Let me ignore any ambitious statement and say that your abelian category $\mathsf{A}$ has a projective generator $p$, this means that $p$ is projective and separates arrows.

Notice that the $p$-points $\mathsf{A}(p,a)$ of an object $a \in \mathsf{A}$ admit a structure of $\mathsf{End}(p)$-module via the multiplication $\phi \cdot c = c \circ \phi.$

Now the representable functor $\mathsf{A}(p,-): \mathsf{A} \to \mathsf{End}(p)\text{-}\mathsf{Mod}$ is faithful (because $p$ is a generator) and exact (because $p$ is projective), as desired.

When $\mathsf{A}$ is the category of abelian groups and $p$ are the integers, this recipe produces the identity functor on abelian groups.

You might be disappointed because I assumed that there is a $1$-object generator, while very often we deal with a family of generators, to take care this problem, notice that when $\{p_i\}_{i \in I }$ is a generator, the object $\prod_{i \in I} p_i$ is a generator, and if $\mathsf{A}$ has enough projectives, its projective cover $Q \twoheadrightarrow \prod_{i \in I} p_i$ is a generator too and is projective. As a result if your abelian category has enogh projectives and products, one can reduce any family of generators to a projective generator.

> **Cor.** Let $\mathsf{A}$ be a complete abelian category with enough projectives and a generator $\{p_i\}_{i \in I }$, then there is a faithful and exact functor in the category of modules over the endomorphism ring of the projective cover of the product of the objects in the generator.