decomposable homomorphisms and full families of groups

Let ${\{G_i\}}_{i\in I},{\{H_i\}}_{i\in I}$ be two families of groups (indexed with the same set $I$).

Definition. We will say that a homomorphism

$$f:\underset{i\in I}{\oplus }{G}_i\to \underset{i\in I}{\oplus }{H}_i$$

is decomposable if there exists a family of homomorphisms ${\{f_i:G_i\to H_i\}}_{i\in I}$ such that

$$f=\underset{i\in I}{\oplus }{f}_i.$$

Remarks. For each $j\in I$ and $g\in {\oplus }_{i\in I}{G}_i$ we will say that $g\in G_j$ if $g(i)=0$ for any $i\ne j$. One can easily show that any homomorphism

$$f:\underset{i\in I}{\oplus }{G}_i\to \underset{i\in I}{\oplus }{H}_i$$

is decomposable if and only if for any $j\in I$ and any $g\in {\oplus }_{i\in I}{G}_i$ such that $g\in G_j$ we have $f(g)\in H_j$. This implies that if $f$ is an isomorphism and $f$ is decomposable, then each homomorphism in decomposition is an isomorphism and

$${\left(\underset{i\in I}{\oplus }{f}_i\right)}^{-1}=\underset{i\in I}{\oplus }{f}_i^{-1}.$$

Also it is worthy to note that composition of two decomposable homomorphisms is also decomposable and

$$\left(\underset{i\in I}{\oplus }{f}_i\right)\circ \left(\underset{i\in I}{\oplus }{g}_i\right)=\underset{i\in I}{\oplus }{f}_i\circ g_i.$$

Definition. We will say that family of groups ${\{G_i\}}_{i\in I}$ is full if each homomorphism

$$f:\underset{i\in I}{\oplus }{G}_i\to \underset{i\in I}{\oplus }{G}_i$$

is decomposable.

Remark. It is easy to see that if ${\{G_i\}}_{i\in I}$ is a full family of groups and $I_0\subseteq I$, then ${\{G_i\}}_{i\in I_0}$ is also a full family of groups.

Example. Let $𝒫=\{p\in \mathbb{N}|p\text{ is prime}\}$. Then ${\{{\mathbb{Z}}_p\}}_{p\in 𝒫}$ is full. Indeed, let

$$f:\underset{p\in 𝒫}{\oplus }{\mathbb{Z}}_p\to \underset{p\in 𝒫}{\oplus }{\mathbb{Z}}_p$$

be a group homomorphism. Then, for any $q\in 𝒫$ and $a\in {\oplus }_{p\in 𝒫}{\mathbb{Z}}_p$ such that $a\in {\mathbb{Z}}_q$ we have that $|a|$ divides $q$ and thus $|f(a)|$ divides $q$, so it is easy to see that $f(a)\in {\mathbb{Z}}_q$. Therefore (due to first remark) $f$ is decomposable.

Counterexample. Let $G_1,G_2$ be two copies of $\mathbb{Z}$. Then $\{G_1,G_2\}$ is not full. Indeed, let

$$f:\mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$$

be a group homomorphism defined by

$$f(x,y)=(0,x+y).$$

Now assume that $f=f_1\oplus f_2$. Then we have:

$$(0,1)=f(1,0)=(f_1(1),f_2(0))$$

and so $f_2(0)=1$. Contradiction, since group homomorphisms preserve neutral elements.