Product-coproduct duality 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:

*

*There is an isomorphism $B \rightarrow \coprod_{t\in T} S_t$ (coproduct, i.e. direct sum)

*$B=\sum_{t\in T}S_t$ 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.
 A: 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:

*

*$j$ is surjective, that is, $\mathrm{Im}(j)=\sum_TS_t=B$;

*$j$ 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:

*

*$\pi$ is injective, that is, $\mathrm{Ker}(\pi)=0$;

*$\pi$ 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:

*

*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.

*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 $0$).

*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.

*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).

