# $R, L$ are left and right adjoint in an equivalence of categories.

In my category theory course, we've defined an equivalence of categories to be a pair of adjoint functors $$(L, R)$$, i.e. $$\text{Hom}_{\mathcal D}(LC, D) \cong \text{Hom}_{\mathcal C}(C, RD)$$ in a natural way, such that $$L$$ and $$R$$ are fully faithfull. An equivalent definition is that $$(L, R)$$ is an adjunction and the unit and the counit $$\eta, \varepsilon$$ are natural isomorphisms.

It follows from the second definition, by reversing $$\eta$$ and $$\varepsilon$$, that $$(R, L)$$ is still an adjunction therefore $$R, L$$ are both left and right adjoint. Indeed $$\text{Hom}_{\mathcal D}(D, LC) \overset{R}{\longrightarrow} \text{Hom}_{\mathcal C}(RD, RLC)\overset{\eta^{-1}_C \circ -}{\longrightarrow}\text{Hom}_{\mathcal C}(RD, C)$$ and we can go the other way around by using $$L$$ and $$\varepsilon_D^{-1}$$. Now I have some troubles to show the fact that $$(R, L)$$ is still an adjunction but only by using the first definition of equivalence of category, i.e. the fully faithfullness of $$R, L$$. Does anyone know how to do that ?

• I think first you need to prove the invertibility of the unit and counit. Commented Jan 7, 2021 at 7:50
• It is not possible to show this without using the unit and the counit, just with the fully faithfulness of $R, L$? Commented Jan 7, 2021 at 7:52
• Well, there might be alternative ways to do this, but I guess the implication def.1$\implies$def.2$\implies R\dashv L$ is the cleanest. Commented Jan 7, 2021 at 8:33

Observe that $$L$$ and $$R$$ being full and faithful is equivalent to there being natural bijections: $${\rm Hom}_{\mathcal{C}}(C,C') \cong {\rm Hom}_{\mathcal{D}}(LC,LC') \ \ (1), \ \ \ \ {\rm Hom}_{\mathcal{D}}(D,D') \cong {\rm Hom}_{\mathcal{C}}(RD,RD') \ \ (2).$$ Then we have the following natural bijection: $$$$\begin{split} {\rm Hom}_{\mathcal{D}}(D,LC) & \cong {\rm Hom}_{\mathcal{C}}(RD,RLC) \ \text{ by (1),}\\ &\cong {\rm Hom}_\mathcal{D}(LRD,LC) \ \text{ since L \dashv R,}\\ &\cong {\rm Hom}_\mathcal{C}(RD,C) \ \text{ by (2).} \end{split}$$$$ Thus $$R \dashv L$$.