Direct sum of free abelian group and quotient of abelian group by subgroup

Question

I'm currently studying abelian groups in Kurosh's The Theory of Groups. I'm trying to understand the proof of the theorem:

Let $B \leqq A$ be abelian groups. If $A/B \cong C$ and $C$ is a free group, then $A\cong B \oplus C$

Kurosh gives a proof on page 144, though I don't quite follow these steps, and am looking for some more detail. I have looked at Manipulating quotients and direct sums of abelian groups, and the answer is certainly helpful, but still does not include a full proof. Both seem to suggest constructing a subgroup of generating elements, but I can't seem to figure out what the isomorphism would look like. Any help would be greatly appreciated!

From Kurosh:

Answer 1

Since $C$ is free abelian, $C = \mathbb{Z}^{(I)}$ for some index set $I$. For each $i \in I$, let $e_i \in \mathbb{Z}^{(I)} = C$ be the element that’s $1$ at $i \in I$ and $0$ everywhere else. Since $A/B = C$, we may pick $a_i \in A$ s.t. $a_i + B = e_i$. Then the desired isomorphism $\pi: B \oplus C \to A$ is given by,

$$\pi(b \oplus (z_i)_{i \in I}) = b + \sum_{i \in I} z_ia_i$$

It is clearly a group homomorphism. To see it is injective, note that $\pi(b \oplus (z_i)_{i \in I}) = 0$ implies $\varphi(\pi(b \oplus (z_i)_{i \in I})) = 0$ where $\varphi: A \to A/B = C$ is the natural quotient map. But $B = \ker(\varphi)$ and $\varphi(a_i) = e_i$, so $0 = \varphi(\pi(b \oplus (z_i)_{i \in I})) = \sum_{i \in I} z_ie_i$. Thus, $z_i = 0$ for all $i$. But then $0 = \pi(b \oplus (z_i)_{i \in I}) = b$. Therefore, $\ker(\pi) = \{0\}$.

For surjectivity, let $a \in A$ be arbitrary. Then $\varphi(a) = (z_i)_{i \in I}$ for some $(z_i)_{i \in I} \in \mathbb{Z}^{(I)} = C$. Thus, $b = a - \sum_{i \in I} z_ia_i$ satisfies that $\varphi(b) = \varphi(a) - \sum_{i \in I} z_ie_i = 0$, i.e., $b \in B$. Now by definition of $\pi$, we see that $a = \pi(b \oplus (z_i)_{i \in I})$, so $a \in \text{ran}(\pi)$, whence $\pi$ is surjective.