= Solution
{c}
Let $e_1,\ldots,e_n$ denote the standard <basis of a module> of the <free module> $R^n$. Since the <module homomorphism> $\phi:R^m\to M$ is <surjective>, choose $v_j\in R^m$ such that $\phi(v_j)=\psi(e_j)$ for each $j$. Define
$$
\boxed{q:R^n\longrightarrow R^m,\qquad q\left(\sum_{j=1}^n r_je_j\right)=\sum_{j=1}^n r_jv_j}.
$$
Uniqueness 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 $x=\sum_jr_je_j$,
$$
\phi(q(x))=\sum_jr_j\phi(v_j)=\sum_jr_j\psi(e_j)=\psi(x).
$$
Thus $\boxed{\phi\circ q=\psi}$. This is the <lifting property of a free module>; only the <surjectivity> of $\phi$ is needed for this part.
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