Quotient group of free abelian group 
Let $A$ be a free abelian group, i.e. $A=\bigoplus_\alpha \mathbb Z$. Also let $B$ be a subgroup of $A$. Prove that $A/B\cong\mathbb Z$ implies $A=B\oplus \mathbb Z$.

p.s. Actually this appears in Hatcher's Algebraic Topology where he says $H_0(X)/\widetilde{H_0}(X)\cong\mathbb Z$ implies $H_0(X)\cong \widetilde{H_0}(X)\oplus\mathbb Z.$
 A: We have an exact sequence of abelian groups (with $\Sigma$ another index set)
$$
0\to B\xrightarrow{\phi}\mathbb{Z}^{\oplus \Lambda}\xrightarrow{\kappa}\mathbb{Z}\to 0
$$
We want to construct a map $\psi: \mathbb{Z}\to \mathbb{Z}^{\oplus \Lambda}$  such that $\kappa\circ \psi=\mathrm{id}_\mathbb{Z}$. Since $\kappa$ is surjective there exists some $F\in \mathbb{Z}^{\oplus \Lambda}$ such that $\kappa(F)=1$. Define $\psi(n)=nF$, then $\psi$ clearly satisfies what we wanted. Also $\psi$ is also injective.
We now have a (unique map by UMP of direct sum) map $\theta: B\oplus \mathbb{Z}\to \mathbb{Z}^{\oplus\Lambda}$ given by $\theta(b,n)=\phi(b)+\psi(n)=\phi(b)+nF$. We claim $\theta$ is an isomorphism. 
Let $\theta(b,n)=0=\phi(b)+nF$, then $\phi(b)=-nF$. Then $0=\kappa \phi(b)=-n\kappa(F)=-n$. Which means $n=0$ and therefore $\phi(b)=0$. Since $\phi$ is injective $b=0$ and therefore $\theta$ is injective.
For surjectivity; let $E_\mu=(e_\mu)\in \mathbb{Z}^{\oplus\Lambda}$ be such that for all $\lambda\neq \mu$ we have $e_\lambda=0$ and for $\lambda=\mu$ we have $e_\mu=1$. Also let $m_\mu=\kappa(E_\mu)$. But now $E_\mu-m_\mu F\in \mathrm{Ker}\kappa=\mathrm{Im}(\phi)$ (last part is because of exact sequence). Hence there exists $b_\mu\in B$ such that $\phi(b_\mu)=E_\mu - m_\mu F$. Then $\theta(b_\mu, m_\mu)=E_\mu$. This shows surjectivity.
Consequently $\mathbb{Z}^{\oplus \Lambda}\simeq B\oplus \mathbb{Z}$. (There is a shorter but less constructive proof using universal mapping properties by the way)
A very strong generalization of this theorem (with almost the same proof) is if $R$ is a commutative ring with unity, $M$ an $R$-module and $P$ a projective $R$-module and $K$ another $R$-module such that
$$
0\to K\to M\to P\to 0
$$
is exact, then $M\simeq K\oplus P$ (more precisely the above exact sequence is split). In fact any $R$-module $P$ such that the above exact sequence is split is projective (so if and only if). In your case $R=\mathbb{Z}$, and $P=\mathbb{Z}$ too (any free module is projective).
