Ideals obtained through ring homomorphisms

Question

Let $I$ be an ideal of S, and let $f$: $R \rightarrow S$ be a ring homomorphism. Show that $f^{-1}(S)$ is an ideal of $R$.

The solution is,

Let $g: S \rightarrow S$/$I$ be the quotient map. The kernel of $gf$ is $f^{-1}(I)$, and kernels are ideals.

I do not understand the solution at all. What does $gf$ "look" like, and whs it its kernel $f^{-1}(I)$? Does it involve the First Isomorphism Theorem? Really, any explanation to help me understand this problem would be tremendously helpful. Thank you.

Answer 1

$gf$ means $g \circ f$. That is $gf(r) = g(f(r))$. Does this clear things up for you?