# 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?

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?
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 $$N=\langle\, x_i\,\rangle_{i\in I} \oplus \ker\varphi\simeq F\oplus M.$$