Let $\prod_{n=1}^{\infty}\mathbb{Z}$ be the Baer-Specker group (infinite direct product of the additive group of integers) and $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ be the natural subgroup which is the infinite direct product with canonical basis $e_n$, $n\geq 1$.

An abelian group $A$ is slender if every homomorphism $f:\prod_{n=1}^{\infty}\mathbb{Z}\to A$ satisfies $f(e_n)=0$ for all but finitely many $n$. Apparently, this is equivalent to the property that every homomorphism $f:\prod_{n=1}^{\infty}\mathbb{Z}\to A$ must factor through $\mathbb{Z}^k$ for some $k$.

For the purpose of this question say an abelian group $A$ is nearly slender if $$\displaystyle Hom\left(\prod_{n=1}^{\infty}\mathbb{Z}\Big/\bigoplus_{n=1}^{\infty}\mathbb{Z},A\right)=0.$$ So $A$ is nearly slender if every homomorphism $f:\prod_{n=1}^{\infty}\mathbb{Z}\to A$ with $f(e_n)=0$ for each $n\geq 1$ is trivial.

Question: It seems that every slender group is nearly slender. Is there a nearly slender group which is not slender?


1 Answer 1


Yes. The displayed condition characterizes cotorsion-free groups. Every slender group is cotorsion-free but not every cotorsion-free group is slender. Take $A=\prod_{n=1}^{\infty}\mathbb{Z}$. Cotorsion-free groups are important in the study of endomorphism rings, see https://mathoverflow.net/a/117860/16678


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