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This question is about comparing the approaches for a formal Wirthmüller isomorphism by Fausk-Hu-May [FHM] in *isomorphisms between left and right adjoints* and by Balmer-Dell'Ambrogio-Sanders [BDS] in *Grothendieck-Neeman duality and the Wirthmüller isomorphism*.


In both articles, a formal Wirthmüller isomorphism is studied: we are given a tensor-exact functor $f^\*: \Dd \to \Cc$ between tensor-triangulated categories that has both a left adjoint $f\_!: \Cc \to \Dd$ and a right adjoint $f\_\*: \Cc \to \Dd$ and we are looking for a `twisted isomorphism' between $f\_!$ and $f\_\*$. In [FHM] this takes the form of an isomorphism $f\_\*(-) \overset{\cong}{\implies} f\_!(- \otimes C)$ for some object $C \in \Cc$, and in [BDS] this takes the form of an isomorphism $f\_\*(- \tensor \omega\_f) \overset{\cong}{\implies} f\_!(-)$ for another object $\omega\_f \in \Cc$.


In [BDS] a comparison is given with the approach of [FHM]: it is stated in proposition 4.4 that if the situations of [BDS] and [FHM] both apply simultaneously, then the object $\omega\_f$ is the dual in $\Cc$ of the object $C$. However, I don't understand the rather short proof: it is claimed that the formal assumptions already give an isomorphism $f\_\* \cong f\_!(- \tensor C)$ by [FHM, Thm. 8.1]. But it seems to me that [FHM, Thm. 8.1] requires some non-formal input, namely one needs to explicitly check that $f\_\*(G) 
\to f\_!(G \tensor C)$ is an isomorphism for some set of generating compact dualizable objects $\{G\}$ of $\Cc$. Indeed, it takes May in the follow-up article *The Wirthmüller isomorphism revisited* quite some time to prove that the latter condition is satisfied in the setting of equivariant homotopy theory. It seems that Proposition 4.4 of [BDS] is cutting this short.


**Question 1:** Why does it follow in [BDS,Prop. 4.4] that $f\_\*(-) \implies f\_!(- \otimes C)$ is a natural isomorphism?


**Question 2:** If it doesn't follow, is it otherwise possible to compare $C$ and $\omega$ without checking that $f\_\*(-) \implies f\_!(- \otimes C)$ is a natural isomorphism on all compact generators?


Let's assume that we know that both isomorphisms hold and additionally that $\omega\_f$ is invertible (with inverse $C$). Then both approaches give a natural isomorphism $f\_\*(-) \implies f\_!(- \otimes C)$.


**Question 3:** (Why) do these two isomorphisms agree?



Comparison: Formal Wirthmüller isomorphism of Fausk-Hu-May vs. Balmer et. al
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Let me write down what I (think I) know so far.


**Answer to 1:** I think that this does not follow, i.e. that the additional assumptions that $f^\*$ has a left adjoint $f\_{(1)}$ (denoted $f\_!$ by [FHM]) and that there is an object $C \in \mathcal{C}$ with $f\_{(1)}(C) \simeq f\_\*(\mathbb{1}\_{\mathcal{C}})$ are not enough to guarantee that the map $f\_\* \to f\_{(1)}(- \otimes C)$ constructed in [FHM] is an equivalence. In fact, it seems that May has explicitly written down a counterexample to this statement in section 2 of *`The Wirthmüller isomorphism revisited'*.


(I should mention though that I think not all compact generators are strongly dualizable in May's example, so not all conditions of [BDS] are satisfied. Still, [BDS] don't give an indication why their extra conditions would make a difference.)


**Edit on 25.03.22**: For a counterexample in which the compact generators *are* strongly dualizable, see Beren Sander's answer below (in which he refers to my other MO-post mentioned in the next paragraph).


**Answer to 2:** Because of section 2 of May's article mentioned above, I expect that the condition $f\_\*(\mathbb{1}\_{\mathcal{C}}) \simeq f\_!(C)$ does not uniquely specify $C$, but I would like to see a counterexample. I have asked this as a [separate question](https://mathoverflow.net/questions/384312).


**Answer to 3:** These isomorphisms do indeed agree after the identification $D(C) \simeq \omega\_f$. Explicitly, letting $\eta: 1\_{\Dd} \to f\_!C$ denote the structure map of $C$ as in [FHM], this identification is given by $$D(C) \to f^{(1)}f\_\*(DC) \simeq f^{(1)}D(f\_!C) \xrightarrow{\eta^\*} f^{(1)}D(1\_{\Dd}) = \omega\_f,$$ and in particular the counit $f\_\*\omega\_f \to 1\_{\Dd}$ becomes under this equivalence simply $$f\_\*D(C) \simeq D(f\_!C) \xrightarrow{\eta^\*} D(1\_{\Dd}) = 1\_{\Dd}.$$
We should thus compare the following two maps:
$$f\_\*(X \tensor DC) \xrightarrow{\eta} f\_\*(X \tensor DC) \tensor f\_!C \simeq f\_!(f^\*f\_\*(X \tensor DC) \tensor C) \to f\_!(X \tensor DC \tensor C) \to f\_!(X)$$
and
$$f\_\*(X \tensor DC) \to f\_\*(f^\*f\_!X \tensor DC) \simeq f\_!X \tensor f\_\*DC \simeq f\_!X \tensor D(f\_!C) \xrightarrow{\eta^\*} f\_!X,$$
where are the ommited maps are either adjunction units/counits or left/right projection formulas. One can now write down a huge commuting diagram that shows the equivalence of these two maps.


