Let $T$ be a set, $R$ be a ring with $1$ and $B, S\_t$ be $R$-modules $\forall t \in T$


My task is to state and prove the dual to the following statement:


Given momomorphisms $j\_t: S\_t \rightarrow B$. Then there are equivalences:


1. There is an isomorphism $B \rightarrow \coprod\_{t\in T} S\_t$ (coproduct, i.e. direct sum)
2. $B=\sum\_{t\in T}S\_t$ and $S\_{t\_0} \cap (\sum\_{t \not= t\_0}S\_t) = 0$ $\forall t\_0 \in T$.


(Here $\sum$ means a sum of modules, not necessarily direct sum).


To state the dual, we first say that we have epimorphisms $\pi\_t :B \rightarrow B/S\_t$.
The dual to 1 is easy:


1') There is an isomorphism $B \rightarrow \prod\_{t\in T} B/S\_t$


To dualize 2, I try to rewrite it in "language of arrows":


2.a) $B=\sum\_{t\in T}S\_t\iff$ the natural map $\coprod\_{t\in T} S\_t \rightarrow B$ is surjective.


2.b) $S\_{t\_0} \cap (\sum\_{t \not= t\_0}S\_t) = 0$ $\forall t\_0 \in T \iff$ the composite $S\_{t\_0} \rightarrow B \rightarrow B/ \sum\_{t\not=t\_0}S\_t$ is injective.


Now it's easy to dualize 2.a:


2'.a) the natural map $B \rightarrow \prod\_{t\in T} B/S\_t$ is surjective (i.e. $\cap\_{t\in T} S\_t = 0$)


I have no idea how to dualize 2.b. What should be the dual object to $B/ \sum\_{t\not=t\_0}S\_t$?


The intuition tells me to simply write $S\_{t\_0} + (\bigcap\_{t\not=t\_0} S\_t) = B, \forall t\_0\in T$ as a dual to 2.b, but then the statement would be true only for finite set T.



Product-coproduct duality
The brief answer is the following:



> 
> the dual of a category of modules (i.e., any category equivalent to $(\mathrm{Mod}(R))^{\mathrm{op}}$ for some ring $R$) is not itself a category of modules, so you have to be careful when you try to dualize statements about modules.
> 
> 
> 


To be more precise, it is true that, given a ring $R$, the category $(\mathrm{Mod}(R))^{\mathrm{op}}$ is Abelian, so categorical statements that just involve *finite* limits and colimits (e.g., kernels, cokernels, finite products=coproducts) tend to dualize smoothly. On the other hand, problems may occur when you try to dualize statements that involve infinite co/limits.


The fact is that categories of modules are (Ab.5) but not (Ab.5$^\*$) Abelian categories, that is, in $\mathrm{Mod}(R)$ directed colimits are exact but inverse limits may fail to be exact.


This easy observation has the following consequence on lattices of submodules: given a right $R$-module $M$, consider a directed family of submodules $\{M\_i\}\_I$ of $M$ and a submodule $K\leq M$, then:
$$
K\cap\sum\_IM\_i=\sum\_I(K\cap M\_i).
$$
On the other hand, given an inverse system of submodules $\{M\_j\}\_J$ of $M$ and a submodule $H\leq M$, it is not difficult to find examples where
$$
H+\bigcap\_{J}M\_j\neq \bigcap\_{J}(H+M\_j).
$$
The failure of this dual equality makes it impossible to characterize products in $\mathrm{Mod}(R)$ in the way you want. Let me try to make you understand why: given your family of morphisms $\{j\_t\colon S\_t\to B\}\_{T}$, by the universal property of the coproduct, there is a canonical morphism $j\colon \coprod\_TS\_t\to B$ and this is an isomorphism if, and only if:


1. *$j$ is surjective*, that is, $\mathrm{Im}(j)=\sum\_TS\_t=B$;
2. *$j$ is injective*, that is, $\mathrm{Ker}(j)=0$. Now note that:
$$
\mathrm{Ker}(j)=\mathrm{Ker}(j)\cap \coprod\_TS\_t=\mathrm{Ker}(j)\cap \sum\_{\text{$T'\subseteq T$ finite}}\left(\coprod\_{T'}S\_t\right)=\sum\_{\text{$T'\subseteq T$ finite}}
\left(\mathrm{Ker}(j)\cap \coprod\_{T'}S\_t\right),$$
where, for the last equality, we have used condition (Ab.5), that is, the exactness of directed colimits. Hence, $\mathrm{Ker}(j)=0$ if, and only if, $\mathrm{Ker}(j)\cap \coprod\_{T'}S\_t=0$ for each finite subset $T'\subseteq T$ (and from this stronger characterization it is not difficult to recover the one you propose considering, instead of finite subsets of $T$, the bigger subsets of the form $T\setminus\{t\_0\}$ with $t\_0$ varying in $T$).


Let us now try to do the same (actually, the dual) with products: the family of morphisms $\{\pi\_t\colon B\to B/S\_t\}\_{T}$ gives you, by the universal property of products, a canonical morphism $\pi\colon B\to \prod\_TB/S\_t$. This $\pi$ is an isomorphism if, and only if:


1. *$\pi$ is injective*, that is, $\mathrm{Ker}(\pi)=0$;
2. *$\pi$ is surjective*, that is, $\mathrm{Im}(\pi)=\prod\_{T}B/S\_t$. Now note that we still have:
$$
\mathrm{Im}(\pi)=\mathrm{Im}(\pi)+ 0=\mathrm{Im}(\pi)+\bigcap\_{\text{$T'\subseteq T$ finite}}\left(\prod\_{T\setminus T'}B/S\_t\right)
$$
but, to go further, one would need to use the (Ab.5$^\*$) condition which, in general, does not hold in $\mathrm{Mod}(R)$.




---


Some final observations:


1. The unique hope I see to find some kind of criterion like the one you propose would be to take into account the derived functor of inverse limits of modules.
2. There are very nice Abelian categories that happen to be (Ab.5$^\*$) but not (Ab.5), for example, consider the category of compact Hausdorff topological Abelian groups, categories of linearly compact vector spaces or linearly compact modules over some linearly topologized ring (e.g., a commutative complete local Noetherian ring, with the unique topology for which the powers of the maximal ideal are a base of neighborhoods of $0$).
3. In the categories described in part 2 you have criteria for "internal product decomposition" along the lines you suggest in your question but, on the other hand, in such categories the criterion for "internal coproduct decomposition" fails.
4. Finally, the unique bicomplete Abelian category that is both (Ab.5) and (Ab.5$^\*$) is the trivial Abelian category $0$. In particular, in an Abelian category with products and coproducts you can hope for at most one of the two "internal decomposition criteria" (but never both if your category is not $0$): either it holds for coproducts (like for categories of modules) or it holds for products (like for compact Hausdorff topological groups).


