Let $R$ and $S$ be commutative rings with unity and $\phi:R\to S$ be a ring homomorphism. Let $\mathfrak{a}\subseteq S$ be an ideal and $\mathfrak{p}$ be a minimal prime ideal containing $\mathfrak{a}$ (in the sense that there is no prime ideal $\mathfrak{p}'\subsetneq\mathfrak{p}$ that contains $\mathfrak{a}$). Must $\phi^{-1}(\mathfrak{p})$ be a minimal prime ideal containing $\phi^{-1}(\mathfrak{a})$?
I know the answer is yes if $\phi$ is surjective, for then if $\phi^{-1}(\mathfrak{a})\subseteq\mathfrak{q}\subseteq\phi^{-1}(\mathfrak{p})$ for a prime ideal $\mathfrak{q}$, then $\ker\phi\subseteq\mathfrak{q}$ and hence $\mathfrak{q}=\phi^{-1}(\phi(\mathfrak{q}))$. By this and surjectivity, $\phi(\mathfrak{q})$ is a prime ideal and hence must be equal to $\mathfrak{p}$.
For the general case, by factoring as $R\to\operatorname{im}(\phi)\hookrightarrow S$, we can see that it suffices to consider $R$ to be a subring of $S$ and $\phi$ to be the inclusion. From this, I am unable to come up with a proof or a counterexample. Any help will be appreciated.