4
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 7 at 1:01
  • 2
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 7 at 1:07
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Paul Plummer Apr 7 at 1:51
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.