Functions of Set

A function assigns to each element of a set, exactly one element of a related set. Functions find their application in various fields like representation of the computational complexity of algorithms, counting objects, study of sequences and strings, to name a few.

Function − Definition

A function or mapping (defined as f: X → Y) is a relationship from elements of one set X to elements of another set Y (X and Y are non-empty sets). X is called the Domain and Y is called the Codomain of function f.

Function f is a relation on X and Y such that for each x ∈ X, there exists a unique y ∈ Y such that (x, y) ∈ R. Here, x is called the pre-image and y is called the image of function f.

A function can be one-to-one or many-to-one, but never one-to-many.

Injective / One-to-One Function

A function f: A → B is injective (one-to-one) if for every b ∈ B, there exists at most one a ∈ A such that f(a) = b. This means f is injective if a₁ ≠ a₂ implies f(a₁) ≠ f(a₂) − different inputs always produce different outputs.

Examples

• f: N → N, f(x) = 5x is injective − if 5x₁ = 5x₂, then x₁ = x₂.
• f: N → N, f(x) = x² is injective − over natural numbers, different inputs give different squares.
• f: R → R, f(x) = x² is not injective since (−x)² = x² − two different inputs give the same output.

Surjective / Onto Function

A function f: A → B is surjective (onto) if the image of f equals its codomain. Equivalently, for every b ∈ B, there exists some a ∈ A such that f(a) = b. This means every element in B is mapped to by at least one element in A.

Examples

• f: N → N, f(x) = x + 2 is surjective.
• f: R → R, f(x) = x² is not surjective since we cannot find a real number whose square is negative.

Bijective / One-to-One Correspondent

A function f: A → B is bijective (one-to-one correspondent) if and only if f is both injective and surjective. A bijective function establishes a perfect pairing between every element in the domain and codomain.

Problem

Prove that a function f: R → R defined by f(x) = 2x − 3 is a bijective function.

Proving injective − If f(x₁) = f(x₂), then −

2x? - 3 = 2x? - 3
2x? = 2x?
x? = x?

Since f(x₁) = f(x₂) implies x₁ = x₂, the function f is injective.

Proving surjective − For any y ∈ R, we need to find x such that f(x) = y −

2x - 3 = y
2x = y + 3
x = (y + 3) / 2

Since x = (y + 3) / 2 belongs to R for every y ∈ R, and f(x) = y, the function f is surjective.

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

Conclusion

Functions map elements from a domain to a codomain. An injective function maps distinct inputs to distinct outputs, a surjective function covers every element in the codomain, and a bijective function does both − creating a perfect one-to-one correspondence between the two sets.