What is a Bijective Function and How To Recognize It?

Relations and Functions 2021 ( Maths Relations and Functions )

Jaya Sharma
Updated on Jul 11, 2025 17:33 IST

By Jaya Sharma, Assistant Manager - Content

One of the important functions amongst all is the bijective function which pairs every input with exactly one unique output. In the case of this function, each possible out comes exactly from one input. Bijective function is both injective and surjective. Considering the fact that it is an important topic, we have also shared an NCERT excercise of Relations and Functions since it is taught in both class 11th and 12th of the CBSE board. Let us now learn about this function in further detaill with examples.

 

Table of content
  • What is a Bijective Function?
  • How to Recognise a Bijective Function?
  • What are the Properties of Bijective Functions?
  • How To Prove That a Function is Bijective?
  • Illustrative Examples on Bijective Function’s
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What is a Bijective Function?

 A function  f : A B  will be considered as bijective if it is both injective (one-to-one) and surjective (onto) in nature.

Injective (one-to-one):

x 1 , x 2 A , f ( x 1 ) = f ( x 2 ) x 1 = x 2 There are no distinct inputs in A that are sharing the same output in B.

Surjective (onto): y B , x A  so that f ( x ) = y

Each element of B is hit by the function.

f is bijective when for every B, y B  there is only one x A with f(x) = y

Let us understand what are injective and surjective functions

  • The injective function is a function that always links the distinct element of its domain to the distinctive element of its co-domain. It is a one-to-one function.
  • The surjective function is a function that maps one or more elements of A to the same element of B and is also called onto function. This can have more than one element in the co-domain. 

The Bijective Function is a combination of both since it is a one-to-one function that works onto the co-domain element.

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How to Recognise a Bijective Function?

To recognize the bijection between two terms, we have first to understand the sets, map f: A → B, and conclude that |A| = |B|.

To prove a function f is bijective, we obtain either the inverse or prove that it is both injective and surjective.

If two sets A and B are not of the same size, then the functions aren’t bijective because bijection is pairing up of the elements in the two sets perfectly. If |A| = |B| = n, then there exists n! bijections between A and B. Let us understand this with an example. Do remember that IIT JAM and NEET are some of the entrance exams that ask questions related to bijective function. 

Example: Show that the function f(x) = 4x – 5 is a bijective function from R to R.
Given, f(x) = 4x – 5

The given function should be both injective and surjective.

(i) To Prove: The function is injective

For this f(a)=c and f(b)=c then a=b must be proved.
Let us take, f(a)=c and f(b)=c
Therefore, it can be written as:
c = 4a-5 and c = 4b-5
Thus, it can be written as: 4a-5 = 4b -5
Simplify the equation; we will get: a = b; hence the given function is injective

(ii) To Prove: The function is surjective
First, we should show that for “a” in R there exists a point “b” in the domain such that f(b) =a
Let, a = 4x -5
Therefore, b must be (a+5)/4

Since this is in R, the function is surjective.

As the function is both injective and surjective, it is bijective.

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What are the Properties of Bijective Functions?

Class 12 students learn this in Relations and Functions,, which has a 40% weightage in the board exams. This topic comes in the functions category and 2-4 mark questions. Even in entrance exams like JEE Main, the chapter holds decent weightage. Now let us move on to understanding the properties of a bijective function:

Properties of Bijective Functions

  1. Invertibility
    Since  f is both one-to-one and onto, there will exist a unique inverse f 1 : B A with f 1 ( f ( x ) ) = x , f ( f 1 ( y ) ) = y .
  2. One-to-One (Injective)
    Unique inputs result in unique outputs: x 1 , x 2 A , f ( x 1 ) = f ( x 2 ) x 1 = x 2 .
  3. Onto (Surjective)
    Each element of codomain is achieved: y B , x A : f ( x ) = y .
  4. Cardinality Preservation
    A and B both have the same number of elements, even for infinite sets.
  5. Composition Closure
    Composition of 2 bijections is a bijection. If f : A B and g : B C are bijections. Then  g f : A C will also be bijective, with g f 1 = f 1 g 1 .
  6. Permutation Behavior (Finite Sets)
    In a finite set, each bijection is a permutation. It rearranges all elements with no repeats and no omissions.
  7. Graph Symmetry
    Graph of f 1 is the reflection of graph of f across the line y = x .
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How To Prove That a Function is Bijective?

Let us understand different methods through which we can prove if a function is bijective:

1. Is the Function Injective (One-to-One)?

Let us prove that

f ( x 1 ) = f ( x 2 ) x 1 = x 2

A. Direct algebraic proof

  1. Let us assume that  f ( x 1 ) = f ( x 2 ) .
  2. Solve the equation using algebra until you get  x 1 = x 2 .
  3. Conclude that no two inputs will ever share the same output.

For Example:

f ( x ) = 3 x + 2

Let us assume that 3 x 1 + 2 = 3 x 2 + 2 .

We will first subtract 2 and then divide it by 3 

Finally, we will get:  x 1 = x 2 .

B. Alternative viewpoints

  • Monotonic functions on : If f ( x ) > 0 everywhere.  In this case, f is injective.
  • Polynomials: A non-constant polynomial of odd degree over might not come as injective only at the turning point. You will be required to check the derivative sign or else use factorization.

2. Is the function Surjective (Onto)?

Let us prove that:

y B , x A : f ( x ) = y

A.  Let us solve for x first

  1. Choose an arbitrary y B .
  2. Solve the equation f ( x ) = y for x .
  3. Verify whether your solution x is lying in the domain A .

For example: f ( x ) = 3 x + 2 .

Given y ,

solve y = 3 x + 2

x = y 2 3

Since y 2 3 , every y has a pre‐image.

 

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Illustrative Examples on Bijective Function’s

Entrance exams like CUET often ask questions about different types of functions. Therefore, we are considering some basic questions so that you can further move on to advanced level questions.

1. Is the mapping injective or surjective?

(i) { (x, y): x is a person, y is the father of x }.

Solution.

(i) Here, y can be a father to two terms in the x domain as it is not specific. Hence, this is a surjective function and not injective.

2. Prove that the function f: A → b is invertible only if f is both one-one and onto.

Solution.

A function f :X → Y is defined to be invertible, if there exists a function g = Y → X such that g(f) = Ix and f(g) = Iy, the function is called the inverse of f and is denoted by f-1. This, if bijective, means f can be invertible.

3. If set a contains five elements, and the set b contains six elements, then the number of bijective mappings from a to b is

Solution.

0. Since the number of elements in B is more than A.

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