$\newcommand{\Cc}{\mathcal{C}}$ $\newcommand{\Dd}{\mathcal{D}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\tensor}{\otimes}$ $\newcommand{\colim}{\rm colim}$ $\newcommand{\Sp}{Sp}$ $\newcommand{\iHom}{\underline{\rm Hom}}$ (This is the follow-up of this question. I have repeated the motivation for the question.)

The following situation frequently arises in abstract category theory: we have symmetric closed monoidal categories $\Cc$ and $\Dd$ and a closed symmetric monoidal functor $f^*: \Dd \to \Cc$ that admits both a left adjoint $f_!: \Cc \to \Dd$ and a right adjoint $f_*: \Cc \to \Dd$, and we are interested in the relation between $f_!$ and $f_*$. This is often called a Wirthmüller context, named after the inspiring example of the Wirthmüller isomorphism in equivariant homotopy theory. (The symbol $f^*$ is just notation: there is not necessarily a morphism $f$ around.)

To compare $f_!$ and $f_*$, Fausk-Hu-May [FHM] assume in the article isomorphisms between left and right adjoints the existence of an object $C \in \Cc$ satisfying $f_!(C) \simeq f_*(1_{\Cc})$, where $1_{\Cc}$ denotes the monoidal unit of $\Cc$. From this data, a natural transformation $f_*(-) \to f_!(- \tensor C)$ is constructed and conditions are given for when this is an isomorphism (the 'formal Wirthmüller isomorphism').

As remarked by Balmer-Dell'Ambrogio-Sanders [BDS] in the article Grothendieck-Neeman duality and the Wirthmüller isomorphism, the object $C$ is a priori not unique and it doesn't come with a `characterizing description'. My main question is:

Question (1): Do we have counterexamples for essentially uniqueness of an object $C$ satisfying $f_!(C) = f_*(1_{\mathcal{C}})$?

I actually want to follow [BDS] and stay in a more restricted setting. We assume that $\Cc$ and $\Dd$ are tensor-triangulated (or, if you prefer, stably symmetric monoidal stable $\infty$-categories) and that they are generated by a set of compact objects and admit internal homs (or equivalently: the tensor products commute with coproducts in both variables). Moreover, we assume that $\Cc$ and $\Dd$ are rigid, i.e. the compact objects are the same as the strongly dualizable objects. Finally, we assume that $f^*: \Dd \to \Cc$ is exact and strong monoidal and that it admits both a left adjoint $f_!: \Cc \to \Dd$ and a right adjoint $f_*: \Cc \to \Dd$.

Examples of rigid tensor triangulated categories are spectra $\operatorname{Sp}$, genuine $G$-spectra $\operatorname{Sp}^G$ for a compact Lie group $G$ and the derived category $D(R)$ for a ring $R$. Several more advanced examples of such categories and functors are in examples 4.6 - 4.8 in the article Grothendieck-Neeman duality and the Wirthmüller isomorphism by Balmer-Dell'Ambrogio-Sanders, based on 'finite group schemes', 'motivic homotopy theory' and 'cohomology rings of classifying spaces'. However, I don't really understand them and in particular I don't know if they give counterexamples to question (1) above. Hence:

Question (2): What are simple examples of functors $f^*: \Dd \to \Cc$ satisfying the above conditions?

Attempts: I expect that it shouldn't be hard to come up with counterexamples for (1), but somehow I cannot make them work. For example:

  • Consider a finite group $G$ and let $f: H \hookrightarrow G$ be a subgroup inclusion. Then the restriction functor $f^*: \Sp^{G} \to \Sp^H$ between (genuine) equivariant spectra is symmetric monoidal with left adjoint $f_!(X) = G \ltimes_H X$ and right adjoint $f_*(X) = F(G/H_+,X)$. In this case we actually have $f_! \simeq f_*$ by the Wirthmüller isomorphism. And I think that if $C \in \Sp^H$ satisfies $f_!(C) \simeq f_*(\mathbb{S}_H) \simeq f_!(\mathbb{S}_H)$, then we already have $C \simeq \mathbb{S}_H$, so this won't give a counterexample.
  • Consider a finite group $G$ and a normal subgroup $N < G$, and let $f: G \to G/N$ be the projection. Then $f^*: \Sp^{G/N} \to \Sp^G$ between (genuine) equivariant spectra is symmetric monoidal with left adjoint $f_!(X) = X/N$ and right adjoint $f_*(X) = X^N$. Now an object $C$ with $f_!(C) \simeq f_!(\mathbb{S}_G)$ is no longer unique since $f_!(\Sigma^{\infty}_+G) \simeq f_!(\Sigma^{\infty}_+ G/N)$. But we don't have the Wirthmüller isomorphism $f_! \simeq f_*$ so it doesn't immediately produce a counterexample.

