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I'm trying to verify that $\prod_{i\in\mathbb Z}\mathbb Z $(the direct product of countably many $\mathbb Z$) is not a coproduct in the category of abelian groups. We know that the coproduct object is $\oplus_{i\in\mathbb Z} \mathbb Z$ (the direct sum of countably many $\mathbb Z$), and since $\oplus_{i\in\mathbb Z} \mathbb Z$ and $\prod_{i\in\mathbb Z}\mathbb Z$ are not isomorphic, the direct product cannot be the coproduct by the uniqueness of universal objects. But I want to check it straightforward, by showing that $\prod_{i\in\mathbb Z}\mathbb Z$ does not safisfiy the universal property, as followed:

Let $C=\oplus_{i\in\mathbb Z}\mathbb Z$, there are natural inclusion maps $f_i:\mathbb Z\rightarrow \oplus_{i\in\mathbb Z}\mathbb Z$, mapping $\mathbb Z$ to the $i^{th}$ component of $\oplus_{i\in\mathbb Z} Z$. There are also inclusion maps $j_i:\mathbb Z\rightarrow \prod_{i\in\mathbb Z} Z$, with whom we assume $\prod_{i\in\mathbb Z}\mathbb Z$ is a coproduct in the category of abelian groups.

I want to show that these $f_i$ cannot be uniquely extended to $\phi:\prod_{i\in\mathbb Z} \mathbb Z\rightarrow \oplus_{i\in\mathbb Z}\mathbb Z$ such that $\phi\circ j_i=f_i$ for all $i$. Clearly any homomorphism $\phi$ which fixes $\oplus_{i\in\mathbb Z} Z$ will suffice (by identifying the direct sum as a subgroup of direct product). My question is: is the problem of $\phi$ being non-exist or non-unique? Is there even any momorphism other than $0$ from $\prod_{i\in\mathbb Z}\mathbb Z$ to $\oplus_{i\in\mathbb Z}\mathbb Z$?

Icer
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    From a purely set-theoretic side, $\prod_{i\in\mathbb{Z}}\mathbb{Z}$ is uncountable, but $\oplus_{i\in \mathbb{Z}}\mathbb{Z}$ is countable, so you cannot have a monomorphism from the former to the latter. – Arturo Magidin Apr 07 '20 at 01:01
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    It is a theorem of Specker that the only homomorphisms from $\prod \mathbb{Z}$ to $\mathbb{Z}$ are the linear combinations of the projections. See this previous post for your over-arching question. – Arturo Magidin Apr 07 '20 at 01:07

2 Answers2

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Here's another way to show there is no surjective homomorphism $\prod_{i \in \mathbb{Z}} \mathbb {Z} \to \bigoplus_{i \in \mathbb{Z}} \mathbb{Z}$. By a theorem of Specker, every homomorphism $\prod_{i \in \mathbb{Z}} \mathbb {Z}\to\mathbb{Z}$ factors through a finite subproduct. In particular, there are only countably many such homomorphisms. However, there are uncountably many different homomorphisms $\bigoplus_{i \in \mathbb{Z}} \mathbb{Z}\to\mathbb{Z}$, since each of the free generators can map anywhere in $\mathbb{Z}$. We could compose these homomorphisms with a surjective homomorphism $\prod_{i \in \mathbb{Z}} \mathbb {Z} \to \bigoplus_{i \in \mathbb{Z}} \mathbb{Z}$ to get uncountably many different homomorphisms $\prod_{i \in \mathbb{Z}} \mathbb {Z}\to\mathbb{Z}$, which is a contradiction. Thus no such surjective homomorphism can exist.

Eric Wofsey
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There is no surjective homomorphism $\phi: \prod_{i \in \mathbb{Z}} \mathbb {Z} \to \bigoplus_{i \in \mathbb{Z}} \mathbb{Z}$.

There is a theorem of Dudley in Continuity of homomorphisms which proves that any homomorphism from a Polish group to a "normable" group is continuous. As an example $\mathbb{Z}$ and direct sums of normable groups are normable. This type of result is known as an automatic continuity result: under what conditions are homomorphism (or whatever) always continuous.

Open subsets of $\prod_{i \in \mathbb{Z}} \mathbb {Z}$ are easy to describe, just coming from the product topology. In particular any map $\phi$ must have open kernel so the kernel must be a subgroup which has all but finitely many coordinates the full coordinate $\mathbb{Z}$ subgroup. This gives you a way to "classify" the maps $\phi$ you can have.

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    As a comment the direct sum is a dense subset of the direct product so if there were a map it would be completely determined by that subgroup. –  Apr 07 '20 at 01:51