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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.

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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.

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  • $\begingroup$ Assuming that by $G(R)^*$ you mean the vector space dual, then $P=G(R)^*$ is not usually projective. Even when $G=\text{Hom}_R(R,-)$ is the forgetful functor, $G(R)^*\cong R^*$ which is not projective unless $R$ is self-injective. $\endgroup$
    – Jeremy Rickard
    Commented May 11, 2020 at 7:02
  • $\begingroup$ @JeremyRickard Thanks, I was missing a dual. I edited the post to fix the mistake-- I think it is correct now. $\endgroup$
    – Phil Tosteson
    Commented May 11, 2020 at 15:28
  • $\begingroup$ Why is the Hom-functor tautologically right exact? I thought since Hom is right adjoint to the tensor functor, it usually just is left exact. $\endgroup$
    – S.Farr
    Commented May 11, 2020 at 19:01
  • $\begingroup$ @S.Farr Sorry, I wrote it backwards-- I am using left exactness in the argument. $\endgroup$
    – Phil Tosteson
    Commented May 11, 2020 at 19:03
  • $\begingroup$ @PhilTosteson, good, then it makes sense. I think you also need to change the 'left exact' to 'right exact' in the last line. $\endgroup$
    – S.Farr
    Commented May 11, 2020 at 19:07

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