Injective Surjective and Bijective Functions

Types of Function Mappings

In mathematics, functions map elements from one set (the domain) to another set (the codomain). This mapping can be classified into several types based on how domain and codomain elements are connected. The three main types are injective, surjective, and bijective functions.

Assume we have a function $f: X \to Y$ .

Injective or One to One Function

A function is called injective or one-to-one if every distinct element in the domain $X$ maps to a distinct element in the codomain $Y$ . In other words, no two different domain elements can have the same image (output) in the codomain.

Formal Definition:

A function $f: X \to Y$ is injective if for every $x_1, x_2 \in X$ , the following holds:

f(x_1) = f(x_2) \implies x_1 = x_2

Or, equivalently (using the contrapositive):

x_1 \neq x_2 \implies f(x_1) \neq f(x_2)

Imagine every student in one school (domain) must have a unique student ID number (codomain). No two students can have the same ID number. The mapping function from students to ID numbers is an injective function.

Examples:

The function $f(x) = 2x$ for $x \in \mathbb{R}$ is injective, because every distinct value of $x$ will produce a distinct $2x$ .
The function $g(x) = x^2$ for $x \in \mathbb{R}$ is not injective, because $g(2) = 4$ and $g(-2) = 4$ . There are two different inputs ( $2$ and $-2$ ) that produce the same output ( $4$ ).

Surjective or Onto Function

A function is called surjective or onto if every element in the codomain $Y$ is the image of at least one element in the domain $X$ . In other words, there are no "unreachable" elements in the codomain that don't have a corresponding element in the domain. The range of a surjective function is equal to its codomain.

Formal Definition:

A function $f: X \to Y$ is surjective if for every $y \in Y$ , there exists at least one $x \in X$ such that:

f(x) = y

Imagine every seat in a movie theater (codomain) must be occupied by at least one audience member (domain) when the movie starts. The mapping function from audience members to seats is surjective if all seats are filled.

Examples:

The function $f(x) = x^3$ from $\mathbb{R} \to \mathbb{R}$ is surjective, because every real number $y$ in the codomain is the cube of some real number $x$ (specifically, $x = \sqrt[3]{y}$ ).
The function $g(x) = x^2$ from $\mathbb{R} \to \mathbb{R}$ is not surjective, because there is no real number $x$ that yields $g(x) = -1$ (or any other negative number). Negative elements in the codomain have no corresponding element in the domain.
However, if we restrict the codomain to $g(x) = x^2$ from $\mathbb{R} \to [0, \infty)$ (non-negative real numbers), then this function becomes surjective.

Bijective Functions

A function is called bijective if it is both injective and surjective. This means that every element in the domain maps to a unique element in the codomain, and every element in the codomain has exactly one corresponding element in the domain.

A bijective function creates a perfect one-to-one correspondence between the elements of the domain and the codomain.

Formal Definition:

A function $f: X \to Y$ is bijective if for every $y \in Y$ , there exists exactly one $x \in X$ such that:

f(x) = y

Imagine two groups with the same number of members. Each member of the first group is paired with exactly one member of the second group, and each member of the second group also has exactly one partner. A mapping like this is bijective.

A function $f:X\to Y$ has a two-sided inverse $f^{-1}:Y\to X$ exactly when it is bijective. An injective function can still be inverted on its actual range after that range is chosen as the codomain.

Examples:

The function $f(x) = 2x$ from $\mathbb{R} \to \mathbb{R}$ is bijective (injective and surjective).
The function $f(x) = x^3$ from $\mathbb{R} \to \mathbb{R}$ is bijective (injective and surjective).
The function $g(x) = x^2$ from $\mathbb{R} \to \mathbb{R}$ is not bijective (neither injective nor surjective).
The function $h(x) = e^x$ from $\mathbb{R} \to \mathbb{R}$ is injective and is not surjective, so it is not bijective.
The function $k(x) = x^3 - x$ from $\mathbb{R} \to \mathbb{R}$ is surjective and is not injective, so it is not bijective.

