Epimorphisms $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ are split Consider the additive group of integer sequences $\mathbb{Z}^{\mathbb{N}}$. Why does every epimorphism of groups $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ split? $(\star)$
Actually this claim is equivalent to the Whitehead problem for countable abelian groups:
"$\Rightarrow$": Recall Specker's result which states that $\mathbb{Z}^{(\mathbb{N})} \to \hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $e_i \mapsto \mathrm{pr}_i$ is an isomorphism, i.e. $\mathbb{Z}^{\oplus \mathbb{N}}$ is reflexive. Now assume that $A$ is countable and $\mathrm{Ext}^1(A,\mathbb{Z})=0$. Choose a presentation $0 \to P \to Q \to A \to 0$ with free abelian groups $P,Q$, w.l.o.g. of rank $\aleph_0$. By assumption $Q^* \to P^*$ is an epi, hence splits. Since $P,Q$ are reflexive, then also $P \to Q$ splits, and $A$ is free.
"$\Leftarrow$": If $f : \mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ is an epimorphism, then $f^* : \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ is a monomorphism, the cokernel $A$ of $f^*$ is countable and satisfies $\mathrm{Ext}^1(A,\mathbb{Z})=0$, since $f^{**} \cong f$ is epi. Hence $A$ is free, which implies that $f^*$ splits and therefore also $f^{**} \cong f$ splits. $~\square$
The countable Whitehead problem was proved by Stein in 1950. He used injective resolutions, i.e. $\mathrm{Ext}^1(A,\mathbb{Z}) = \mathrm{coker}(\hom(A,\mathbb{Q}) \to \hom(A,\mathbb{Q}/\mathbb{Z}))$. In particular, $(\star)$ is true. On the other hand, the equivalence above suggests an alternative proof of the countable Whitehead problem. Therefore my question is: Is there a direct proof for $(\star)$?
By Specker's result an endomorphism of $\mathbb{Z}^{\mathbb{N}}$ corresponds to an endomorphism of $\mathbb{Z}^{\oplus \mathbb{N}}$, and therefore to a column-finite matrix. But I don't know how to characterize surjectivity. And this is where I get stuck.
 A: I'll start by describing the notation that I'll use.
I'll think of elements of $\mathbb{Z}^\mathbb{N}$ as infinite column vectors
$${\bf x}=\begin{pmatrix}
x_0\\x_1\\x_2\\\vdots
\end{pmatrix}$$
of integers, with rows indexed by $\mathbb{N}$.
The $i$th "unit vector" will be denoted by ${\bf e(i)}$. I.e., $e(i)_i=1$ and $e(i)_j=0$ for $j\neq i$.
As described in the question, Specker's result implies that endomorphisms of $\mathbb{Z}^\mathbb{N}$ are of the form ${\bf x}\mapsto A{\bf x}$ where
$$A=\begin{pmatrix}
a_{00}&a_{01}&a_{02}&\cdots\\
a_{10}&a_{11}&a_{12}&\cdots\\
a_{20}&a_{21}&a_{22}&\cdots\\
\vdots&\vdots&\vdots&\ddots
\end{pmatrix}$$
is an infinite matrix of integers, with rows and columns indexed by $\mathbb{N}$, that is row-finite (i.e., each row has only finitely many non-zero entries).
Now suppose that $\alpha:\mathbb{Z}^\mathbb{N}\to\mathbb{Z}^\mathbb{N}$ is a surjective group homomorphism, given by a row-finite matrix $A$.
To prove that $\alpha$ is split, we need to find a right inverse. In fact,  it suffices to find another row-finite matrix $B$ such that $AB$ is lower unitriangular (i.e., diagonal entries are all $1$ and entries above the diagonal are all $0$), since such a matrix is invertible.
Fix $n\geq0$, and decompose $\mathbb{Z}^\mathbb{N}=G_n\oplus H_n$, where $G_n\cong\mathbb{Z}^n$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $i\geq n$, and $H_n\cong\mathbb{Z}^\mathbb{N}$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $i<n$.
Since $\alpha$ is surjective, $\mathbb{Z}^\mathbb{N}=\alpha\left(G_n\right)+\alpha\left(H_n\right)$.
Consider the quotient maps 
$$\mathbb{Z}^\mathbb{N}\stackrel{\theta}{\to} \mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\stackrel{\varphi}{\to} \left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)/T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right),$$ 
where $T(X)$ denotes the torsion subgroup of an abelian group $X$.
Since $\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)/T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$ is a finite rank free abelian group, using Specker's result again implies that $\varphi\theta({\bf e(i)})=0$, or equivalently $\theta({\bf e(i)})\in T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$, for all but finitely many $i$. 
Since $T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$ is finite, for all but finitely many $i$ with $\theta({\bf e(i)})\in T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$, there is $i'>i$ with $\theta({\bf e(i')})=\theta({\bf e(i)})$, or equivalently ${\bf e(i)}-{\bf e(i')}\in\alpha(H_n)$.
So we can choose a sequence $0=t_0<t_1<t_2<\dots$ of integers such that for every $i\geq t_n$ there is some $i'>i$ and some ${\bf b(i)}\in H(n)$ with ${\bf e(i)}-{\bf e(i')}=\alpha\left({\bf b(i)}\right)$. For each $i$, do this with the largest $n$ such that $i\geq t_n$, and let $B$ be the matrix whose $i$th column is ${\bf b(i)}$.
Then $B$ is row-finite, and $AB$ is lower unitriangular.
