Bijective Function

Bijection, or bijective function, is a one-to-one correspondence function between the elements of two sets. In such a function, each element of one set pairs with exactly one element of the other set, and each element of the other set has exactly one paired partner in the first set. While understanding bijective mapping, it is important not to confuse such functions with one-to-one correspondence. This article will help you understand clearly what is bijective function, bijective function example, bijective function properties, and how to prove a function is bijective. 


Surjective, Injective and Bijective Functions

Functions can be one-to-one functions (injections), onto functions (surjections), or both one-to-one and onto functions (bijections). Here is a brief overview of surjective, injective and bijective functions:

  1. Surjective: If f: P → Q is a surjective function, for every element in Q, there is at least one element in P, that is, f (p) = q.

  2. Injective: If f: P → Q is an injective function, then distinct elements of P will be mapped to distinct elements of Q, such that p=q whenever f (p) = f (q).

  3. Bijective: If f: P → Q is a bijective function, for every element in Q, there is exactly one element in P, that is, f (p) = q.


(image will be uploaded soon)


Bijective Function Definition

What is a bijective function? A bijective function has no unpaired elements and satisfies both injective (one-to-one) and surjective (onto) mapping of a set P to a set Q. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. In mathematical terms, let f: P → Q is a function; then, f will be bijective if every element ‘q’ in the co-domain Q, has exactly one element ‘p’ in the domain P, such that f (p) =q.


Bijective Function Properties

Now that you know what is a bijective mapping let us move on to the properties that are characteristic of bijective functions. We know the function f: P → Q is bijective if every element q ∈ Q is the image of only one element p ∈ P, where element ‘q’ is the image of element ‘p,’ and element ‘p’ is the preimage of element ‘q’. Also, 

  1. Each element of P should be paired with at least one element of Q.

  2. No element of P must be paired with more than one element of Q.

  3. Each element of Q must be paired with at least one element of P, and

  4. No element of Q must be paired with more than one element of P.


Bijective Function Example

Example 1: The function f (x) = x2 from the set of positive real numbers to positive real numbers is injective as well as surjective. Thus, it is also bijective.


However, the same function from the set of all real numbers R is not bijective since we also have the possibilities f (2)=4 and f (-2)=4.


Example 2: The function f: {months of a year} {1,2,3,4,5,6,7,8,9,10,11,12} is a bijection if the function is defined as f (M)= the number ‘n’ such that M is the nth month.


How to prove a function is bijective?


If we have defined a map f: P → Q and we have to prove that the function f is a bijection, we have to satisfy two conditions. First of all, we have to prove that f is injective, and secondly, we have to show that f is surjective. Only when we have established that the elements of domain P perfectly pair with the elements of co-domain Q, such that, |P|=|Q|=n, we can conveniently say that there are n bijections between P and Q. Let us understand the proof with the following example:


Bijective Function - Solved Example

Example: Show that the function f (x) = 5x+2 is a bijective function from R to R.


Solution: Given function: f (x) = 5x+2. 


Step 1: To prove that the given function is injective.

To prove injection, we have to show that f (p) = z and f (q) = z, and then p = q.

Say, f (p) = z and f (q) = z. 

Therefore, we can write z = 5p+2 and z = 5q+2 which can be thus written as: 5p+2 = 5q+2. Simplifying the equation, we get p  =q, thus proving that the function f is injective.


Step 2: To prove that the given function is surjective.

To prove surjection, we have to show that for any point “c” in the range, there is a point “d” in the domain so that f (q) = p.

Let, c = 5x+2.

Therefore, d will be (c-2)/5. Since this number is real and in the domain, f is a surjective function. 

Therefore, since the given function satisfies the one-to-one (injective) as well as the onto (surjective) conditions, it is proved that the given function is bijective.

FAQ (Frequently Asked Questions)

1. What are Some Examples of Surjective and Injective Functions?

Ans:

  • The function f: {Indian cricket players’ jersey} N defined as f (W) = the jersey number of W is injective, that is, no two players are allowed to wear the same jersey number.

  • f (x) = x2 from a set of real numbers R to R is not an injective function. This is because: f (2) = 4 and f (-2) = 4. So, even if f (2) = f (-2), 2 and the definition f (x) = f (y), x = y is not satisfied.

  • The function f (x) = 2x from the set of natural numbers N to a set of positive even numbers is a surjection.

  • The function f: {Lok Sabha seats} → {Indian states} defined by f (L) = the state that L represents is surjective since every Indian state has at least one Lok Sabha seat.

2. What are the Fundamental Differences Between Injective, Surjective and Bijective Functions?

Ans:

  • Injective: In this function, a distinct element of the domain always maps to a distinct element of its co-domain.

  • Surjective: In this function, one or more elements of the domain map to the same element in the co-domain.

  • Bijective: These functions follow both injective and surjective conditions.


Injective: The mapping diagram of injective functions:

 

(image will be uploaded soon)


Surjective: The mapping diagram of surjective functions:


(image will be uploaded soon)


Bijective: The mapping diagram of bijective functions:


(image will be uploaded soon)