TITLE: Fully faithful and essentially surjective is an equivalence
QUESTION [16 upvotes]: The question asks to prove the statement in the subject.  
So assume the functor is $F: \mathcal{C} \rightarrow \mathcal{D}$ is fully faithful and essentially surjective.  We need to construct a map $G$, such that $F\circ G$ is naturally isomorphic to $id$.
So far I know how to construct the map $G$.  By definition of $F$ being fully faithful and essentially surjective, $\forall X \in \mathcal{D}$, we can find $A_x$, such that $F(A_x) \cong X$, hence define $G(X)=A_x$.  And for any $f \in Hom(X,Y)$, $X,Y$ being objects in $\mathcal{OB}(D)$, We can find $A_x,A_y$ such that $X \cong F(A_x)$, $Y \cong F(A_y)$, and thus $Hom(X,Y) \cong Hom(F(A_x),F(A_y)) \cong Hom(A_x,A_y)$.  Hence we can map $f$ to something in $Hom(A_x,A_y)$.
My problem is I don't know how to show $G\circ F$ is naturally isomorphic to $id$.  I am supposed to find for each $A,B \in \mathcal{C}$ and for any morphism $f$ from $A$ to $B$, an $\eta_A : A \rightarrow G\circ F(A),\eta_B: B \rightarrow G\circ F(B)$, such that $\eta_A \circ G\circ F(f) = \eta_B \circ f$.    I don't see how to show the diagram commutes.

REPLY [3 votes]: The following gives a direct proof. Assuming you already have constructed (using axiom of choice) the isomorphism of functors $\epsilon:F\circ G\Longrightarrow id$, we can easily construct $\eta:G\circ F\Longrightarrow id$.
First note that given $X\in Ob(C)$, since $F$ is fully faithful, we are able to map $\epsilon_{F(X)}$ to the unique morphism $\eta_X$ such that $F(\epsilon_{F(X)})=\theta_X$. We claim that $\eta$ is the desired natural isomorphism.
Suppose we have $f:X\rightarrow Y$. We have $G(F(f)):G(F(X))\rightarrow G(F(Y))$. Apply $F$ to both morphisms and use $\epsilon$ to obtain the usual commutative diagram. 
Since $F$ is faithful, we obtain $f\circ\eta_X=\eta_Y\circ(G\circ F)(f)$, proving that $\eta$ is indeed a natural transformation. Simply noting that a fully faithful functor is conservative allows us to conclude that $\eta$ is indeed an isomorphism.