Duality for modules I'm doing exercises given in Maclane "Homology", and I have problems with the following exercise.
The task is to state the dual of this proposition: 
"For submodules $S_t\subset B, \ t\in T$ the following conditions are equivalent:
$(i)$ The diagram $\{j_t: S_t\rightarrow B\}, j_t$ the injection, is universal for ends $S_t$,
$(ii)$ $B=\cup S_t$ and, for each $t_0\in T$, $S_{t_0}\cap (\bigcup\limits_{t\neq t_0}S_t)=0$."
Following Maclane, the dual of a submodule is a quotient module.
It's easy to state dual of the beginning:
"For quotient modules $B/S_t, t\in T$ the following conditions are equivalent:
$(i)$ The diagram $\{j_t: B\rightarrow S_t\}, j_t$ the surjection, is couniversal for ends $S_t$,
$(ii)$ ???"
But I can't get what's the dual of $(ii)$.
Firstly, should I change "and" for "or", and "for each" for "exist"?
Suppose, yes. Secondly, I see two ways to state the dual.
In the first one I just change the module for its factor. So, $B$ is changed for $B/B=0$, $\cup S_t$ for $B/\cup S_t$. We get $0=B/\cup S_t$, which means $\cup S_t=B$ and nothing changed. The same thing for the second equality. So, we get:
1) $(ii)$ $B=\cup S_t$ or, there exists $t_0\in T$, such that $S_{t_0}\cap (\bigcup\limits_{t\neq t_0}S_t)=0$.
And here is the another obvious way:
2) $(ii)$ $0=\cap B/S_t$ or, there exists $t_0\in T$, such that $B/S_{t_0}\cup (\bigcap\limits_{t\neq t_0}B/S_t)=B$."
The question is: Is any of these variants correct? Or are they equivalent?
Please, help.
 A: Given submodules $S,T$ of a module $B$, it is more common to write $S+T$ for their sum, so the smallest submodule containing both $S$ and $T$, rather than use Maclane's notation $S\cup T$.
In the original statement (which is I, Proposition 4.6 in Maclane), we have inclusions $S_t\to B$. Then


*

*$B=\sum_tS_t$ is equivalent to saying that the natural map $\coprod_tS_t\to B$ is surjective.

*$S_{t_0}\cap\sum_{t\neq t_0}S_t=0$ is equivalent to saying that the composite $S_{t_0}\to B\to B/\sum_{t\neq t_0}S_t$ is injective.
The conclusion is then that the natural map $\coprod_tS_t\to B$ is an isomorphism.
We can now dualise this. We start with epimorphisms $B\to S_t$, and the dual conditions are


*

*The natural map $B\to\prod_tS_t$ is injective, equivalently the intersection of the kernels of the maps $B\to S_t$ is zero.

*The composite $\mathrm{Ker}(B\to\prod_{t\neq t_0}S_t)\to B\to S_{t_0}$ is surjective.
(I assume the conclusion should be that the natural map $B\to\prod_tS_t$ is an isomorphism, but I don't see this at the moment.)
