I am curious about the statement mentioned in the title and tried to prove it. The proof looks fine to me but it would be nice if someone could confirm it to be correct.
Let $\phi: R\to S$ be a homomorphism of rings and let $I=(g_1,\dots, g_r)\subset S$ be an ideal. Then, $$\phi^{-1}(I)=(\phi^{-1}(g)\mid g\in I)=(\phi^{-1}(\sum a_ig_i)\mid a_i\in S)=(\sum\phi^{-1}(a_i)\phi^{-1}(g_i)\mid a_i\in S)\\ =(\sum b_i\phi^{-1}(g_i)\mid b_i\in R)=(\phi^{-1}(g_1),\dots,\phi^{-1}(g_r)).$$
Feel free to point at any mistake. Thanks!
There are two problems essentially boiling down to $\phi$ being possibly not invertible:
$\phi^{-1}(g)$ does not make sense if $g$ has more than one preimage.
Since $\phi$ is not given surjective, for $g$ not in the image of $\phi$,$\phi^{-1}(g)$ does not make sense.
Therefore, you need to change your hypotheses. Making $\phi$ surjective, and letting $\phi^{-1}(a)$ denote the set of all preimages of $a$ rather than a single one, should make your proof work.