= Solution
{c}
The induced map on the <spectrum of a commutative ring> is <contraction of an ideal>:
$$
\boxed{f(\mathfrak q)=\varphi^{-1}(\mathfrak q).}
$$
To see that it is well defined, let $\mathfrak q$ be a <prime ideal> of $S$. Its inverse image is an <ideal> of $R$, and it is proper because $\varphi(1)=1\notin\mathfrak q$. If $ab\in\varphi^{-1}(\mathfrak q)$, then $\varphi(a)\varphi(b)\in\mathfrak q$, whence $\varphi(a)\in\mathfrak q$ or $\varphi(b)\in\mathfrak q$. This proves that the inverse image is a <prime ideal>.
For any <ideal> $I\subseteq R$, its <extension of an ideal> to $S$ is $IS=\varphi(I)S$. We have
$$
f^{-1}(V_R(I))=\{\mathfrak q:I\subseteq\varphi^{-1}(\mathfrak q)\}=V_S(IS).
$$
Thus inverse images of <Zariski-closed sets> are <Zariski-closed sets>, which proves that \b[$f$ is a continuous map].
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