Is the unit/counit property automatically satisfied for an equivalence of categories. The following may be a basic question.
Let $\mathscr{C}, \mathscr{D}$ be categories. Assume that $F: \mathscr{C}\to \mathscr{D}$ and $G: \mathscr{D}\to \mathscr{C}$ define an equivalence of categories, i.e. there are natural isomorphisms
$$\epsilon: FG \to \operatorname{id}_\mathscr{D}, \quad \eta: \operatorname{id}_{\mathscr{C}} \to GF.$$
Are the identities
$$F(\eta_U)= \epsilon_{FU}^{-1}, \quad G(\epsilon_X) = \eta_{GX}^{-1}$$
automatically satisfied?
I think we can choose $\epsilon, \eta$ to be natural isomorphisms satisfying these identities, but I'm not sure if they are automatic. After trying a little bit, I am starting to believe that these identities are not automatically satisfied. Can someone comfirm this hunch?
 A: Let $F:\sf C\to \sf D$ be an equivalence, coming with a quasi-inverse $ G:\sf D\to C$ and natural isos $\eta:\mathrm{id}_\mathsf C\Rightarrow GF$, $\epsilon:GF\Rightarrow\mathrm{id}_\mathsf D$.
Suppose to have an arrow $f:c\to Gd$, for $c\in\mathsf C,d\in \mathsf D$. Since $\eta_c$ is an iso, you get an arrow $f\circ \eta_c^{-1}:GFc\to Gd$. As $G$ is fully faithful, there is a unique arrow $g:Fc\to d$ such that $Gg=f\circ \eta_c^{-1}$; i.e. there is a unique $g:Fc\to d$ such that $Gg\circ \eta_c=f$, proving $F\dashv G$. Call $\xi$ the counit of this adjunction: it is a natural iso, because, for every $d\in \sf D$, one has $\mathrm {id}_{Gd}=G\xi_d\circ \eta_{Gd}$, meaning that $G\xi_d$ is an iso, and so that $\xi_d$ is an iso, for a fully faithful functor reflects isos. (But even if the unit is not an iso, you can prove that if $F\dashv G$ and $G$ is fully faithful, the counit is already an iso).
Hence you can recover an adjoint equivalence $F\dashv G$ with unit $\eta$, and the same is true for the counit, by duality.
