
    – Gabriel
    Commented Sep 24, 2021 at 19:57
  • $\begingroup$ @Gabriel I think that the paper of Fausk-Hu-May abstracted the various cases, and when I was reading it, was trying to find concrete examples in which this formalism can be applied. So, on the one hand, you need to specify suitable geometry (in order the projection formula map would be an iso.) As far as it concerns the left adjoint of $f^{*}$, then this is defined as $f_{!}:=f_{*}(f^{!}(\mathcal{O}_{Y}) \otimes -)$. $\endgroup$
    – Christos
    Commented Sep 24, 2021 at 22:53
  • $\begingroup$ the "best" concrete examples of this formalism is Verdier duality, the six functors in étale cohomology, in D-modules (and, I think, in some motivic stuff). In the context of Grothendieck duality, we have that $f_!=f_*$ (in the notation of Fausk-Hu-May, which disagrees with yours), so this is somewhat a particular case. And in this particular case we have a formal projection formula. $\endgroup$
    – Gabriel
    Commented Sep 25, 2021 at 8:38
  • $\begingroup$ Surely we need something more in order for the projection formula. I'm just asking what this may be. For example, perhaps it suffices to consider an axiom of proper base change? Or perhaps it suffices to give a morphism $f_!\to f_*$ (which is compatible with composition)? (The latter, as simple as it may seem, is already a very non-trivial result in the context of D-modules.) $\endgroup$
    – Gabriel
    Commented Sep 25, 2021 at 8:39

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