Given two modules over a commutative ring with unity , can we define any module homomorphism between them using functors between categories? Let $G$ be a group  , let $\mathcal C_G$ be the category whose object consist of only one member namely $G$ , and the morphisms are elements of $G$ i.e. $Mor(G,G)=G$ , where for $x,y\in Mor(G,G)$ , define $x \circ y:=xy$ (the group operation ) . Then we notice that for groups $G,H$ , any group homomorphism $f:G\to H$ defines a natural functor(co-variant) from  $\mathcal C_G$ to  $\mathcal C_H$ and conversely any functor from  $\mathcal C_G$ to  $\mathcal C_H$ gives rise to a group homomorphism $f:G\to H$ . So if $M,N$ are $R$-modules then just considering the additive group structure , we can get an additive map between the modules via a functor as described , but can we get the $R$-homogeneity of a map $f:M\to N$ i.e. $f(rm)=rf(m)$ via some functor between some suitably defined categories which does not assume any knowledge of module homomorphism ? 
I am just beginning to learn category theory , so sorry if I am in error at some point . Please feel free to tell me and correct them . 
 A: I don't think there's a natural way to make modules into categories and module morphisms into functors between them. Modules don't feel like the right kind of object to do this (rings do, as I'll write down below).
That said, there is a way to turn modules into functors, and module homomorphisms into natural transformations between them.
First, you have to work with preadditive categories. As the article points, out a preadditive category with a single object is the same thing as a ring. It has a collection of morphisms (= elements) which form an abelian group, and they can be biadditively composed (= multiplied).
Now let $\mathcal R$ be such a category, its single object $\bullet$, and $R = \mathrm{Hom}(\bullet, \bullet)$ the corresponding ring.
An additive functor $F$ from $\mathcal R$ to $\mathrm{Ab}$ picks out a single Abelian group $M = F\bullet$, and comes down to a ring morphism $R = Hom(\bullet, \bullet) → Hom(M, M)$, which makes $M$ into a left $R$-module. Furthermore, module homomorphisms are exactly the natural transformations between such functors.
This might seem like a rather roundabout way to talk about modules, and in and of itself is just a rewording of the standard definitions, but it generalizes smoothly, and can be elegant to work with.
