Fully faithful and essentially surjective is an equivalence

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.

Solution 1:

For $A$ in $C$ there is a $B$ in $C$ such that $F(B)\simeq F(A)$. But this $B=GF(A)$ will generally be distinct from $A$, so it's not a priori clear which isomorphism to take. I suggest that you go the other direction and find an isomorphism from $FG(X)\to X$, as such an iso is already given by hypothesis. So it is the best choice to make use of this assumed iso $FG(X)\to X$, let's call it $\varepsilon_X$.
Recall the lemma:

Given a functor $F:\mathcal C\to\mathcal D$, if for each $X\in\mathcal D$ there is a $G_0(X)\in\mathcal C$ and an arrow $ε_X:FG_0(X)\to X$ which is universal from $F$ to $X$, then $G_0$ is the object function of a unique functor $G:\mathcal D\to\mathcal C$ such that the $ε_X$ form a natural transformation $FG\to\mathbb I$. This $G$ is then right adjoint to $F$ and $ε$ is the counit of this adjunction.

In particular, this $G$ sends an arrow $g:X\to Y$ to the arrow $g':GX\to GY$ such that $ε_Y\circ Fg'=g\circε_X$, which is unique by universality of $ε_Y$.

Now, given $h:FA\to X$ we can compose $ε_X^{-1}\circ h:FA\to FGX$, which by fully faithfulness of $F$ yields a unique $h':A\to GX$ so that $Fh'=ε^{-1}_X\circ h$, or in other words $ε_X\circ Fh'=h$. Hence $\langle GX,ε_X\rangle$ is universal from $F$ to $X$.

In particular, $G$ has the compact form: $$Gg=F^{-1}\left(ε^{-1}_Y\circ g\circε_X\right)$$ To show that the unit $\eta:\mathbb I\to GF$ is an isomorphism, too, use the triangular identity $εF\circ F\eta=\Bbb I_F$.

Solution 2:

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.