The following theorem is well-known:

Let $A$ be a ring. Let $C$ be a cocomplete $Ab$-category and $X$ an $A$-module-object in $C$, i.e. an object endowed with a ring homomorphism $\sigma : A \to \text{End}(X)$. Then this yields a cocontinuous $Ab$-functor $F : \text{Mod}(A) \to C$ with $F(A)=X$ and the action of $F$ on endomorphisms of $A$ is given by $\sigma$. Actually this yields an equivalence of categories between the category of cocontinuous $Ab$-functors $\text{Mod}(A) \to C$ and the category of $A$-module objects in $C$.

The crucial step is the construction of $F$. It works as follows: Construct first the right adjoint of $F$ (which has to exist by abstract nonsense). For an object $Y$, obviously the abelian group $\text{Hom}(X,Y)$ carries the structure of a right $A$-module; just precompose with the action of $X$. The functor $\text{Hom}(X,-) : C \to \text{Mod}(A)$ has a left adjoint $F = X \otimes -$: Namely it is enough to define $F$ on objects and verify the corresponding universal property; *the action on morphisms, functoriality and preservation of colimits is automatic*. Well, $X \otimes A = X$, it is then clear what to do for free $A$-modules, and choosing a presentation of an arbitrary $A$-module $M$, it is clear how to define $X \otimes M$.

**Question** Is there a direct construction of the functor $X \otimes -$ which does not involve universal properties?

The background is that I want to prove some variants of this theorem where we cannot just enrich the hom-functor over $\text{Mod}(A)$ and use adjoints. So this question is not just out of curiosity. I hope that an affirmative answer to this question makes the other theorems accessible.

It's clear how to start: For every free $A$-module $M$, fix a basis $B_M$, and put $X \otimes M := X^{\oplus B_M}$ (I think if we do not want to use the global Axiom of Choice, we should use anafunctors instead). Thinking of matrices, it is easy to make $X \otimes -$ a functor on the free modules. Now if $M$ is an arbitrary $A$-module, fix once and for all a presentation

$M_1 \to M_2 \to M \to 0$

with free $A$-modules $M_1,M_2$, and define $F(M)$ to be the cokernel of $F(M_1) \to F(M_2)$, i.e. the sequence

$F(M_1) \to F(M_2) \to F(M) \to 0$

is exact (at least if $C$ is abelian, which you may assume if necessary). Now in order to show that $F$ is functorial, let $N$ be another $A$-module and $N_1 \to N_2 \to N \to 0$ its presentation. Suppose $f : M \to N$ is a homomorphism. Since $M_2$ is projective, we may lift it to a homomorphism $g : M_2 \to N_2$. However, I don't know how to go on.

Even if this works, I'm curious if we can do it without using that free objects in the presentation are projective. The motivation for this comes from this unanswered question.