Every free abelian group is slender, why?

This follows fairly straightforwardly from the fact that every map from $A=\mathbb{Z}^\mathbb{N}$ to $\mathbb{Z}$ factors through a finite subproduct (let me call this "Specker's theorem"). Suppose $F$ is a free abelian group and $f:A\to F$ is a homomorphism. Since a subgroup of a free abelian group is free, we may assume $f$ is surjective. If $F$ is finitely generated, the conclusion follows from Specker's theorem, so we may assume $F$ is infinitely generated. Since $F$ is free, the map $f$ splits and so $F$ is a direct summand of $A$. But there are uncountably many homomorphisms $F\to\mathbb{Z}$, and hence uncountably many homomorphisms $A\to \mathbb{Z}$. This contradicts Specker's theorem.

The short article by Nunke is available, with its own bibliography. It seems everything relevant happened from the late 1950's to the early 1960's.

I'm looking at Kap's book Infinite Abelian Groups. I do not see the word slender. However, Kap admired Baer, and one of nine references by Baer is Die Torsionsuntergruppe... , plus Baer is thanked in the Introduction. So I think it is a matter of figuring out how the concept "slender" is discussed here.

Hmmm; originally 1954, but second edition 1968. Should also check Fuchs, who has more than one edition.