Idea 10.5.2.
As in the figure above, \(\funcinvimg{f}{C} \) collects together all those elements of \(A \) whose images under \(f \) land inside \(C \text{.}\)
Section 10.5 Inverses
Suppose \(\funcdef{f}{A}{B} \) is a function. By definition, \(f \) associates an element of \(B \) to each element of \(A \text{.}\) Sometimes we want to reverse this process: given an element \(b \in B \text{,}\) can we determine an element \(a \in A \) such that \(f(a) = b \text{?}\) We’ll begin to answer this question by first finding all possible “reverse results” from elements in subsets of \(B \text{.}\)
- inverse image (of a subset \(C \) of the codomain \(B \))
-
the set of all domain elements \(a \in A \) for function \(\funcdef{f}{A}{B} \) for which the corresponding output element \(f(a) \) lies in the subset \(C \) of the codomain
- \(\funcinvimg{f}{C} \)
-
the inverse image of the subset \(C \subseteq B \) under the function \(\funcdef{f}{A}{B} \text{,}\) so that\begin{equation*} \funcinvimg{f}{C} = \setdef{a \in A}{f(a) \in C} \end{equation*}
Example 10.5.3. Some inverse images under sine.
Consider \(\funcdef{\sin}{\R}{\R} \text{,}\) the familiar sine function in trigonometry.
Then
\begin{equation*}
\inv{\sin}\bbrac{\{-1,0,1\}} = \setdef{\frac{m \pi}{2}}{m \in \Z}
\end{equation*}
because \(\sin(m \pi / 2) \) will equal \(0 \) when \(m \) is even and will equal \(1 \) or \(-1 \) when \(m \) is odd, and no other input values will produce outputs of \(0 \text{,}\) \(1 \text{,}\) or \(-1 \text{.}\)
Aside: Careful.
However,
\begin{equation*}
\inv{\sin}\bbrac{\setdef{y \in \R}{y \gt 1}} = \emptyset
\end{equation*}
because there are no input values for sine that will produce an output value greater than \(1 \text{.}\)
Now let’s return to the question of trying to reverse an input-output relationship \(f(a) = b \text{:}\) the set \(\inv{f}\bbrac{\{b\}} \) collects together all possible candidates for the inverse image of \(b \text{.}\)
- inverse image (of an element \(b \) of the codomain \(B \))
-
the inverse image \(\inv{f}\bbrac{\{b\}} \text{,}\) which consists of all domain elements \(a \in A \) for which \(f(a) = b \)
- \(\funcinvimg{f}{b} \)
-
simplified notation to mean the inverse image of element \(b \)
This gives us a way to associate to an element \(b \in B \) a set \(\funcinvimg{f}{b} \) of elements of \(A \text{.}\)
Question 10.5.4.
When does this association \(b \mapsto \funcinvimg{f}{b} \) give us a function \(\funcdef{\inv{f}}{B}{A} \text{?}\)
There are two possible ways that this will fail to give us a function.
-
Suppose there is an element \(b \in B \) such that the set \(\funcinvimg{f}{b} \) contains (at least) two distinct elements \(a_1,a_2 \text{.}\) Then in general there is no way to choose between \(\funcinvimg{f}{b} = a_1 \) and \(\funcinvimg{f}{b} = a_2 \text{.}\) Therefore, if \(f \) is not injective, the function \(\funcdef{\inv{f}}{B}{A} \) is not well-defined.
-
Suppose there is an element \(b \in B \) such that \(\funcinvimg{f}{b} = \emptyset \text{.}\) Then there is no element of \(A \) which we can assign to \(\funcinvimg{f}{b} \text{.}\) Therefore, if \(f \) is not surjective, the function \(\funcdef{\inv{f}}{B}{A} \) is undefined on some elements of \(B \text{.}\)
So it seems we will need a function to be bijective in order to be able to reverse the input-output rule to obtain an inverse function.
- inverse function
-
for a bijective function \(f \text{,}\) the inverse function associates to each codomain element of \(f \) the corresponding unique domain element that produces it through \(f \)
- \(\inv{f} \)
-
the inverse function \(\funcdef{\inv{f}}{B}{A} \) for bijective function \(\funcdef{f}{A}{B} \text{,}\) so that for \(b \in B \) we have \(\funcinvimg{f}(b) \) defined to be the unique element \(a \in A \) such that \(f(a) = b \)
Example 10.5.5. An invertible single-variable, real-valued function.
The function \(\funcdef{f}{\R}{\R} \text{,}\) \(f(x) = x^3 \text{,}\) is bijective and has inverse \(\inv{f}(x) = x^{\frac{1}{3}} \text{.}\)
Example 10.5.6. Inverting a numerical encoding of the alphabet.
Returning again to the bijection \(\funcdef{\varphi}{\Sigma}{B} \) encountered in Example 10.2.6 and Example 10.2.8, where
\begin{align*}
\Sigma \amp = \ShortEngAlphabet, \amp B \amp = \{ 1, 2, \dotsc, 26 \},
\end{align*}
the inverse function \(\funcdef{\inv{\varphi}}{B}{\Sigma} \) associates to each number \(1 \le b \le 26 \) the corresponding letter at that position of the alphabet. For example, \(\inv{\varphi}(11) = \mathrm{k} \text{.}\)
Example 10.5.7. A non-invertible function.
The function \(\funcdef{g}{\R}{\R} \text{,}\) \(g(x) = x^2 \text{,}\) does not have an inverse since it is not bijective. However, the function \(\funcdef{h}{\nnegset{\R}}{\nnegset{\R}} \text{,}\) \(h(x) = x^2 \text{,}\) so that \(h = \funcres{g}{\nnegset{\R}} \) but with codomain also restricted down to the image of \(g \text{,}\) has inverse \(\inv{h}(x) = \sqrt{x} \text{.}\)
Note 10.5.8.
If \(f \) is bijective, then so is \(\inv{f} \text{,}\) and \(\inv{f} \) is the unique function \(B \to A \) such that both
\begin{align*}
\inv{f} \funccomp f \amp = \id_A, \amp f \funccomp \inv{f} \amp = \id_B \text{.}
\end{align*}
Checkpoint 10.5.9.
Prove that if \(f \) is bijective then so is \(\inv{f} \text{,}\) and \(\inv{(\inv{f})} = f \text{.}\)
