Invertible functions

To invert a function means to undo it: if $f$ takes $x$ to $y$, then $f^{-1}$ takes $y$ back to $x$. Not every function admits an inverse , only bijections do. The aim of this subtopic is to define $f^{-1}$ precisely, prove it exists when $f$ is bijective, and give you a recipe for computing it.

Definition

A function $f : A \to B$ is invertible if there exists a function $g : B \to A$ such that

$g \circ f = I_A \quad \text{and} \quad f \circ g = I_B.$

When it exists, $g$ is called the inverse of $f$ and is denoted $f^{-1}$.

Notice that the two conditions are not redundant. Without the first, $g$ may merely be a right inverse; without the second, it may merely be a left inverse. Both together force $f$ and $g$ to be bijections that exactly cancel.

Theorem: invertible iff bijective

Theorem. A function $f : A \to B$ is invertible if and only if it is bijective. The inverse, when it exists, is unique.

Proof. ($\Leftarrow$) Suppose $f$ is bijective. Define $g : B \to A$ by setting $g(y) =$ the unique $x \in A$ such that $f(x) = y$. (Uniqueness comes from injectivity; existence from surjectivity.) Then $g(f(x)) = x$ for all $x$, so $g \circ f = I_A$. And $f(g(y)) = y$ by construction, so $f \circ g = I_B$.

($\Rightarrow$) Suppose $g \circ f = I_A$ and $f \circ g = I_B$. From $g \circ f = I_A$ we see $f$ is one-one (because $g \circ f$ is). From $f \circ g = I_B$ we see $f$ is onto (because $f \circ g$ is). So $f$ is bijective.

Uniqueness. If $g_1$ and $g_2$ both satisfy the inverse conditions, then $g_1 = g_1 \circ I_B = g_1 \circ (f \circ g_2) = (g_1 \circ f) \circ g_2 = I_A \circ g_2 = g_2$. $\blacksquare$

The uniqueness justifies the article the in "the inverse". You can speak about $f^{-1}$ unambiguously.

Computing an inverse

For a bijective $f$ given by a formula $y = f(x)$, the inverse is found by solving for $x$ in terms of $y$:

1. Write $y = f(x)$.
2. Solve algebraically for $x$ to get $x = g(y)$.
3. Swap symbols (optional): rename $y$ as $x$ to write $f^{-1}(x) = g(x)$.
4. State the domain of $f^{-1}$, which equals the range of $f$.

The graphical view: the graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$.

Important: domain restriction

If $f$ is not one-one on all of its natural domain, you must restrict the domain to make $f$ one-one before defining the inverse. This is exactly the device used to define the principal branches of $\arcsin$, $\arctan$, etc., in Chapter 2.

The reversal rule for inverses

Theorem. If $f : A \to B$ and $g : B \to C$ are bijective, then $g \circ f$ is bijective and $(g \circ f)^{-1} = f^{-1} \circ g^{-1}.$

Proof. Both compositions $(f^{-1} \circ g^{-1}) \circ (g \circ f)$ and $(g \circ f) \circ (f^{-1} \circ g^{-1})$ reduce to identities by associativity and the inverse relations. $\blacksquare$

The pattern reverse and flip is also true for matrix inverses and group elements, so commit it to long-term memory.

Worked examples

Example 1. $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 5x + 2$. Find $f^{-1}$.

Solve $y = 5x + 2$ for $x$: $x = (y - 2)/5$. So $f^{-1}(y) = (y - 2)/5$.

Verification: $f(f^{-1}(y)) = 5 \cdot \tfrac{y - 2}{5} + 2 = y - 2 + 2 = y$. And $f^{-1}(f(x)) = ((5x + 2) - 2)/5 = x$. $\checkmark$

Example 2. Find the inverse of $f : \mathbb{R} \setminus \{2\} \to \mathbb{R} \setminus \{1\}$, $f(x) = \tfrac{x + 1}{x - 2}$.

Solve $y = (x + 1)/(x - 2)$: $y(x - 2) = x + 1 \Rightarrow x(y - 1) = 2y + 1 \Rightarrow x = (2y + 1)/(y - 1)$.

