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 \(\funcdef{f}{\R}{\R}\text{,}\) \(f(x) = \sin x\text{.}\)
Then
\begin{equation*}
\inv{f}\bbrac{\{-1,0,1\}} = \setdef{\frac{m \pi}{2}}{m \in \Z}
\end{equation*}
because
\begin{equation*}
\sin\left(\frac{m \pi}{2}\right)
\end{equation*}
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 \(-1\text{.}\)
However,
\begin{equation*}
\inv{f}\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 \(\inv{(\inv{f})} = f\text{.}\)
