TITLE: Any exact faithful functor is represented by a unique projective generator
QUESTION [7 upvotes]: In the book 'Tensor Categories' by Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik on page 10 it says:
'Conversely, it is well known (and easy to show) that any exact faithful functor $F :  \mathcal{C} \rightarrow \text{Vec}$ is represented by a unique (up to a unique isomorphism) projective generator $P$.'
But I could not find any proof of that fact. Can someone tell me how to prove it or where I can find a proof?
Note: Here $\mathcal{C}$ is a finite k-linear abelian category for some field k.

REPLY [5 votes]: Let $\mathcal C$ be the category of finite dimensional left modules over a finite dimensional ring $R$.   Let $G: \mathcal C \to \mathrm{Vec}$ be an exact and faithful functor to finite dimensional vector spaces. We use $V^*$ to denote the dual vector space. For motivation, notice that if we had a representing object $M$,  we would have $$G(R^*) = Hom_R(M,R^*) = Hom_R(R, M^*) = M^*.$$
Now $G(R^*)$ is a right $R$ module via the action of $R$ by left multiplication. So we define $P:= G(R^*)^*$ to be the dual left module and consider the functor $Hom_R(P,-)$.  
This functor is tautologically left exact, and it takes the injective left module $R^*$ to $Hom_R(P,R^*) = Hom_R(R, G(R^*)) = G(R^*)$.  Any other finite left module $M$ admits an injective presentation $$0 \to M \to R^{* \oplus a} \to R^{* \oplus b} \to $$
dual to the presentation of $M^*$ as a right $R$ module.  So by left exactness, we see that there is a natural isomorphism  $Hom_R(P,-) \simeq G(-)$. 
Thus $G$ is represented by $P$. Since $G$ is right exact $P$ is projective, and since it is faithful $P$ is a generator.