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Let $(E_i)_{i \in I}$ be a non-empty collection of modules over a given ring $R$ (this could be generalized to other suitable categories, but I won't consider these here). We define the countably-direct sum by taking all the elements of the product $\prod_{i \in I} E_i$ having at most countably many non-zero components: $$ \underset{i \in I}{\Large \boxplus}\; E_i := \left\{x \in \prod_{i \in I} E_i \;\middle|\; \#\{i \in I : x_i \neq 0\} ≤ \aleph_0 \right\}$$

My question is:

Is there a universal property (using objects and morphisms in $\mathsf{Mod}_R$) characterizing the countably-direct sum, as there is for the direct sum?


I recall the universal property for direct sums, a.k.a. coproducts :

For any object $M$ and any collection of morphisms $f_i : E_i \to M$ there exists a unique morphism $f : \bigoplus\limits_{i \in I} E_i \to M$ such that $f_i = f \circ r_i$ for all $i \in I$ (where $r_j : E_j \to \bigoplus\limits_{i \in I} E_i$ denotes the canonical inclusion).

$$\begin{array}{center} \quad\qquad \bigoplus_{j \in I} E_j \;\dashrightarrow M \\ \quad\qquad \uparrow r_i \, \, \, \, \, \, \nearrow f_i \\ E_i \\ \end{array}$$

The definition of universal property can be more general than this one, because we are dealing with a family of objects (see for instance these diagrams). Related questions: What things can be defined in terms of universal properties?, How to define $A\uparrow B$ with a universal property as well as $A\oplus B$, $A\times B$, $A^B$ in category theory?.

Following this question and this MO answer, I tried to show that the countably-direct sum doesn't satisfy the universal property of direct sums, but I was stuck.

Thank you for your help!

Watson
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    In case $I$ is countable this is just the product. By a cardinally argument we know that a infinite countable products are not always the direct sum, so it can not satisfy the universal property of direct sums. What we know is that the countably-direct sum is kind of the union of the countable products. Thus the countably-direct sum should be the direct limit of all those countable products. Since products and direct limits have universal properties you get at least some kind of a universal property. – user60589 Aug 17 '16 at 13:17
  • Well. Kind of, but a colimit of a limit doesn't have a universal property in its own right. – Kevin Carlson Aug 17 '16 at 15:33
  • The finiteness thing in the coproduct comes in because the operations in hand are finitary. If we considered "$\sigma$-modules" (over "$\sigma$-rings"), their coproduct would be just the countably direct sum. – Berci Aug 18 '16 at 00:46
  • @Berci : what do you mean by "$\sigma$-modules"? – Watson Aug 18 '16 at 08:33

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