Submodules in a preadditive category with one object. I'm trying to establish the full link between this generalization of rings on preadditive categories and the basic case where one considers a category $\mathcal{A} = \{X\}$ with one object. Is it easy to see that (How to see that a one-object pre-additive category is a ring?) a preadditive category with one object defines a ring. 
I was able to see that the following definition of $\mathcal{A}$-module:

Let $\mathcal{A}$ be a preadditive category. A left
  $\mathcal{A}$-module is an additive functor from $\mathcal{A}$ to
  $Ab$. That is a functor $F:\mathcal{A} \to Ab$ such that the
  corresponding map $F:\mathcal{A}(X,Y) \to \mathcal{B}(F(X),F(Y))$ is a
  group homomorphism for any pair of objects $X,Y$.

is consistent with the notion of module in $\mathcal{A} = \{X\}$. In fact, if $M$ is the additive functor, then $M(X)$ is a module with external operation given by $f \cdot m = M(f)(m)$ defined from $\mathcal{A}(X,X) \times M(X) \to M(X)$. 
Linking the two views in the case of submodules seems more delicate:

A submodule of a left $\mathcal{A}$-module $M$ is a subobject of $M$
  seen as an object of $\mathcal{A}$-mod.

here the notion of subobject is:

Let $X,Y,Z$ be objects in a category. Morphisms $f:X\to Z$ and $g:Y
 \to Z$ are called isomorphic if there exists an isomorphism $h:X \to
 Y$ such that $g \circ h = f$. A subobject of $Z$ is the isomorphism
  class of a monomorphism into $Z$.

According to the above definition I should be taking subobjects of a functor $F$ as $\mathcal{A}$-mod. So I should be considering morphisms between functors $Y,Z,K$. 
I'm new to category theory so I would appreciate if you could point out the way to go from here in order to show that the submdules of a module of the above ring coincide with the submodules in this new definition.
 A: $\newcommand{\A}{\mathcal{A}} \DeclareMathOperator{\mod}{-mod} \DeclareMathOperator{\Im}{Im}$
I think that the best way to understand how this functorial description capture the right notion of submodule is by observing how we can identify a submodule (in the classical sense) with an isomorphism class of injective module-homomorphisms (i.e. monomorphisms of $\A$-modules, also in the classical sense).
If $M$, $N$ and $N'$ are three $\A$-modules and let $f\colon N \to M$ and $g \colon N' \to M$ be two $\A$-module monomorphisms. 
They are isomorphic if and only if there is an $\A$-module isomorphism $\gamma \colon N \to N'$ such that 
$$f=g\circ\gamma\ .$$
It is not really hard to see that this happens if and only if $f$ and $g$ have the same images: i.e. $\Im(f)=\Im(g)$. From this it follows that there is a bijection between the isomorphism-classes of monomorphisms in $M$ and submodules of $M$, namely the mapping sending every monomorphism-class into the (common) image of all the elements of the said isomorphism-class.
Using the fact that the natural transformations in $\A\mod$ naturally correspond to the $\A$-module homomorphisms it shouldn't be hard to see how the functorial definition of submodule relates with the classical one.
Feel free to ask for any additional detail.
