 Solution

Let $e_1,\dots ,e_n$ denote the standard basis of a module of the free module $R^n$. Since the module homomorphism $\varphi :R^m\to M$ is surjective, choose $v_j\in R^m$ such that $\varphi (v_j)=\psi (e_j)$ for each $j$. DefineUniqueness of coordinates in a basis of a module makes this well-defined, and the formula preserves addition and scalar multiplication, so $q$ is a module homomorphism. For every $x={\sum }_j{r}_j{e}_j$,Thus $\boxed{\varphi \circ q=\psi }$. This is the lifting property of a free module; only the surjectivity of $\varphi$ is needed for this part.