Mutual Left/Right Adjoints and the Triangle Identities

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.

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.
• 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. Jan 13, 2022 at 17:58
• 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. Jan 13, 2022 at 18:14
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.