# Every homomorphism from the Baer–Specker group into a slender group factors through ${\bf Z}^n$, why?

Wikipedia states that every homomorphism from the Baer–Specker group ${\bf Z}^{\bf N}$ into a slender group factors through ${\bf Z}^n$ for some natural number $n$. Where can I find a proof?

If this is not trivial, then I would also need a reference to insert in my paper.

• This seems to be the definition given in the same article. – Alex Degtyarev Nov 15 '14 at 18:36
• @AlexDegtyarev Sorry I can't see how the fact follows from the definition. Could you please provide more details? – Alexander Gelbukh Nov 15 '14 at 18:50
• Perhaps I don't see something very trivial: if $\varphi:{\bf Z}^{\bf N}\to G$ sends all $(0, ..., 0, 1, 0, ...)$ to $1$, why does it send $(1, 1, ...)$ to $1$? – Irina Nov 15 '14 at 21:05
• This is certainly not immediate from the definition. If we put $Q=\prod_n\mathbb{Z}/\bigoplus_n\mathbb{Z}$ then the quotient map $\pi:\mathbb{Z}^{\mathbb{N}}\to Q$ sends all $e_i$ to zero, but does not factor through $\mathbb{Z}^n$ for any $n$ (by a cardinality argument). If Wikipedia is correct then $Q$ must not be slender then there must be a different $\varphi:\mathbb{Z}^{\mathbb{N}}\to Q$ with $\varphi(e_i)\neq 0$ for infinitely many $i$. One can make such a map using $\mathbb{N}\simeq\mathbb{N}\times\mathbb{N}$, but certainly some work is required. – Neil Strickland Nov 15 '14 at 21:23

## 1 Answer

The key word is every in the definition.

Assume that the image of $f\colon\Bbb Z^{\Bbb N}\to G$ is infinitely generated (as otherwise we are done). Denote by $p_i$ the projections of $\Bbb Z^{\Bbb N}=\prod\Bbb Z$ to its factors and let $\inf a:=\min\{i\,|\,p_i(a)\ne0\}$ for $a\in\Bbb Z^{\Bbb N}$. Then, playing a bit, one can find an infinite sequence $a_i\in\Bbb Z^{\Bbb N}$ such that $f(a_i)\ne0$ and $\inf a_i>\inf a_j$ whenever $i>j$. (In particular, say, $\inf a_i\ge i$.) Each $a_i$ defines a homomorphism $h_i\colon\Bbb Z\to\Bbb Z^{\Bbb N}$, $1\mapsto a_i$, and, hence, compositions $g_{ij}:=p_j\circ h_i\colon\Bbb Z\to\Bbb Z$. By the assumption above, $g_{ij}=0$ if $i>j$; hence, we get well defined maps $u_j\colon\Bbb Z^{\Bbb N}\to\Bbb Z$, $(n_i)\mapsto\sum g_{ij}(n_i)$. Then, the composition of $\prod u_j\colon\Bbb Z^{\Bbb N}\to\prod\Bbb Z=\Bbb Z^{\Bbb N}$ and $f$ violates the definition.

I don't know where it is written but, as I mentioned in my comment, this is a direct consequence of the definition.