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.
 A: 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.
