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Let $\mathcal{M}$ be the category of left $R$-modules. Then the product (direc product) and co-product (direct sum) of any family of objects $\{A_i\}_i$ in this category exists, denote by $\bigoplus_i A_{i\in I}$ and $\prod_{i\in I} A_i$.

When set $I$ is infinite, I have seen their difference only on the level of some counting arguments, and now I wanted to know, in the category theory, how they are really different.

In other words, to show that $\bigoplus_i A_i$ is not the co-product of $\{A_i\}_i$, I have to show that it violates a universal property of the product/co-product. But, I am confused here, how to proceed for it.

Can one give some hint to prove that the co-product and product are not isomorphic in the category of modules for an infinite family?

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For simplicity, let's consider the category $Ab$ of abelian groups. Now, the product of countably many copies of $\mathbb Z$ is simply their direct product: the set of all sequences of integers with point-wise operation. The coproduct though is the set of all eventually $0$ such sequences with point-wise operation. To verify these claims, simply establish the universal property. For the product it is completely straightforward, while for the coprouct there is a tiny little but to pay attention to.

Now, these two groups are not isomorphich, for instance by counting their cyclic subgroups. You can show directly that the product does not satisfy the universal property of the coproduct, and vice versa. Basically any attemp will be successful, so try to keep the choice of morphisms simple.

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  • $\begingroup$ Thanks for the suggestions; but still I have not solved. What I tried is to prove that $(\oplus_i A_i, \{\varphi_j\}_j)$ is not the product of $\{A_i\}$, where $\varphi_j\colon \oplus_i A_i\rightarrow A_j$ are morphisms, for this I assumed that this is product, and for $p_j:\prod_i A_i\rightarrow A_j$, the natural projections, there exists unique $\theta:\prod_i A_i\rightarrow \oplus_i A_i$ such that $\varphi_j\circ\theta=p_j$ for all j. But, taking an element $(a_i)_i \in \prod_i A_i$ with all $a_i$'s non-zero leads to contradiction. ...What I could not prove is that $\prod_i A_i$ is not sum. $\endgroup$ – p Groups Sep 23 '16 at 9:27
  • $\begingroup$ Just show directly that the coproduct universal property does not hold for the infinite product. $\endgroup$ – Ittay Weiss Sep 23 '16 at 9:40
  • $\begingroup$ sorry, I didn't get any idea, it may be very simple, but.... $\endgroup$ – p Groups Sep 24 '16 at 3:11
  • $\begingroup$ If you have a cone $\mathbb Z \to A$, how will you construct a function from the product to $A$? Where will you send the element $(1,1,1,1,\cdots)$ to? $\endgroup$ – Ittay Weiss Sep 24 '16 at 3:19
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It's possible for the product and coproduct to be isomorphic as modules.

For example, the product and coproduct of a countable collection of vector spaces of continuum dimension are vector spaces with the same (continuum) dimension.

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