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

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

• 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 7 at 1:01
• 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 7 at 1:07

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.
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.