When is a fully faithful functor an equivalant functor?

Let $\mathcal{A},\mathcal{B}$ be Abelian categories, and let $R:\mathcal{A}\to \mathcal{B}$, $L:\mathcal{B} \to \mathcal{A}$ be fully faithful functors. Moreover, we assume $R$ is right adjoint to $L$, and $R$ is also left adjoint to $L$. Are these conditions enough to ensure $R$ is an equivalent functor between $\mathcal{A},\mathcal{B}$ ?

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Yes. In fact this is true for any categories $\mathcal{A}$ and $\mathcal{B}$, only assuming that $L$ and $R$ are fully faithful functors that are in adjunction $L \dashv R$. It is not hard to show the following:

Lemma. Let $L \dashv R : \mathcal{A} \to \mathcal{B}$ be an adjunction.

1. $R$ is fully faithful if and only if the adjunction counit $\epsilon : L R \Rightarrow \textrm{id}_{\mathcal{A}}$ is a natural isomorphism.

2. $L$ is fully faithful if and only if the adjunction unit $\eta : \textrm{id}_{\mathcal{B}} \Rightarrow R L$ is a natural isomorphism.

The claim is this:

Proposition. A functor $F : \mathcal{A} \to \mathcal{B}$ is (half of) an equivalence of categories if and only if $F$ is fully faithful and has a fully faithful left or right adjoint.

Proof. Suppose $F$ is fully faithful and has a fully faithful right adjoint $G : \mathcal{B} \to \mathcal{A}$. Then the lemma implies $F G \cong \textrm{id}_{\mathcal{B}}$ and $\textrm{id}_{\mathcal{A}} \cong G F$, so $F$ is indeed (half of) an equivalence of categories. ◼

Here is an example of what goes wrong when you drop full-faithfulness. The terminal category $\mathbb{1}$ with exactly one object is an abelian category (for trivial reasons), and the canonical functor $\mathcal{A} \to \mathbb{1}$ has both a left adjoint and a right adjoint if $\mathcal{A}$ is a category with an initial object and a terminal object. Of course, if $\mathcal{A}$ is an abelian category then the initial and terminal objects coincide, so the left and right adjoints of $\mathcal{A}$ also coincide.

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Great! Thank you so much! – Li Yutong Mar 5 '13 at 19:03
In your lemma, the unit and counit should be switched. – Justin Campbell Jul 20 '14 at 14:30