Get a free home demo of LearnNext

Available for CBSE, ICSE and State Board syllabus.
Call our LearnNext Expert on 1800 419 1234 (tollfree)
OR submit details below for a call back


Inverse of a Bijective Function

Have a doubt? Clear it now.
live_help Have a doubt, Ask our Expert Ask Now
format_list_bulleted Take this Lesson Test Start Test

Inverse of a Bijective Function - Lesson Summary

Let f: A → B be a function. If, for an arbitrary x ∈ A we have f(x) = y ∈ B, then the function, g: B → A, given by g(y) = x, where y ∈ B and x ∈ A, is called the inverse function of f.


Define f: A → B such that

f(2) = -2, f(½) = -2, f(½) = -½, f(-1) = 1, f(-1/9) = 1/9

Here f one-one and onto.

g(-2) = 2, g(-½) = 2, g(-½) = ½, g(1) = -1, g(1/9) = -1/9

g is the inverse of f.

A function, f: A → B, is said to be invertible, if there exists a function, g : B → A, such that g o f = IA and f o g = IB.

The function, g, is called the inverse of f, and is denoted by f -1.


Let 2 ∈ A.Then gof(2) = g{f(2)} = g(-2) = 2

Let -2 ∈ B.Then fog(-2) = f{g(-2)} = f(2) = -2.


Let P = {y ϵ N: y = 3x - 2 for some x ϵN}. Show that a function, f : N → P, defined by f (x) = 3x - 2, is invertible, and find f-1.


Let us consider an arbitrary element, y ϵ P.

⇒ y=3x-2

⇒ x = (y+2)/3

Let us define g : P → N by g(y) = (y+2)/3

Let x ∈ N.

Then g o f (x) = g (f (x)) = g (3x - 2)

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

This shows g o f = IN …(i)

Let y ∈ P.

Then fog (y) = f (g (y)) = f((y+2)/3)

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

This shows fog = IP …(ii)

Hence, f is invertible and g is the inverse of f.


Let f : X → Y and g : Y → Z be two invertible (i.e. bijective) functions. Then g o f is also invertible with (g o f)-1 = f -1o g-1.


Given, f and g are invertible functions.

To prove that g o f is invertible, with (g o f)-1 = f -1o g-1.

It is sufficient to prove that:

i. (f -1 o g-1) o (g o f) = IX, and

ii. (g o f)o( f -1o g-1) = IZ.

Now, ( f -1 o g-1) o (g o f) = {( f -1 o g-1) o g} o f {'.' l o (m o n) = (l o m) o n}

= { f -1 o (g-1 o g)} o f

= ( f -1oIY)of {'.' g-1og = IY}

= f-1of { '.' f-1oI = f-1}

= IX

Hence, (f -1o g-1)o(g o f) =IX …… ( i)

Similarly, (g o f)o( f -1o g-1)

={(g o f) o f--1} o g-1

={g 0 (f o f--1} o g-1

=(g 0 Ix) o g-1

=g o g-1 = Iz

(g o f)o( f -1o g-1) =IZ ……. (ii)

From equations (i) and (ii),

(g o f)-1 = f -1 o g-1

Hence, the composition of two invertible functions is also invertible.


keyboard_arrow_downJoin Koda and learn coding in a fun way. Click Here
Feel the LearnNext Experience on App

Download app, watch sample animated video lessons, and get free trial to learn on the go.

Desktop App Download Now
Tablet App
Mobile App