During a lesson we proved that a functor $F:A\text{-Mod} \rightarrow\text{Set}$, where $A$ is a Commutative unital ring, and which preserves finite products -- in the sense that
- it maps the terminal element $0$ to the terminal element $\{*\}$,
- for each $M,N$ in $A\text{-Mod}$ there exist a map $\varphi_{M,N}: F(M \oplus N) \rightarrow F(M) \times F(N)$ which is an isomorphism in Set
factorizes over $\text{Ab}$, in the sense that $F(M)$ has a natural abelian sum induced over it for each $M$.
Now, the proof is clear to me, but it assumes that the isomorphism $\varphi_{M,N}$ is natural in $M$ and $N$, while asking to prove it as an exercise.
I tried to prove it using the universal property of the product in Set, but I can't find a reason why the induced map, let's call it $\psi : F(M\oplus N) \rightarrow F(M) \times F(N)$ should be exactly $\varphi$. I'll explain this better:
(sorry for the quality of the diagram but this site doesn't seem to handle diagonal arrows well)
$$\require{AMScd} \begin{CD} F(M) @<{\pi_{F(M)}}<< F(M)\times F(N) @>{\pi_{F(N)}}>> F(N) \\ @| @A{\exists!}AA @| \\ F(M) @<{F(\pi_M)}<< F(M\oplus N) @>{F(\pi_N)}>> F(N) \end{CD}$$
This is the diagram that I expect should generate $\varphi$ and it indeed would imply its naturality. The problem is that I don't know yet that $\varphi$ is natural, thus that it is the map that makes the diagram commute, and if I forget about $\varphi$ and call this map $\psi$, I can't find a reason why it should necessarily be an isomorphism. I tried to define an inverse map but I wasn't very successful, and even in that case I couldn't seem to find a reason why the composition of the two maps would be the identity and not an automorphism.
EDIT: In a lapsus I wrote $\otimes$ instead of $\oplus$, needless to say that the result did not make sense. I now corrected it.