Set-Theoretic Concepts: Inverses, Images, and Pre-Images

A function $f(x)$ maps the elements of some set $S$ to elements in another set $T$. Something helpful to consider is whether there is any way to "restore" the original set, i.e. whether there is a function $g(x)$ from $T$ to $S$ so that for all $s \in S$, $g(f(s))=s$. If such a function exists, $g$ is called the inverse of $f$, written as $f^{-1}(x)$ and $f$ is an invertible function.

The problem is, I lied. $g(x)$ isn't a function from $T$ to $S$, as we have defined it only for those $t\in T$ for which there is some $s \in S$ such that $t=f(s)$. The domain of $f^{-1}$, the values in $T$ that can actually be obtained by plugging something into $S$, is called the image of the function $f$. These are the same if $f$ is what is know as a surjective function.

Inverses have some nice properties. For example, if $g$ is the inverse of $f$, then by definition for all $s \in S$, $g(f(s))=s$. But this implies that $f(g(f(s)))=f(s)$, and every element $t$ in the image of $f$ is by definition equal to $f(s)$ for some $s \in S$, so $g(f(t))=t$ and $f$ is the inverse of $g$. What this means is that $(f^{-1})^{-1}=f$.

Finally, it can be helpful to, instead of taking a subset of $S$ and seeing what it is mapped to in $T$, look at a subset $I$ of $T$ and find the set of all $s\in S$ for which $f(s) \in I$. This set is called the preimage of $I$. As an example of this, by definition the preimage of the image of $S$ under any function equals $S$