Injective, surjective and bijective functions and inverses

Domain, codomain and range

A function $f: A \to B$ has three sets attached to it.

The range is always inside the codomain, but it may be smaller. For $f:\mathbb{R}\to\mathbb{R}$, $f(x) = x^2$, the codomain is all of $\mathbb{R}$, but squares are never negative, so the range is $[0, \infty)$.

Injective, surjective, bijective

Compare two functions using a mapping diagram.

In the left diagram, $-2$ and $2$ share the output 4, so it is not injective. In the right diagram, the function $x + 1$ has domain $\{-1, 0, 1\}$ and codomain $\{0, 1, 2, 3\}$. Nobody maps to 3, so with this codomain it is not surjective. Surjectivity depends on the codomain chosen: $x + 1$ is bijective as a function $\mathbb{R}\to\mathbb{R}$.

Two tests

Horizontal line test. Draw horizontal lines across the graph. If some line cuts the graph more than once, two inputs share an output and $f$ is not injective.

Algebraic test. Assume $f(x_1) = f(x_2)$ and try to prove $x_1 = x_2$. If you can, $f$ is injective. To show a function is not injective, you only need one counter-example: two different inputs with the same output.

Drag the slider $a$: the horizontal line $y = a$ cuts $y = x^2$ twice for every $a > 0$, so $x^2$ fails the test on $\mathbb{R}$.

The inverse function

An inverse "undoes" a function. If $f$ sends $x$ to $y$, then $f^{-1}$ sends $y$ back to $x$.

Why bijective? Injective means each output came from only one input, so going back is unambiguous. Surjective means every element of $B$ has something to go back to, so $f^{-1}$ is defined on all of $B$.

The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

Method

1. Write $y = f(x)$.

2. Swap roles: solve the equation for $x$ in terms of $y$.

3. Replace $y$ by $x$ in the result to get $f^{-1}(x)$.

4. State the domain of $f^{-1}$ (the range of $f$).

The graph of the inverse

Swapping $x$ and $y$ reflects the graph in the line $y = x$. In the picture below, the lines $y = ax + b$ and its inverse are mirror images across $y = x$.

Rational functions

The same two solutions come from $f(x) = x$: here $x = \dfrac{2x+1}{x-3}$ gives $x^2 - 5x - 1 = 0$ directly, so both meeting points lie on the mirror line $y = x$.

Restricting the domain

$f(x) = x^2$ is not injective on $\mathbb{R}$, so it has no inverse there. But we can choose a smaller domain on which it is injective. Taking only $x \ge 0$ gives a one-to-one function with the inverse $\sqrt{x}$.

For a quadratic, the turning point splits the parabola into two one-to-one halves. Completing the square tells you where the split is.

Using the derivative

For cubics and other curves, use $f'(x)$. Take $f(x) = x^3 - 3x$. It is not injective on $\mathbb{R}$, since $f(0) = 0 = f(\sqrt3)$. Its derivative is $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$. This is positive for $x > 1$ and $x < -1$, and negative between $-1$ and $1$. So $f$ is strictly increasing on $[1, \infty)$, hence injective there. Its range on that interval is $[f(1), \infty) = [-2, \infty)$.

Exponentials and general rational functions

Exponentials and logs

$f(x) = e^{2x} + 1$ is injective because $f'(x) = 2e^{2x} > 0$ for all $x$, so it is strictly increasing. Since $e^{2x}$ takes every positive value, the range is $(1, \infty)$. Taking this as the codomain makes $f$ bijective.

Equations such as $f(x) = f^{-1}(x+2)$ mix $e^{2x}$ with $\ln$, so they cannot usually be solved exactly. Checking that $f^{-1}(f(x)) = x$ is the useful skill.

The general rational function

Summary