If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general? Let $F : \mathbf{C} \rightleftarrows \mathbf{D} : G$ be an equivalence with natural isos $\alpha : 1_\mathbf{C} \to GF$ and $\beta : 1_\mathbf{D} \to FG$ witnessing the referred equivalence. I wonder if it holds generally that $\alpha_{GD} = G(\beta_D)$?
I think the answer is negative:
The reason is that, if it does hold necessarily, it seems that $\alpha$ is generally the unit of the adjunction
$F \dashv G$, since for any $f : C \to GD$ we have a unique $g : FC \to D$ given by $$g := \beta_{D}^{-1} \circ F(f) $$ such that
$$f = ((\alpha_{GD})^{-1} \circ GF(f)) \circ \alpha_C = ((G\beta_{D})^{-1} \circ GF(f)) \circ \alpha_C = G(\beta_{D}^{-1} \circ F(f)) \circ \alpha_C$$
And since I heard that it is not generally the case that if we have an equivalence between categories the natural isos witnessing it are the unit and counit, then it must follow that
$\alpha_{GD} = G(\beta_D)$ is not necessarily true.
Is this reasoning correct?
Thanks!
 A: If I understand correctly your argument try to prove that the claim is false proving that every equivalence satisfying the relation $\alpha_{G}=G(\beta)$ should give an adjoint equivalence $\langle F,G,\alpha,\beta\rangle$, where $\alpha$ should be the unit and $\beta^{-1}$ the counit of the adjunction.
Your proof shows that if the relation given, namely $\alpha_{G(D)}=G(\beta_D)$, holds for every $D \in \mathbf D$ then $\alpha$ is the unit of an adjunction between $F$ and $G$.  Unfortunately your proof doesn't show that $\beta^{-1}$ is the counit of the adjunction and so it doesn't prove that the equivalence provided is an adjoint equivalence.
Indeed is true (take a look at the remarks after proposition 3.4.3 of Borceux's Handbook of categorical algebra-Basic category theory) that if $\langle F,G,\alpha,\beta\rangle$ is an equivalence $\alpha$ is the unit of an adjunction between $F$ and $G$.
So if I've understood correctly your argument it seems that it doesn't disprove (or prove) that in an equivalence it should be $\alpha_{G(D)}=G(\beta_D)$.
Nevertheless I think too that the claim is false, although I've no counterexample at hand.....at least for the moment. Need some time to come up with something
Edit: here's a counterexample (based on an exercise of Borceux's book).
Consider the category $\mathbf C$ with only one object and two morphisms:
$1_C$ and $f$, both having as source and target the only object $C \in \mathbf C$, where $f^2=1_C$. 
In this category the morphisms $1_C$ and $f$ induce two natural auto-equivalences on the functor $1_{\mathbf C}$, hence the quadruple $\langle 1_{\mathbf C}, 1_{\mathbf C}, (1_C), (f)\rangle$ is an equivalence of categories, nonetheless $1_{\mathbf C}(f) \ne 1_C$ (in this context $G=1_{\mathbf C}$, $\alpha_{G(D)}=1_C$ and $\beta_D=f$).
Hope this helps.
