Consider a short exact sequence of left $R$-modules $\DeclareMathOperator{\id}{id}$
$$0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$$
I want to show the following that the following statements are equivalent:
(1) The sequence above splits, i.e. $f(A)$ is a direct summand of $B$, i.e. there is a submodule $B' \leq B$ such that $B = f(A) \oplus B'.$
(2) There is a module homomorphism $\alpha: B \to A$ such that $\alpha \circ f = \id_A.$
(3) There is a module homomorphism $\beta: C \to B$ such that $g \circ \beta = \id_C$.
Here is my attempt for $(1) \implies (2),(3)$. Is this correct?
Assume (1) holds with $B = f(A) \oplus B'$. Define
$$\alpha: B \to A: b = f(a) + b' \mapsto a$$
Because the sum is direct and $f$ is injective, $\alpha$ is well defined. For $a \in A$, we have
$$\alpha \circ f (a) = \alpha(f(a)) = \alpha(f(a)+0) = a$$
and thus $\alpha \circ f = \id_A$. This shows $(2)$.
Define $\beta: C \to B$ in the following way.
Given $c \in C$, we can choose $b \in B$ such that $g(b) = c$ and we can decompose $b$ uniquely as $b = f(a) + b'$. We then define $\beta(c) := b'.$
Perhaps more clearly, $\beta: C \to B$ is defined by $\beta(g(b' +f(a)) = b'.$
This is well defined:
Assume $c= g(b_1) =g(b_2)$ with $b_1 = f(a_1) + b_1', b_2 = f(a_2) + b_2'.$ Then
$$b_1 - b_2 \in \ker g = f(A)$$
then $$b_1 -b_2 = f(a_1) + b_1'-f(a_1) - b_2' \in f(A) $$$$\implies b_1' - b_2' \in B' \cap f(A) = 0 \implies b_1' = b_2'$$
Now, let $c \in C$. Choose $b = b' + f(a)\in B$. Then $g(\beta(c)) = g(b') = g(b'+f(a)) = g(b) = c$ since $f(a) \in \ker g$. Hence, $g \circ \beta = \id_C$ and $(3)$ follows.
Is this correct?
This is fundamentally correct for me. However, you didn't prove that α and β are homomorphisms. Also, I would make the proof of (3) a bit clearer, along the following lines:
In the proof that any $b$ such that $g(b)=c$ has a decomposition $b=b'+f(a)$, where $b'\in B'$, add the remark that $b'$ is unique, in the sense that it is independent of $b$, and that $g(b')=c$.