Divisible principle of the exact sequence involved in vector spaces Given an exact sequence $...\to \mathbb C^m \to X\xrightarrow{\phi} Y \to \mathbb C^n\to...$, where $X,Y$ are $\mathbb Z$-modules.  Now given $k\in \mathbb Z$, if $\phi(x)$ is divisible by $k$ in $Y$, why does that imply $x$ is divisible by $k$ in $X$?
I think it should be related to the two vector spaces $\mathbb C^m$ and $\mathbb C^n$ but can't figure out the reason.
 A: In fact, $\mathbb{C}^m$ can be replaced with any divisible $\mathbb{Z}$-module and $\mathbb{C}^n$ can be replaced with any torsion-free $\mathbb{Z}$-module and the result will still hold. Indeed, suppose $$Z\xrightarrow{\alpha} X\xrightarrow{\phi} Y\xrightarrow{\beta}W$$ is an exact sequence of $\mathbb{Z}$-modules, where $Z$ is divisible and $W$ is torsion-free, and suppose that $\phi(x)$ is divisible by $k\in\mathbb{Z}\setminus\{0\}$ for some $x\in X$. Let $y\in Y$ be such that $\phi(x)=ky$. Since $\beta\circ\phi=0$, we know $$k\beta(y)=\beta(ky)=(\beta\circ\phi)(x)=0.$$ Since $W$ is torsion-free and $k\neq 0$, this forces $\beta(y)=0$, whence, by exactness at $Y$, we have $y\in\operatorname{im}\phi$. Thus let $x_0\in X$ be such that $\phi(x_0)=y$. Now we have $$\phi(x-kx_0)=\phi(x)-k\phi(x_0)=\phi(x)-ky=0,$$ and so, by exactness at $X$, we must have $x-kx_0\in\operatorname{im}\alpha$. Thus let $z\in Z$ be such that $\alpha(z)=x-kx_0$. Since $Z$ is divisible, there exists $z_0\in Z$ with $z=kz_0$, and we then have $$x=\alpha(z)+kx_0=k\alpha(z_0)+kx_0=k(\alpha(z_0)+x_0),$$ whence $x$ is divisible by $k$, as desired.
