What conditions ensure that a functor has a left / right inverse. A functor $F : C \to D$ has a left inverse $G : D \to C$ if $G \circ F : C \to C$ is naturally isomorphic to $\mathrm{id}_C$ and ditto for a right inverse. Are there nice criteria for when a functor admits a left inverse or a right inverse?
I know that a functor is an equivalence of categories (i.e. admits both a left and a right inverse) if and only if it is fully faithful and essentially surjective.
Is it true that a functor admits a left inverse if and only if it is faithful and a right inverse if and only if it is full and essentially surjective?
It is clear that if $G \circ F \simeq \mathrm{id}_{\mathrm{C}}$ then $G$ must be full and essentially surjective and $F$ must be faithful.
However, I can't show that these conditions are sufficient.
 A: Usually what you ask for is that a fully faithful functor $F$ has a left adjoint $L$ (and in that case, the counit $LF \to 1$ of the adjunction is an isomorphism), or a right adjoint $R$, and in that case the unit $1\to RF$ of the adjunction is an isomorphism.
Point is, you need $F$ to be fully faithful both times!
A: Neither are sufficient.
If $F$ is faithful, it need not admit a left inverse.
For example, $F:\varnothing\to\mathbf1$ is faithful, but no functor $\mathbf1\to\varnothing$ exists!
Even if we are mapping between nonempty categories, faithfulness is insufficient: just take $F$ to be the natural inclusion $\{\bullet\phantom\to\bullet\}\hookrightarrow\{\bullet\to\bullet\}$.
If $G:\def\cD{\mathcal D}\cD\to\def\cC{\mathcal C}\cC$ is essentially surjective and full, then it need not have an inverse either.
For example, let $\cD := \mathbf B\mathbb N_0$ be the one-object category with a morphism $\phi_n$ for every $n\geq0$ such that $\phi_n\phi_m=\phi_{n+m}$, and let $\cC$ be a one-object category with a non-identity endomorphism $\psi$ where $\psi\psi=\psi$.
Then, we have a functor $G:\cD\to\cC$ sending
$$
\phi_n \mapsto \begin{cases} \def\id{\operatorname{id}}\id, & n = 0 \\ \psi, & n > 0 \end{cases}
$$
This is full and surjective. However, it does not have a left inverse: if $GF\simeq\id_\cC$, then $GF=\id_\cC$ because the only isomorphism in $\cC$ is the identity.
In particular, $GF\psi=\psi$ means that $F\psi=\phi_n$ for some $n>0$. This is impossible because this would mean $\phi_n=F\psi=F(\psi\psi)=\phi_{2n}$ and thus $n=2n$, which is impossible.

You could say that the "problem" in both cases is the lack of faithfulness (hence @fosco's answer): being fully faithful helps control any of the regularity issues you might have when managing functors on the level of morphisms, since handling objects is mostly already fine (one way to see this is to just retrace the proof that a fully faithful and essentially surjective functor has a weak inverse and note where full and faithfulness are used).
Without being fully faithful, I am not sure if there is a "nice" way to describe when a functor has a weak left/right inverse besides just providing such an inverse.
