A functor $F \colon C \rightarrow D$ is an equivalence of categories if and only if $F$ is essentially surjective and fully faithful. I’m having problems understanding the proof for the $(\leftarrow)$ direction.
I’m following Christian Kassels “Quantum Groups”, page 278:
Proof of $(\leftarrow)$:
Let $F$ be an essentially surjective and fully faithful functor. For any object $W \in \mathrm{ob}(D)$, we choose an object $G(W) \in \mathrm{ob}(C)$ and an isomorphism $\nu(W) \in \hom_D(W,FG(W))$. We can do this because $F$ is essentially surjective.
If $g \colon W \rightarrow W'$ is a morphism of $D$, we may consider: $$\nu(W') \circ g \circ \nu(W)^{-1} \colon FG(W) \rightarrow FG(W') \,.$$
Since $F$ is fully faithful, there exists a unique morphism $G(g) \colon G(W) \rightarrow G(W')$ such that $$FG(g)=\nu(W') \circ g \circ \nu(W)^{-1} \colon FG(W) \rightarrow FG(W') \,.$$
This defines a functor $G \colon D \rightarrow C$ and $\nu \colon \mathrm{id}_D \rightarrow FG$ is a natural isomorphism.
In order to show that $F$ and $G$ are equivalences of categories, we have only to find a natural isomorphism $\theta \colon GF \rightarrow \mathrm{id}_C$.
Define $\theta(V) \colon GF(V) \rightarrow V$ for any object $V \in \mathrm{Ob}(C)$ as the unique morphism such that $F(\theta(V)) = \nu(F(V))^{-1}$. It is easily checked that this formula defines a natural isomorphism.
Can somebody explain to me why this map $\theta(V)$ exists, and how we know it’s unique? And then help me to do the easy check to show that this formula defines a natural isomorphism. Thanks!
Best Answer
Existence and uniqueness follow by virtue of the fact that $F$ is fully faithful: there is exactly one solution for $f$ in the equation $F(f)=\nu(F(V))^{-1}$ because $\nu(F(V))^{-1}:FGF(V)\to F(V)$ is a morphism whose domain and codomain lie in the image of $F$. You then define $\theta(V) := f$ to be said unique solution.
As for naturality, you are trying to see that $\require{AMScd}$ \begin{CD} GF(V) @>\theta(V)>> V \\ @VGF(\varphi)VV @VV\varphi V \\ GF(W) @>>\theta(W)> W \end{CD} commutes for any $\varphi:V\to W$. However, since $F$ is fully faithful, we can apply $F$ to the entire diagram and instead check commutativity of \begin{CD} FGF(V) @>\nu(F(V))^{-1}>> F(V) \\ @VFGF(\varphi)VV @VVF(\varphi)V \\ FGF(W) @>>\nu(F(W))^{-1}> F(W) \end{CD} but $FG(F(\varphi)) = \nu(F(W))\circ F(\varphi)\circ\nu(F(V))^{-1}$ by how $G$ is defined (take $g=F(\varphi)$ where you define the action of $G$ on morphisms), which completes the proof.