So $f^{-1}(y) = (2y + 1)/(y - 1)$ for $y \neq 1$.

Example 3. Let $f : [0, \infty) \to [0, \infty)$, $f(x) = x^2$. Find $f^{-1}$.

$y = x^2$ with $x \ge 0$ gives $x = \sqrt{y}$. So $f^{-1}(y) = \sqrt{y}$ on $[0, \infty)$.

Example 4. Show that the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is invertible and find $f^{-1}$.

$f$ is strictly increasing (since $f'(x) = 3x^2 \ge 0$, zero only at one point), so $f$ is one-one. $f$ is onto $\mathbb{R}$ because cube roots exist for all real numbers. So $f$ is bijective, and $f^{-1}(y) = y^{1/3}$.

Example 5. Let $f : \mathbb{N} \to \mathbb{N}$ be $f(n) = n + 1$. Show $f$ is not invertible. What if we change codomain?

$f$ is not onto $\mathbb{N}$ (taking $\mathbb{N} = \{1, 2, 3, \ldots\}$): $1$ is not in the range. So $f$ is not bijective. If we replace the codomain by $\mathbb{N} \setminus \{1\}$, $f$ becomes bijective with inverse $n \mapsto n - 1$.

Example 6. $f : [0, \pi/2] \to [0, 1]$, $f(x) = \sin x$. Find $f^{-1}$.

$f$ is strictly increasing and onto on this domain. The inverse is $\arcsin$, restricted: $f^{-1}(y) = \arcsin y$ for $y \in [0, 1]$. This launches the next chapter.

Try it yourself

1. Find the inverse of $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 7 - 3x$.
2. Find the inverse of $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3 + 1$.
3. Find the inverse of $f : \mathbb{R} \setminus \{-1\} \to \mathbb{R} \setminus \{1\}$, $f(x) = \tfrac{x - 1}{x + 1}$.
4. Find the inverse of $f : [1, \infty) \to [0, \infty)$, $f(x) = \sqrt{x - 1}$.
5. Show that $f(x) = \tfrac{ax + b}{cx + d}$ on appropriate domain has inverse $f^{-1}(y) = \tfrac{dy - b}{a - cy}$, provided $ad - bc \neq 0$.
6. Let $f, g : \mathbb{R} \to \mathbb{R}$ be invertible. Express $(f \circ g)^{-1}$ in terms of $f^{-1}$ and $g^{-1}$.
7. Given $f(x) = \log x$ (base $e$) on $(0, \infty)$, find $f^{-1}$.
8. Given $f(x) = 2^x$ on $\mathbb{R}$, find $f^{-1}$.
9. Show that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x|x|$, is bijective and find $f^{-1}$.
10. If $f, g$ are mutually inverse, show $f$ is one-one and $g$ is onto.
11. If $f$ is invertible and $f \circ f = I$, what can $f$ be? (Such functions are called involutions.)
12. Find the inverse of $f : [0, \pi/2] \to [0, 1]$, $f(x) = \sin x$, and compute $f^{-1}(\tfrac{1}{2})$.
13. Prove that if $f : A \to B$ is bijective, then $f^{-1}$ is also bijective and $(f^{-1})^{-1} = f$.
14. Compute the inverse of $f : (0, \infty) \to \mathbb{R}$, $f(x) = \ln(x) + x$. (Hint: this cannot be solved in elementary form; conclude that not every inverse has a closed form, though it exists.)

Pitfalls / Tricks

• Do not confuse $f^{-1}(y)$ (the inverse map) with $1/f(y)$ (the reciprocal).
• A common error is to write $f^{-1}$ before checking $f$ is one-one and onto. If $f$ is one-one but not onto, the inverse exists only as a map from the range of $f$.
• When restricting domain to obtain a bijection, you must explicitly state both new domain and new codomain.
• $(f^{-1})^{-1} = f$ always. $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ , the order reverses.
• Graphically, $f^{-1}$ is the reflection of $f$ across $y = x$. If $f$ has a horizontal tangent, $f^{-1}$ has a vertical one.

The chapter closes with a tour of binary operations, which generalise composition.