Prove that if $X/A$ is a free module and if $A$ is a submodule of $X$, then $A$ is a direct summand of $X$

Question

Prove that if $X/A$ is a free module and if $A$ is a submodule of $X$, then $A$ is a direct summand of $X$. I need to prove it without sequences.

Answer 1

Let $p:X\rightarrow X/A$ the quotient map, If $X/A$ is a free module, there. exists $g:X/A\rightarrow X$ such that $p\circ g=Id$, to show that, take a basis of $(e_i)_{i\in I}$ of $X/A$, and write $g(e_i)=x_i$ where $p(x_i)=e_i$.

$g(X/A)$ is a summand, let $u\in X, p(u-g(p(u))=0, u-g(p(u))\in A$, $u=u-g(p(u))+g(p(u))$.

$u\in g(X/A)\cap A$ implies that $p(u)=0, u=g(v), p(g(v))=p(u)=0=v$ implies that $u=0$.