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

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.

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.