Morphism to an infinite coproduct on an abelian category given a family with only finite non-zero morphisms In abelian categories we know finite products and coproducts are isomorphic, but they are not necessarily the same when it comes to infinite products and coproducts, so you can't really get a unique morphism from an object $Z$ to $\coprod_{i \in I} A_i$ given an arbitrary family $\{f_i\}_{i \in I}$ of morphisms $f_i$ from $Z$ to $A_i$ if $I$ is infinite, while you can do so when $I$ is finite (using the fact that it is a coproduct).
However, in the category of abelian groups if only finitely many of those morphisms $f_i$ are non-zero, then there is such a special morphism (namely, the one that takes $z \in Z$ and sends it to $\sum_{i \in I} f_i(z)$).
So I was wondering, is it true that, given a family $\{f_i\}_{i \in I}$ of morphisms $f_i$ from $Z$ to $A_i$, there is a unique morphism $f$ from $Z$ to $\coprod_{i \in I}$ such that $p_i \circ f = f_i$ for every $i \in I$, with $p_i$ being the projection from $\coprod_{i \in I} A_i$ to $A_i$ (more explicitly, the unique morphism such that $p_j \circ u_i = 1_{A_i}$ if $i = j$ and $p_j \circ u_i = 0$ if $i \neq j$ if $u_i$ is the inclusion of $A_i$ on $\coprod_{i \in I} A_i$) if, and only if, $f_i \neq 0$ for finitely many $i \in I$?
 A: No. For example, $Z$ could itself be the infinite coproduct $\bigsqcup_{i \in I} A_i$, with the maps $f_i : Z \to A_i$ given by the projections $p_i$. Then there is a morphism $f : Z \to \bigsqcup_{i \in I} A_i$ such that $p_i \circ f = f_i$, namely the identity, and none of the $f_i$ are zero (assuming the $A_i$ are nonzero).
Even in the category of abelian groups you can see that in order for $\sum_i f_i(z)$ to make sense it's not necessary that only finitely many of the $f_i$ are nonzero, only that for any $z \in Z$ only finitely many of the $f_i(z)$ are nonzero, and that condition is satisfied by the infinite coproduct as above.
However, this is true if $Z$ is compact, meaning that $\text{Hom}(Z, -)$ preserves filtered colimits. Since the infinite coproduct is the filtered colimit of the finite coproducts, it follows that any map $Z \to \bigsqcup_{i \in I} A_i$ factors through a finite coproduct.
In $\text{Mod}(R)$ the compact objects are exactly the finitely presented modules. In a general abelian category they are the objects which "behave as if they are finitely presented."