I also tried a little outside the range of equivariant homotopy theory, but there I fail to even come up with functors satisfying the required criteria:

  • Let $f: \Z \hookrightarrow \Q$ and let $f^*: \Dd = D(\Q) \to \Cc = D(\Z)$ be the forgetful functor. This is symmetric monoidal (since $\Q$ is a (derived) localization of $\Z$). However, it does not preserve compact objects: $\Q$ is not finitely generated over $\Z$. Edit: as remarked in the comments, it is actually only lax symmetric monoidal, since the monoidal unit is not sent to the monoidal unit.
  • Let $f^*: \Dd = D(\Z/2\Z) \to \Cc = D(\Z)$ be the forgetful functor. This preserves compact objects, but it is not symmetric monoidal: we have $\Z/2Z \tensor^L_{\Z/2\Z} \Z/2\Z \simeq \Z/2\Z$, but $\Z/2\Z \tensor^L_{\Z} \Z/2\Z$ has higher Tor-groups so cannot be $\Z/2$.
  • If $\Cc$ is cartesian closed and pointed, and $\Dd = 1$ is the trivial category, then the functor $f^*: 1 \to \Cc: * \mapsto *$ is symmetric monoidal, has both adjoints $f_!$ and $f_*$, and any object $C$ satisfies $f_!(C) \simeq f_*(1_{\Cc})$. But $\Cc$ cannot be tensor-triangulated (as $- \times -$ doesn't preserve coproducts in either variable.)
  • $\begingroup$ As with your other question, please say from what discipline this arises and what are the leading examples. $\endgroup$ Feb 18, 2021 at 12:11
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    $\begingroup$ @PaulTaylor I'm not sure I agree that these two questions need to be edited in the way you suggest. As someone who thinks about triangulated categories, these are very natural questions, and already well motivated. The Wirthmuller isomorphism is already mentioned in both as a leading example. $\endgroup$ Feb 18, 2021 at 13:24
  • $\begingroup$ Note that $f^* : D(\mathbb Q)\to D(\mathbb Z)$ is not strong symmetric monoidal, it's only lax symmetric monoidal, as $\mathbb Q\neq \mathbb Z$. $\endgroup$ Feb 18, 2021 at 14:15
  • $\begingroup$ @MaximeRamzi $\mathbb{Q}$-vector spaces are flat and their tensor products over $\mathbb{Q}$ and $\mathbb{Z}$ coincide, so it is strong monoidal. $\endgroup$ Feb 18, 2021 at 14:56
  • 2
    $\begingroup$ @FernandoMuro : strong monoidality includes a comparison of units $\endgroup$ Feb 18, 2021 at 15:00

1 Answer 1


Consider the projective line $\mathbb{P}^1_k$ over a field $k$. The structure morphism $f:\mathbb{P}^1_k\rightarrow \mathrm{Spec}(k)$ induces a functor $f^*:D(k) \rightarrow D_{qc}(\mathbb{P}^1_k)$ which is fully faithful (equivalently, $f_*(1) \simeq 1$) and which satisfies Grothendieck--Neeman duality. Since $f_*(1)\simeq 1$ is (trivially) a rigid separable commutative algebra, there is a canonical map $\theta:1 \to \omega_f$ such that $f_*(\theta)$ an isomorphism (by Lemma 4.5 of my paper "A characterization of finite étale morphisms in tensor triangular geometry"). Thus, $f_*(1) \simeq f_*(\omega_f) \simeq f_!(1)$ where the last isomorphism is the Wirthmüller isomorphism provided by [BDS]. However, $\omega_f = \Sigma \mathcal O_{\mathbb{P}^1_k}(-2)$ is not isomorphic to $1=\mathcal O_{\mathbb{P}^1_k}$.

Thus $C=\omega_f^{-1}$ and $C=1$ both satisfy $f_!(C)\simeq f_*(1)$ showing that $C$ need not be unique.


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