Let $$I: C \rightleftarrows D: F$$ be biadjoint [1] functors between categories $C, D$. That is, $I$ is the left and also the right adjoint of $F$ (thus vice versa). Put in notations, it's

$$ \cdots \vdash I \vdash F \vdash I \vdash F \cdots$$

Then since $I$ is left adjoint to $F$, for each object $X$ in $D$ there exists a unit map

$$ \eta_X: X \to IFX.$$

Similarly, since $F$ is left adjoint to $I$, there exists a unit map

$$ \eta'_{FX}: FX \to FIFX.$$

Question: Given their type, one naturally asks if $F(\eta_X): FX \to FIFX$ equals to $\eta'_{FX}$, or is there any general relation?

(Or if you have some interesting examples of such functors, please let me know. I'm happy to check if they satisfy this condition or not.)

Attempt: This is a special case of an ambidextrous adjunction. The corresponding (co)monads are Frobenius monads. I've looked in some paper about them, and couldn't find this structure discussed.

I've also tried to prove it myself.. but these unit maps come from different adjunctions, so a priori there is no reason for them to relate. But I can't be sure.

Footnote [1] "biadjunction" is not a standard term. see Mike Shulman's comment below

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    $\begingroup$ If $I$ is left adjoint to $F$, that is $I\dashv F$, then you have a unit $X\rightarrow FI(X)$ and a counit $IF(Y)\rightarrow Y$. $\endgroup$ Nov 21, 2020 at 23:17
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    $\begingroup$ Just a note that this isn't usually called a "bi-adjunction"; that word is more commonly used for a weak 2-adjunction. As you note, this situation is usually called an ambidextrous adjunction. $\endgroup$ Nov 22, 2020 at 0:52
  • $\begingroup$ I do understand Fernando's comment.. but how does that relate? @Shulman thanks for your comment! I thought ambidextrous is slightly weaker.. so I invented a name. Turned out to be bad :( let me at least add a warning! $\endgroup$
    – Student
    Nov 22, 2020 at 3:26
  • $\begingroup$ I think the best you can obtain is a commutative square made from two zig-zag identities: if $G\underset{\eta'}{\overset{\epsilon'}\dashv} F \underset{\eta}{\overset{\epsilon}\dashv} G$, then $\epsilon F \circ F\eta = F\epsilon'\circ \eta' F$. Now if for some reason $\epsilon F, F\epsilon'$ are invertible, the two arrows are "isomorphic in a certain comma category"; does this satisfy you? $\endgroup$
    – fosco
    Nov 22, 2020 at 8:59

2 Answers 2


(This is not an answer to your question, just a long comment.)

There is something to be careful about with ambidextrous adjunctions. When we work with an ordinary adjunction we can rest assured that there's no harm in just specifying one of the functors, because it having an adjoint is a property in the sense that the data of the adjoint functor, the unit, and counit are together unique up to unique isomorphism. (Original phrasing of this corrected following Peter LeFanu Lumsdaine's comment.)

Having an ambidextrous adjoint is not a property in this sense; the problem is that there is no natural map from the left adjoint to the right adjoint, so saying that $F \vdash G \vdash F$ amounts to saying that $G$ has both a left and a right adjoint and that they happen to be isomorphic, and that you are fixing an isomorphism between them, which you have conjured up somehow. This choice is not unique up to unique isomorphism; the set of such choices is a torsor over $\text{Aut}(F)$ (which is canonically isomorphic to $\text{Aut}(G)$), and an answer to a question about how the left and right adjoint structures interact may a priori depend nontrivially on this choice.

As an illustrative example, if $f : H \to G$ is a morphism of finite groups it induces a pullback functor $f^{\ast} : \text{Rep}(G) \to \text{Rep}(H)$ which has both a left and a right adjoint (induction and coinduction) $f_L, f_R : \text{Rep}(H) \to \text{Rep}(G)$ given by $k[G] \otimes_{H} (-)$ and $\text{Hom}_{H}(k[G], -)$ respectively. These functors are isomorphic, and the set of isomorphisms between them is a torsor over

$$\text{Aut}(f^{\ast}) \cong \text{Aut}_{H \times G}(k[G]) \cong Z_{k[G]}(k[H])^{\times}$$

and $Z_{k[G]}(k[H])$ has dimension the number of orbits of $H$ acting by conjugation on $G$. ($f^{\ast}$ is also naturally a symmetric monoidal functor and it would have fewer automorphisms if we ask for symmetric monoidal automorphisms; I think then we get $N_G(H)$ but I haven't checked this.)

As it happens, in this case, for reasons I don't entirely understand there is a preferred isomorphism between $f_L$ and $f_R$ called the Nakayama isomorphism, but I don't know how to get the Nakayama isomorphism out of purely the fact that $f^{\ast}$ has isomorphic left and right adjoints; it seems to me that more input is needed.

  • $\begingroup$ Thank you for the example. I think this answers my question directly -- it shows that it doesn't always satisfy my condition.. perhaps I will wait for slightly longer to see if someone else has something to say. $\endgroup$
    – Student
    Nov 22, 2020 at 3:29
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    $\begingroup$ “being adjoint is a property in the sense that the data of the unit and counit are unique up to unique isomorphism.” — Do you mean that, for a functor $F$, having a right/left adjoint is a property, in that $(G,\eta,\varepsilon)$ together are suitably unique? For given $F$, $G$, the data $(\eta,\varepsilon)$ are not in general unique (unless I’m severely undercaffeinated), and if they were, then “being biadjoint” would be a property too, surely? (But this is just a local possible-bug, the rest of your excellent answer still holds up.) $\endgroup$ Nov 22, 2020 at 10:16
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    $\begingroup$ I also don't fully understand the privileged role of the Nakayama isomorphism. But maybe it is interesting to note that the isomorphism of induction and coinduction generalizes to the setting where $f : B \rightarrow A$ is a homomorphism of finite dimensional algebras, $A$ is a free right $B$-module and $\mathrm{Hom}_B(A_B,B_B)$ is isomorphic to $A$ as a $B-A$-bimodule (with the usual action: $(b \cdot \theta)x = b (\theta x)$ and $(\theta \cdot a)x = \theta(ax)$). $\endgroup$ Nov 22, 2020 at 11:58
  • $\begingroup$ @Peter: oops, yes, you're right, otherwise the same argument applies. I'll edit. $\endgroup$ Nov 22, 2020 at 22:10

Qiaochu Yuan’s answer excellently explains the general phenomenon. Here is another slightly simpler concrete example:

Let $\newcommand{\Z}{\mathbb{Z}}\newcommand{\BZ}{{\mathbf{B}\Z}}\BZ$ be the group $\Z$ viewed as a one-object groupoid, with arrows $a^n : \ast \to \ast$ for $n \in \Z$. Then its identity functor $\newcommand{\id}{\mathrm{id}}\id_\BZ$ is self-adjoint in $\Z$-many ways: for any $n$, we can take $a^n$ as the unit and $a^{-n}$ as the counit. (Any identity functor is trivially self-adjoint, and then as Qiaochu explains, we can conjugate it by natural automorphisms of the identity functor, which in this case are precisely $\Z$.)

So now we can choose units+counits for the biadjunction $\id_\BZ \dashv \id_\BZ \dashv \id_\BZ$ in $\Z^2$-many ways. Writing $a^n$ and $a^m$ for the two choices of unit, your condition will hold just when $m=n$.


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