Injective, Surjective and Bijective Functions

This lesson works through what it means for a function to be injective (one to one), surjective (onto), and bijective (both at once), then proves these properties for two worked examples.

The three definitions

• A function is injective (one to one) when different inputs always give different outputs.
• A function is surjective (onto) when every element of the codomain is hit by some input.
• A function is bijective when it is both injective and surjective.

Example 1: one to one but not onto

Let $f : \mathbb{N} \to \mathbb{N}$ be defined by

$f(x) = x^2 + x + 1.$

We prove $f$ is one to one but not onto.

Proving it is one to one

Take $x_1, x_2 \in \mathbb{N}$ with $f(x_1) = f(x_2)$. Then

$x_1^2 + x_1 + 1 = x_2^2 + x_2 + 1.$

Cancel the $+1$ and group:

$(x_1^2 - x_2^2) + (x_1 - x_2) = 0.$

Factor, using $x_1^2 - x_2^2 = (x_1 - x_2)(x_1 + x_2)$:

$(x_1 - x_2)(x_1 + x_2 + 1) = 0.$

Since $x_1, x_2 \in \mathbb{N}$, the factor $x_1 + x_2 + 1 \neq 0$. Therefore

$x_1 - x_2 = 0 \quad\Longrightarrow\quad x_1 = x_2.$

So $f(x_1) = f(x_2)$ forces $x_1 = x_2$, and $f$ is injective.

Why it is not onto

For every $x \in \mathbb{N}$ we have

$f(x) = x^2 + x + 1 \geq 3,$

since the smallest value is at $x = 1$, where $f(1) = 1 + 1 + 1 = 3$. So the range of $f$ starts at $3$. But the codomain is $\mathbb{N}$, which contains $1$ and $2$. No input maps to $1$ or $2$, so the range is not equal to the codomain. Therefore $f$ is not onto.

Example 2: a bijective function

Let $A = \mathbb{R} \setminus \{3\}$ and $B = \mathbb{R} \setminus \{1\}$, and let $f : A \to B$ be

$f(x) = \frac{x - 2}{x - 3}, \qquad x \in A.$

We show $f$ is bijective.

Proving it is one to one

Suppose $f(x_1) = f(x_2)$:

$\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}.$

Cross multiply:

$(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3).$

Expand both sides:

$x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6.$

Cancel $x_1 x_2$ and $6$, then collect terms:

$-3x_1 + 2x_1 = -3x_2 + 2x_2 \quad\Longrightarrow\quad -x_1 = -x_2.$

So $x_1 = x_2$, and $f$ is injective.

Proving it is onto

Set $y = f(x) = \dfrac{x - 2}{x - 3}$ and solve for $x$ in terms of $y$. Cross multiplying,

$y(x - 3) = x - 2 \quad\Longrightarrow\quad xy - 3y = x - 2.$

Gather the $x$ terms:

$xy - x = 3y - 2 \quad\Longrightarrow\quad x(y - 1) = 3y - 2,$

so

$x = \frac{3y - 2}{y - 1}.$

To confirm this $x$ really maps to $y$, substitute it back into $f$:

Numerator: $\dfrac{3y - 2}{y - 1} - 2 = \dfrac{3y - 2 - 2(y - 1)}{y - 1} = \dfrac{y}{y - 1}.$

Denominator: $\dfrac{3y - 2}{y - 1} - 3 = \dfrac{3y - 2 - 3(y - 1)}{y - 1} = \dfrac{1}{y - 1}.$

Therefore

$f\!\left(\frac{3y - 2}{y - 1}\right) = \frac{\,y/(y-1)\,}{\,1/(y-1)\,} = y.$

Every $y \in B$ has a preimage in $A$, so $f$ is onto.

Since $f$ is both injective and surjective, $f$ is bijective.

Key takeaways

• To prove a function is one to one, assume $f(x_1) = f(x_2)$ and show this forces $x_1 = x_2$.
• To show a function is not onto, find elements of the codomain that the range never reaches.
• To prove a function is onto, solve $y = f(x)$ for $x$ in terms of $y$ and check that this input maps back to $y$.
• A function that is both injective and surjective is bijective.