Homomorphisms from $\prod_{i\in\mathbb Z}\mathbb Z $ to $\oplus_{i\in\mathbb Z}\mathbb Z$ that fixes $\oplus_{i\in\mathbb Z}\mathbb Z$

Question

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 $\oplus_{i\in\mathbb Z}\mathbb Z$?

Answer 1

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.

Answer 2

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.