Condition for an additive functor to be an equivalence

Question

Consider an additive functor $F : \mathcal{A} \longrightarrow \mathcal{B}$ between abelian categories and suppose that $F$ is a dense functor, that is, for every object $B$ in $\mathcal{B}$ there is some object $A$ in $\mathcal{A}$ such that $F(A) \simeq B$. Well, to prove that $F$ is an equivalence of categories we need to show that it is full and faithful, and my question is: are there other ways to prove that $F$ is an equivalence, other than showing that $F$ is full and faithful?

In particular, if we know that $F$ is exact, under which hypothesis can we conclude that $F$ is an equivalence?

Answer 1

I hope this may be useful:

LEMMA. Let $F\colon \mathcal A\to \mathcal A'$ be an exact functor between Abelian categories. If $F$ reflects $0$-objects (i.e., $F(X)=0$ implies $X=0$), then $F$ is faithful.

Proof. Consider a morphism $\phi\colon A\to B$ in $\mathcal A$ such that $F(\phi)=0$. Let $\pi\colon A\to A/\ker(\phi)$ be the obvious projection and let $\bar\phi\colon A/\ker(\phi)\to B$ be the unique morphism such that $\bar\phi\pi=\phi$. Note that $\pi$ is an epimorphism and $\bar\phi$ is a monomorphism so, by the exactness of $F$, $F(\pi)$ is an epimorphism and $F(\bar \phi)$ is a monomorphism. One can now conclude as follows: $0=F(\phi)=F(\bar\phi\pi)=F(\bar\phi)F(\pi)$ implies that $F(\bar\phi)$ is $0$ as we can cancel the epimorphism $F(\pi)$, but $F(\bar\phi)$ is also a monomorphism, so $F(A/\ker(\phi))=0$, which means that $A=\ker(\phi)$, since $F$ reflects $0$-objects. ///