Let $A$ be an Artin algebra, $\text{mod}\,A$ the category of finitely generated $A$-modules and $\text{Ab}$ the category of abelian groups. Is every additive, covariant, left-exact functor $F:\text{mod}\,A \rightarrow \text{Ab}$ (natural) isomorphic to $\text{Hom}(X,-)$ for some $X\in \text{mod}\,A$? How do we obtain $X$?



Is every additive, left exact functor isomorphic to a hom functor?
One can prove the following result: let $F \colon \mathrm{Mod}\_R \to \mathrm{Mod}\_S$ be left-exact and preserve small products (equivalently, a continuous functor). Then $F$ is of the form $\mathrm{Hom}(M,-)$.


Let me first take a step back. The context of your question is (as I think you know) the Eilenberg-Watts theorem, a form of which states that if $G:{}\_R\mathrm{Mod}^{\mathrm{op}} \to \mathrm{Mod}\_S$ is left exact and preserves products, then there is an $(R,S)$-bimodule $M$ such that $G \simeq \mathrm{Hom}(-,M)$. In fact applying the functor to $R$ shows that $M =G(R)$, so the candidate module was obvious all along. And in a dual form, if $H:\mathrm{Mod}\_R \to \mathrm{Mod}\_S$ is right exact and preserves coproducts, then there is an $(R,S)$-bimodule $M$ such that $H \simeq -\otimes M$, and necessarily $M \cong H(R)$.


But in this case it's not as immediate what to plug into the functor $F$ to get a candidate module. However one can apply the special adjoint functor theorem. Module categories are complete, well powered, and have a cogenerator. So $F$ has a left adjoint $H$. You can apply Eilenberg-Watts to $H$ to deduce that $H \simeq - \otimes M$ for a bimodule $M$. Then $F \simeq \mathrm{Hom}(M,-)$.


In your case you have a left-exact additive functor $\mathrm{mod}\_R \to \mathrm{Mod}\_S$, where $\mathrm{mod}\_R$ denotes the compact objects (finitely presented modules). It extends uniquely to a functor $\mathrm{Mod}\_R = \mathrm{Ind}(\mathrm{mod}\_R) \to \mathrm{Mod}\_S$ preserving filtered colimits. If we make the additional assumption that the extended functor preserves all products, then by the above argument it is of the form $\mathrm{Hom}(M,-)$. Since it commutes with filtered colimits, $M$ is moreover compact.


