In this section, we will look at the bijective function and understand it in the different forms of function.

## Defining a Bijective Function

Suppose we have two sets, \(A\) and \(B\), and a function \(f\) points from \(A\) to \(B\) \((f:A\to B)\). If every element in codomain \(B\) is pointed to by at least one element in domain \(A\), the function is called a bijective function.

A function \(f:A\to B\) is bijective if, for every \(y\) in \(B\), there is exactly one \(x\) in \(A\) such that \(f(x)=y\).

A bijective function is both injective (one-one function) and surjective (onto function) in nature.

If every element of the range is mapped to exactly one element from the domain is called the **injective function. **That is, no element of the domain points to more than one element of the range.

In a **surjective function**, every element of the co-domain is an image of at least one element of the domain.

So, it logically follows that if a function is both injective and surjective in nature, it means that every element of the domain has a unique image in the co-domain, such that all elements of the co-domain are also part of the range (have a corresponding element in the domain). Such a function is called a **bijective function**.

You can consider a bijective function to be a perfect one-to-one correspondence. Every element in the domain has exactly one corresponding image in the co-domain, and vice-versa.

Note that the onto function is not bijective, as it needs to be a one-one function to be bijective. Let's take a look at the difference between these two to understand it better.

## Difference between Bijective and Surjective functions

We will see the difference between bijective and surjective functions in the following table.

Bijective function | Surjective function |

A function \((f:A\to B)\) is bijective if, for every \(y\) in \(B\), there is exactly one \(x\) in \(A\) such that \(f(x)=y\). | A function \((f:A\to B)\) is surjective if for every \(y\) in \(B\) there is at least one \(x\) in \(A\) such that \(f(x)=y\). |

A bijective function is both one-one and onto function. | A surjective function is onto function. |

The domain and co-domain have an equal number of elements. | A co-domain can be an image for more than one element of the domain. |

Bijective graphs have exactly one horizontal line intersection in the graph. | Surjective graphs have at least one horizontal line intersection in the graph. |

Example - \(f:\mathbb{R}\to \mathbb{R}, f(x)=2x\) | Example - \(f:\mathbb{R}\to \mathbb{R}, f(x)=x^{3}-3x\) |

## Composition of Bijective functions

Consider the functions \(f:A\to B , g:B\to C\). Then the composition of the function \((g\circ f)(x)=g(f(x))\) from function \(A\) to \(C\). The composition of the bijective function is derived from the composition of injective and surjective functions.

The function \(f:A\to B , g:B\to C\) are injective function, then the composition \(g\circ f\) is also injective. Similarly, for the two surjective functions \(f\) and \(g\), their composition \(g\circ f\) is also surjective.

Suppose both \(f:A\to B\) and \( g:B\to C\) are bijective. This implies that both \(f\) and \(g\) are both injective and surjective as well. The composition of the functions \(g\circ f\) is both injective and surjective. Hence, the composition of function \(g\circ f\) is bijective.

Note that if \(g\circ f\) is bijective, then it can only be possible that \(f\) is injective and \(g\) is surjective.

## Bijective function graph

We can determine a bijective function based on the plotted graph too. To identify a bijective function graph, we consider a horizontal line test based on injective and surjective functions. For a function to be bijective both the test for injective and surjective should be satisfied.

**Horizontal line test**

This test is used to check the injective, surjective, and bijective functions. We determine the type of function based on the number of intersection points with the horizontal line and the given graph.

To check this, draw horizontal lines from different points. If each horizontal line intersects the graph at most one point then, it is an injective function. If the function is surjective, then a horizontal line should intersect at at least one point. So, when checking for bijective function, there should be **exactly one** intersecting point with a horizontal line.

## Bijective function examples

Show bijection for the function \(f:\mathbb{R}\to \mathbb{R}, f(x)=x\).

**Solution:**

Consider the function \(f(x)=x\), where the domain and co-domain are the set of all real numbers.

All values in the co-domain correspond to a unique value in the domain. Thus, the function is bijective in nature.

It is injective because every value of \(x\) leads to a different value of \(y\). It is surjective because any possible real number \(r\) can have a corresponding value \(x\) such that \(f(x)=r\).

When a bijective function is drawn on a graph, a horizontal line parallel to the X-axis must intersect the graph at exactly one point (horizontal line test).

The following graph demonstrates this for the function \(f(x)=x\).

Verify if the function \(f:\mathbb{R}\to \mathbb{R}, f(x)=x^{2}\) is bijective or not.

**Solution:**

Here for the given function, the range of the function only includes values \(\ge 0\). But the co-domain includes all negative real numbers too. And the members of the co-domain can be images of multiple members of the domain, for example \(f(2)=f(-2)=4\). Hence, the function \(f(x)=x^{2}\) is not injective. So, \(f(x)=x^{2}\) is not bijective.

When we draw the function on a graph, we can notice how it fails the horizontal line test as it intersects at two different points.

Is the function \(f(x)=2x\) bijective? Also, show for which domain and co-domain.

**Solution:**

When we set the domain and co-domain of the function to the set of all real numbers, it was a bijective function.

Hence, for \(f:\mathbb{R}\to \mathbb{R}, f(x)=2x\) is bijective.

However, if we restrict the domain and co-domain of the function to the set of all natural numbers, this no longer remains a bijective function. Since the range would include all even numbers but exclude all odd numbers, but they remain part of the co-domain. For example, it is impossible to get \(f(x)=3\), for any natural number value of \(x\). Thus, the function is not surjective, and consequently not bijective.

So, \(f:\mathbb{N}\to \mathbb{N}, f(x)=2x\)$f:\mathrm{\mathbb{N}}\to \mathrm{\mathbb{N}},f\left(x\right)=2x$ is not bijective.

## Bijective Functions - Key takeaways

- A bijective function is both injective and surjective in nature.
- A function \(f:A\to B\) is bijective if, for every \(y\) in \(B\), there is exactly one \(x\) in \(A\) such that \(f(x)=y\).
- A bijective function is one-one and onto function, but an onto function is not a bijective function.
- The composition of bijective functions is again a bijective function.
- For a bijective function, there should be exactly one intersecting point with a horizontal line.

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##### Frequently Asked Questions about Bijective Functions

What is a bijective function?

A function that is both injective and surjective is called a bijective function.

How to prove a function is bijective?

To prove that a function is bijective, first prove that it is injective and then prove that it is surjective.

Are continuous functions bijective?

All bijective functions are continuous but not all continuous functions are bijective.

Are all linear functions bijective?

All linear continuous functions are bijective. For example f(x)=2x.

What is bijective function with example?

The function f(x)=x is an example of a bijective function as it is both injective and surjective,

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