Let $\mathcal{F} : \mathcal{A} \rightarrow \mathcal{B}$ be a functor between abelian categories. Assume $\mathcal{F}$ is exact, essential surjective and faithful. Can I say that $\mathcal{F}$ is equivalence of categories?
As I know that essential surjective, faithful and full functor is equivalence. But Can exactness of functor can be helpful in my situation. If not what extra condition(of course not full functor) should I put on $\mathcal{F}$ so that it becomes equivalence of categories.
No. For instance, let $\mathcal{A}$ be the category of $\mathbb{Z}^2$-modules and $\mathcal{B}$ be the category of $\mathbb{Z}$-modules, with $F$ the forgetful functor. Then $F$ is exact and faithful, and essentially surjective since every $\mathbb{Z}$-module can be given a $\mathbb{Z}^2$-module structure. Buf $F$ is not an equivalence.
I doubt there is any condition you could add that is more useful than just directly asking for $F$ to be full.
One common definition of an equivalence of categories is an essentially surjective, full and faithful functor. But a stronger sense is an adjoint equivalence, which is a pair of functors each inverse to the other up to natural isomorphism -- sometimes called quasi-inverse. Different people prefer different ones of these.
Given enough choice the two definitions agree. For small categories the usual axiom of choice suffices to prove they agree (though note one essentially surjective, full and faithful functor will generally have many quasi-inverses). For class-sized categories you need some version of global choice dpending on your exact choice of foundation.
