This has to do with definition 4.3.1 and exercise 4.3.i in Riehl's "Category Theory in Context".

I am trying to determine the triangle identities for a pair of mutually left (or right) adjoint funtors, but I am running into some trouble. I tried dualizing the unit and counit definition of adjoint functors by replacing one of the categories with its opposite. Below is my attempt.

Let $F:C^{op}\rightarrow D$ and $G:D\rightarrow C^{op}$ such that $F\dashv G$. So, there must be a unit and counit, $\eta:1_{C^{op}}\Rightarrow GF$ and $\epsilon:FG:\Rightarrow 1_D$ satisfying the triangle identities. I.e., $(\epsilon F)\cdot(F\eta)=1_F$ and $(G\epsilon)\cdot(\eta G)=1_G$.

Now I attempt to interpret this back into the category $C$. The natural transformation $\eta:1_{C^{op}}\Rightarrow GF$ can be thought of as a natural transformation $\eta:GF\Rightarrow 1_C$ since each component of $\eta_c:c\rightarrow GFc$ in $C^{op}$ determines a component $\eta_c:GFc\rightarrow c$ in $C$. Naturality is preserved since composition of morphisms is reversed.

However, I did not make it much farther than this. Since the direction of $\eta$ is swapped, I see no interesting way to compose $\eta$ and $\epsilon$, even after whiskering. Does anybody spot where I have gone wrong or have any tips on how to proceed? Any input is appreciated.

EDIT I realized I may have messed up how the whiskering works with a contraviant functors. After reviewing it, I think it should be that $F\eta:F\Rightarrow FGF$ since $F$ can only be applied to $\eta:1_{C^{op}}\Rightarrow GF$, not $\eta:GF\Rightarrow 1_C$. But the components of this natural transformation are in $D$ so there is no reversing to do. (I think the notation I am using confused me). But, then we have the first triangle identity, $(\epsilon F)\cdot(F\eta)=1_F$, given by the assumption that $F\dashv G$.

Now, for the other triangle. We have $\eta G:G\Rightarrow GFG$. Since the components of this natural transformation are in $C^{op}$, this determines $\eta G:GFG\Rightarrow G$. We also have $\epsilon:FG\Rightarrow 1_D$. Applying $G$ lend $G\epsilon:GFG\Rightarrow G$ in $C^{op}$. This determines $G\epsilon:G\Rightarrow GFG$ in $C$. Finally, we have that $(\eta G)\cdot(G\epsilon)=1_G$ as an application of the reversal of composition applied to each component.

Does this seem correct? Sorry for the edits.


2 Answers 2


If you redefine $\eta:1_{C^\text{op}}\Rightarrow GF$ as $\eta:GF\Rightarrow 1_C$ (components are in $C$), you also need to redefine $F:C^\text{op}\rightarrow D$ and $G:D\rightarrow C^\text{op}$ as $F:C\rightarrow D^\text{op}$ and $G:D^\text{op}\rightarrow C$. There is no reversing the arrows of $C$ without reversing the arrows of $D$. After reversing all the arrows, the adjunction also reverses, in the sense that one gets $G \dashv F$ with $\eta$ as the counit and $\epsilon$ as the unit.

Update: The pair of mutual left adjoints would actually be $F:C^\text{op}\rightarrow D$ and $G:D^\text{op}\rightarrow C$. Writing $F^\text{op}$ and $G^\text{op}$ for the functors obtained by reverting all arrows in the categories, the two counits are $\eta:GF^\text{op}\Rightarrow 1_C$ and $\epsilon:FG^\text{op}\Rightarrow 1_D$. Reverting arrows again, we get $\eta^\text{op}: 1_{C^\text{op}}\Rightarrow G^\text{op}F$ with components in $C^\text{op}$ (we do analogously with $\epsilon$). Then we have the pair of adjoints $F\dashv G^\text{op}$ with unit $\eta^\text{op}$, which we will now call $\eta$, and counit $\epsilon$. This gives the triangle identities $(\epsilon F)\circ(F \eta)=1_F$ and $(G^\text{op}\epsilon)\circ(\eta G^\text{op})=1_{G^\text{op}}$. The latter gives $(\eta G)\circ(G \epsilon)=1_{G}$. The case of mutual right adjoints is analogous.

  • 1
    $\begingroup$ Thank you for the response. I see that taking $(-)^{op}$ of everything reverses the adjunction. I'm not trying to take the $(-)^{op}$ of everything and reverse the adjuction. Instead I am trying to define mutually left/right adjoints in terms of units and counits. This is definition 4.3.1 in Riehl's "Category Theory in Context". She leaves dualizing the triangle identities to the reader. $\endgroup$ Jan 13, 2022 at 17:58
  • $\begingroup$ Yes. I should have read the book first. I think now I understand, but shouldn't it be instead $G:D^{op}\rightarrow C$ in your third paragraph for you to have $F, G$ for a pair of mutual left adjoints? I realize that might have been just a typo. Hence, my confusion. $\endgroup$ Jan 13, 2022 at 18:14
  • $\begingroup$ @IsAdisplayName I added my solution. $\endgroup$ Jan 13, 2022 at 19:36
  • $\begingroup$ Thank you! I probably should have included the problem in the original question. $\endgroup$ Jan 14, 2022 at 21:31

I figured out the answer to this. The proper triangle identities are $(\epsilon F)\cdot (F\eta)=1_F$ and $(\eta G)\cdot (G\epsilon)$ for mutually left adjoints. Dually, they are $(G\epsilon)\cdot(\eta G)=1_G$ and $(F\eta)\cdot(\epsilon F)=1_F$ for mutually right adjoints. Since I don't think anybody is intrested, I won't type up the proof.

I'm not sure how to close the question without deleting it, though I might end up doing just that so no one spends time formulating a response.


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