= Solution
{c}
Put $K=\ker\phi$, $H=\ker\psi$, and $Q=\operatorname{Im}q$. The <kernel of a module homomorphism> $H$ is a <submodule>, so its image $L=q(H)$ is a <submodule>. For $x\in H$, $\phi(q(x))=\psi(x)=0$, showing $L\subseteq K$. Thus the <quotient module> $K/L$ exists. Also $L\subseteq Q$.
Consider the restriction to $K$ of the natural <module homomorphism> $R^m\to R^m/Q$:
$$
\alpha:K\longrightarrow R^m/Q,\qquad y\longmapsto y+Q.
$$
Its <kernel of a module homomorphism> is $K\cap Q$. We have $L\subseteq K\cap Q$. Conversely, if $y\in K\cap Q$, choose $x\in R^n$ with $y=q(x)$. Then
$$
\psi(x)=\phi(q(x))=\phi(y)=0,
$$
so $x\in H$ and $y\in L$. Hence $\boxed{K\cap Q=L}$.
The map $\alpha$ is <surjective>. Given $v+Q\in R^m/Q$, use the <surjectivity> of $\psi$ to choose $x\in R^n$ with $\psi(x)=\phi(v)$. Set $y=v-q(x)$. Then
$$
\phi(y)=\phi(v)-\psi(x)=0,\qquad y+Q=v+Q.
$$
Thus $y\in K$ is a preimage of the arbitrary class $v+Q$.
By the <first isomorphism theorem for modules>, $\alpha$ induces the <module isomorphism>
$$
\boxed{\overline\alpha:\ker\phi/L\xrightarrow{\ \cong\ }R^m/\operatorname{Im}q,\qquad
y+L\longmapsto y+\operatorname{Im}q}.
$$
Explicitly, replacing $y$ by $y+\ell$ with $\ell\in L$ leaves its class modulo $Q$ unchanged, proving well-definedness. Its formula preserves <addition> and <scalar multiplication>. If its value is zero, $y\in K\cap Q=L$, so $y+L=0$, proving <injectivity>; <surjectivity> was proved above. These checks give all the requested properties of the induced map.
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