10.5: Inverses

Suppose $f: A\rightarrow B$ 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 $B\text{.}$

Definition: inverse image (of a subset $C$ of the codomain $B$)

the set of all domain elements $a \in A$ for function $f: A\rightarrow B$ for which the corresponding output element $f(a)$ lies in the subset $C$ of the codomain

Definition: $f^{-1}(C)$

the inverse image of the subset $C \subseteq B$ under the function $f: A\rightarrow B\text{,}$ so that

$$\begin{equation*} f^{-1}(C) = \{a \in A \vert f(a) \in C\} \end{equation*}$$

Figure $\PageIndex{1}$: A Venn diagram of a function inverse image.

Idea $\PageIndex{1}$

As in the figure above, $f^{-1}(C)$ collects together all those elements of $A$ whose images under $f$ land inside $C\text{.}$

Example $\PageIndex{1}$: Some inverse images under sine.

Consider $f: \mathbb{R} \rightarrow \mathbb{R}\text{,}$ $f(x) = \sin x\text{.}$

Then

$$\begin{equation*} f^{-1}(\{-1,0,1\}) = \left \{\dfrac{m \pi}{2} \vert m \in \mathbb{Z} \right \} \end{equation*}$$
because

$$\begin{equation*} \sin\left(\dfrac{m \pi}{2}\right) \end{equation*}$$
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,

$$\begin{equation*} f^{-1}(\{y \in \mathbb{R} \rightarrow y \gt 1\}) = \emptyset \end{equation*}$$
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 $f^{-1}\bbrac{\{b\}}$ collects together all possible candidates for the inverse image of $b\text{.}$

Definition: inverse image (of an element $b$ of the codomain $B$)

the inverse image $f^{-1}(\{b\})\text{,}$ which consists of all domain elements $a \in A$ for which $f(a) = b$

Definition: $f^{-1} (b)$

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 $f^{-1} (b)$ of elements of $A\text{.}$

Question $\PageIndex{1}$

When does this association $b \mapsto f^{-1} (b)$ give us a function $f^{-1} : B \rightarrow A\text{?}$

There are two possible ways that this will fail to give us a function.

1. Suppose there is an element $b \in B$ such that the set $f^{-1} (b)$ contains (at least) two distinct elements $a_1,a_2\text{.}$ Then in general there is no way to choose between $f^{-1} (b) = a_1$ and $f^{-1} (b) = a_2\text{.}$ Therefore, if $f$ is not injective, the function $f^{-1} : B \rightarrow A$ is not well-defined.
2. Suppose there is an element $b \in B$ such that $f^{-1} (b) = \emptyset \text{.}$ Then there is no element of $A$ which we can assign to $f^{-1} (b)\text{.}$ Therefore, if $f$ is not surjective, the function $f^{-1} : B \rightarrow A$ 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.

Definition: 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 $f$

Definition: $f^{-1}$

the inverse function $f^{-1} : B \rightarrow A$ for bijective function $f: A\rightarrow B\text{,}$ so that for $b \in B$ we have $f^{-1} (b)$ defined to be the unique element $a \in A$ such that $f(a) = b$

Example $\PageIndex{2}$: An invertible single-variable, real-valued function

The function $f: \mathbb{R} \rightarrow \mathbb{R}\text{,}$ $f(x) = x^3\text{,}$ is bijective and has inverse $f^{-1}(x) = x^{\dfrac{1}{3}}\text{.}$

Example $\PageIndex{3}$: Inverting a numerical encoding of the alphabet

Returning again to the bijection $\varphi: \Sigma \rightarrow B$ encountered in Example 10.2.4 and Example 10.2.6, where

$$\begin{align*} \Sigma & = \{a, b, \ldots, z\} , & B & = \{ 1, 2, \ldots, 26 \}, \end{align*}$$
the inverse function $\varphi ^{-1} : B \rightarrow \Sigma$ associates to each number $1 \le b \le 26$ the corresponding letter at that position of the alphabet. For example, $\varphi^{-1} (11) = \text{k} \text{.}$

Example $\PageIndex{4}$: A non-invertible function

The function $g:\mathbb{R} \rightarrow \mathbb{R}\text{,}$ $g(x) = x^2\text{,}$ does not have an inverse since it is not bijective. However, the function $h:\mathbb{R}_{\geq 0}: \mathbb{R}_{\geq 0}\text{,}$ $h(x) = x^2\text{,}$ so that $h = g\vert _{\mathbb{R}_{\geq 0}}$ but with codomain also restricted down to the image of $g\text{,}$ has inverse $h^{-1}(x) = \sqrt{x}\text{.}$

Note

If $f$ is bijective, then so is $f^{-1}\text{,}$ and $f^{-1}$ is the unique function $B \to A$ such that both

$$\begin{align*} f^{-1} \circ f & = \id_A, & f \circ f^{-1} & = \id_B \text{.} \end{align*}$$

Checkpoint

Prove that if $f$ is bijective then so is $f^{-1}\text{,}$ and $(f^{-1})^{-1} = f\text{.}$