An Introduction to Proof via Inquiry-Based Learning

There are two important types of sets related to functions.

The image of a subset \(S\) of the domain is simply the subset of the codomain we obtain by mapping the elements of \(S\text{.}\) It is important to emphasize that the function \(f\) maps elements of \(X\) to elements of \(Y\text{,}\) but we can apply \(f\) to a subset of \(X\) to yield a subset of \(Y\text{.}\) That is, if \(S\subseteq X\text{,}\) then \(f(S)\subseteq Y\text{.}\) Note that the image of the domain is the same as the range of the function. That is, \(f(X)=\range(f)\text{.}\)

When it comes to preimages, there is a real opportunity for confusion. In Section 8.3, we introduced the inverse relation \(f^{-1}\) of a function \(f\) (see Definition 8.70) and proved that this relation is a function exactly when \(f\) is a bijection (see Theorem 8.74). If \(f^{-1}:Y\to X\) is a function, then it is sensible to write \(f^{-1}(y)\) for \(y\in Y\text{.}\) Notice that we defined the preimage of a subset of the codomain regardless of whether \(f^{-1}\) is a function or not. In particular, for \(T\subseteq Y\text{,}\) \(f^{-1}(T)\) is the set of elements in the domain that map to elements in \(T\text{.}\) As a special case, \(f^{-1}(\{y\})\) is the set of elements in the domain that map to \(y\in Y\text{.}\) If \(y\notin \range(f)\text{,}\) then \(f^{-1}(\{y\})=\emptyset\text{.}\) Notice that if \(y\in Y\text{,}\) \(f^{-1}(\{y\})\) is always a sensible thing to write while \(f^{-1}(y)\) only makes sense if \(f^{-1}\) is a function. Also, note that the preimage of the codomain is the domain. That is, \(f^{-1}(Y)=X\text{.}\)

We can generalize the results above to handle arbitrary collections of sets.