Check Your Understanding

The formula $q(x)=x^2$ is used in each mapping below. Classify each function as injective, surjective, both, or neither. Only the codomain changes between the items, so compare the answers to see how the codomain decides the classification.

$q:\mathbb{R}\to\mathbb{R}$
$q:\mathbb{R}\to[0,\infty)$
$q:[0,\infty)\to[0,\infty)$

Solution

It is neither injective nor surjective. It is not injective because $q(2)=q(-2)=4$ , and it is not surjective because no negative real number is reached.
It is surjective but not injective. Every $y\ge0$ is reached by choosing $x=\sqrt{y}$ , but opposite inputs still have the same square.
It is bijective. The restricted domain removes the duplicate negative input, and every value in $[0,\infty)$ is reached exactly once. Its inverse is $q^{-1}(x)=\sqrt{x}$ .

The formula did not change. The classification changed because the domain and codomain changed.

Types of Function Mappings

Assume we have a function $f: X \to Y$ .

Injective or One to One Function

Formal Definition:

A function $f: X \to Y$ is injective if for every $x_1, x_2 \in X$ , the following holds:

f(x_1) = f(x_2) \implies x_1 = x_2

Or, equivalently (using the contrapositive):

x_1 \neq x_2 \implies f(x_1) \neq f(x_2)

Examples:

The function $f(x) = 2x$ for $x \in \mathbb{R}$ is injective, because every distinct value of $x$ will produce a distinct $2x$ .
The function $g(x) = x^2$ for $x \in \mathbb{R}$ is not injective, because $g(2) = 4$ and $g(-2) = 4$ . There are two different inputs ( $2$ and $-2$ ) that produce the same output ( $4$ ).

Surjective or Onto Function

Formal Definition:

A function $f: X \to Y$ is surjective if for every $y \in Y$ , there exists at least one $x \in X$ such that:

f(x) = y

Examples:

The function $f(x) = x^3$ from $\mathbb{R} \to \mathbb{R}$ is surjective, because every real number $y$ in the codomain is the cube of some real number $x$ (specifically, $x = \sqrt[3]{y}$ ).
The function $g(x) = x^2$ from $\mathbb{R} \to \mathbb{R}$ is not surjective, because there is no real number $x$ that yields $g(x) = -1$ (or any other negative number). Negative elements in the codomain have no corresponding element in the domain.
However, if we restrict the codomain to $g(x) = x^2$ from $\mathbb{R} \to [0, \infty)$ (non-negative real numbers), then this function becomes surjective.

Bijective Functions

A bijective function creates a perfect one-to-one correspondence between the elements of the domain and the codomain.

Formal Definition:

A function $f: X \to Y$ is bijective if for every $y \in Y$ , there exists exactly one $x \in X$ such that:

f(x) = y

Examples:

The function $f(x) = 2x$ from $\mathbb{R} \to \mathbb{R}$ is bijective (injective and surjective).
The function $f(x) = x^3$ from $\mathbb{R} \to \mathbb{R}$ is bijective (injective and surjective).
The function $g(x) = x^2$ from $\mathbb{R} \to \mathbb{R}$ is not bijective (neither injective nor surjective).
The function $h(x) = e^x$ from $\mathbb{R} \to \mathbb{R}$ is injective and is not surjective, so it is not bijective.
The function $k(x) = x^3 - x$ from $\mathbb{R} \to \mathbb{R}$ is surjective and is not injective, so it is not bijective.

Check Your Understanding

$q:\mathbb{R}\to\mathbb{R}$
$q:\mathbb{R}\to[0,\infty)$
$q:[0,\infty)\to[0,\infty)$

Solution

It is neither injective nor surjective. It is not injective because $q(2)=q(-2)=4$ , and it is not surjective because no negative real number is reached.
It is surjective but not injective. Every $y\ge0$ is reached by choosing $x=\sqrt{y}$ , but opposite inputs still have the same square.
It is bijective. The restricted domain removes the duplicate negative input, and every value in $[0,\infty)$ is reached exactly once. Its inverse is $q^{-1}(x)=\sqrt{x}$ .

The formula did not change. The classification changed because the domain and codomain changed.