Injection (Injective Function) — Definition, Formula & Examples

An injection (or injective function) is a function where no two different inputs produce the same output. If

f(a) = f(b)

, then it must be the case that

a = b

A function $f: A \to B$ is injective (one-to-one) if and only if for all $a_1, a_2 \in A$ , $f(a_1) = f(a_2)$ implies $a_1 = a_2$ . Equivalently, $a_1 \neq a_2$ implies $f(a_1) \neq f(a_2)$ .

Key Formula

f(a_1) = f(a_2) \implies a_1 = a_2

Where:

$f$ = A function from set A to set B
$a_1, a_2$ = Arbitrary elements in the domain A

How It Works

To prove a function is injective, assume

f(a) = f(b)

for arbitrary elements

a

and

b

in the domain, then show algebraically that

a = b

. To disprove injectivity, find a single counterexample: two distinct inputs that map to the same output. Graphically, a function from

\mathbb{R}

\mathbb{R}

is injective if and only if it passes the horizontal line test — every horizontal line intersects the graph at most once.

Worked Example

Problem: Prove that

f(x) = 3x + 5

is injective over

\mathbb{R}

Assume equal outputs: Suppose

f(a) = f(b)

for some

a, b \in \mathbb{R}

3a + 5 = 3b + 5

Solve for the inputs: Subtract 5 from both sides, then divide by 3.

3a = 3b \implies a = b

Conclude: Since

f(a) = f(b)

forces

a = b

, the function satisfies the definition of injectivity.

Answer:

f(x) = 3x + 5

is injective because equal outputs always imply equal inputs.

Why It Matters

Injectivity is essential for defining inverse functions — a function has a left inverse precisely when it is injective. In discrete math courses, counting the number of injections from one finite set to another yields the permutation formula

P(n, k)

. Cryptographic hash functions and database key constraints also rely on injectivity to guarantee that distinct inputs remain distinguishable.

Common Mistakes

Mistake: Confusing injective (one-to-one) with surjective (onto).

Correction: Injective means no two inputs share an output. Surjective means every element in the codomain is hit by at least one input. A function can be one without the other.