Exact sequence with $F$ free splits Prove that every exact sequence of $R$ modules $$0 \rightarrow M \rightarrow N \xrightarrow{\phi} F \rightarrow 0$$ splits. This is an exercise from Algebra chapter 0, which does not go over projective modules before giving this exercise. It says to use the following prior exercise

Let $R$ be a ring, $F$ a nonzero free $R$-module, and let $\varphi : M \to N$ be a homomorphism of $R$-modules. Prove that $\varphi$ is onto if and only if for all $R$-module homomorphisms $\alpha : F \to N$ there exists an $R$-module homomorphism $\beta : F \to M$ such that $\alpha=\varphi \circ \beta$.

To solve it. So I will try to use this.
Attempt: Since $\phi$ is onto by definition of short exact sequence, for $\text{id}:F \rightarrow F$, there exists $\beta:F \rightarrow N$ such that $\text{id}=\phi \circ \beta$. Thus, $\beta$ is a splitting homomorphism for the sequence and it splits. Is this ok? I tried to come up with an explicit isomorphism $M \oplus F \cong N$ but could not figure this out. Can help me construct an isomorphism?
 A: For "Aluffi Algebra Chapter 0" context only:
Using the hint "6.9 th Exercise"

Let $R$ be a ring, $F$ a nonzero free $R$-module, and let $\varphi : M \to N$ be a homomorphism of $R$-modules. Prove that $\varphi$ is onto if and only if for all $R$-module homomorphisms $\alpha : F \to N$ there exists an $R$-module homomorphism $\beta : F \to M$ such that $\alpha=\varphi \circ \beta$.

Since $$0 \rightarrow M \rightarrow N \xrightarrow{\phi} F \rightarrow 0$$ exact, $\phi : N\to F$ is onto.
$$\require{AMScd}\begin{CD} F @>\exists ! \psi >> N  \\ @V Id VV \swarrow_{\phi}  \\ F \end{CD}$$
Such that the diagram commutative, so it means $Id_F=\phi\psi$. Means $\phi$ has right inverse, and using the splitting lemma we are done.
A: Hint. A splitting homomorphism $\beta$ yields a equality $N=M\oplus\beta(F)$ (internal direct sum) . This is a good exercise and a general fact about split sequences. Note now that $\beta$ is necessarily injective (why?).
It follows from that you have almost one only natural choice for your isomorphism $M\times F\to N$, in view of the equality $N=M\oplus\beta(F)$. Which one?
A: You can obtain an explicit isomorphism considering a basis $(e_i)_{i\in I}$ of $F$ and an element $n_i\in \varphi^{-1}\bigl(\{e_i\}\bigr)$ for each $i\in I$ (of course, if $I$ is not finite, this requires the axiom of choice).
Then set $\beta(e_i)=n_i,\: i\in I$.It is easy to check that $$N=\langle\, x_i\,\rangle_{i\in I} \oplus \ker\varphi\simeq F\oplus M.$$
