Onetoone function or one–one function
A function f: A→B where A,B are two nonempty sets is called a onetoone function if no two distinct
elements of A have the same image in B. That is, f: A→B is a onetoone (or one–one) function
if and only if (i f f).
x_{1}, x_{2}∈A and x_{1}≠ x_{2} then f(x_{1}) ≠f(x_{2}) or
if f(x_{1}) = f(x_{2}) x_{1}= x_{2}
Example 1
f = {(a,x), (b,y), (c,z)}
is a one–one function.
Example 2
Let f: R→R be defined by f(x) = x^{2}. Is f one–one?
Consider f(1) = (1)^{2}
= 1
f(1) = (1)^{2} = 1
f(x_{1}) = f(x_{2}) = 1
But x_{1}≠ x_{2}
1 ≠1
So f is not a one–one function.
A one–one function or onetoone function is also called an injection.
Onto function
A function f: A→B, A,B are nonempty sets, is called an onto function if f(A)=B. That is f is onto
if every element of the
codomain B is the image of at least one element of the domain A.
f: A→B is onto if and only if (iff) for every y ∈B there exists at least one x∈ A such that f(x) = y.
Example 3
Let f: {a, b, c} →{1,2,3} such that f(a) = 3, f(b) = 2, f(c) = 1.
f is an onto function, since each element of {1,2,3} is an image of an element of {a, b, c}.
Example 4
Let f: {2, 4, 7} →{p, q, r}
such that f(2) = q, f(4) = r, f(7)= r
f is not onto as p is not the image of any element of {2, 4, 7}
Example 5
Let f: N→ {1, 1} defined by
f(n) = 1 if f is odd,
= 1 if n is even.
f is onto.
Example 6
Let f: R→R be defined by f(x) = 3x5. Show that f is onto.
Solution:
Let y = f(x) = 3x  5
y = 3x  5
y + 5 = 3x
For every y ∈R there is an x ∈R
such that
An onto function is also called a surjection.
One–one and onto functions
A function f: A→B, A,B are nonempty, is called a one–one and onto
function if it is both one–one and onto. This type of function is also called a bijection.
Example 7
If f = { (p,1), (q,2), (r,3) }
f is one–one and onto
Since f (p) = 1
f(q) = 2
f(r) = 3
For every element of {1,2,3} is the image of element of {p,q,r}.
So f is onto.
Also, f is one–one since two distinct elements of {p,q,r} have two distinct images in {1,2,3}.
Example 8
Let f: R →R be defined by f(x) = 2x + 3
f is a one–one and onto function.
Consider 1≠ 1
f(1) ≠f(1)
as 2 (1) +3 ≠2 * 1+3
2+3 ≠2+3
1 ≠5
Let y = f(x) = 2x + 3
y  3 = 2x
So for every y ∈R there exists an x ∈R such that
So f is onto.
Example 9
Let f: R→R be defined by f (x) = x^{2}  1
Consider f (2) = 2^{2} 1
= 41
= 3.
f(2) = (2)^{2}1
= 41
= 3
f(2) = f(2)
But 2 ≠2
f is not one–one.
So f is not a bijection.
Try these questions
State if the following functions are oneone
a. f1(x) = x^{2} f1: R→ R
b. f2(x) = 3x f2: R→ R
Solutions:
a. f1 : R →R
f1(x) = x^{2}
f1is not One–One since
f1(2) = (2)^{2}
= 4
f1(2) = (2)^{2}
= 4
f1(2) = f1(2)
= 4
but 2 ≠2
b. f2: R →R
f2(x) = 3x
f2(x_{1}) ≠ f(x_{2})
3x_{1} ≠3x_{2}
x_{1} ≠x_{2}
So f_{2} is one–one.
 State if the following functions are onto
g_{1}(x) = 2x^{3} g_{1}: R→ R
g_{3}: Z→ Z defined by g_{3} (x) = x1
Solutions:
a. g_{1}: R→ R
g_{1}(x) = 2x^{3}
Let y = g_{1} (x) = 2x^{3}
y = 2x^{3}
y/2 = x3
g1 is onto.
b. g_{3}: Z→ Z g_{3}(x) = x1
g_{3}is an onto function since g_{3} (Z) = Z.
Or
for every x ∈ Z there exists a y ∈ Z such that g(x) = y.

State if the following functions are bijections. (One–One and onto functions).
a. f_{1} = {(a,1), (b,2), (c,3), (d, 4)}
f_{2} : R→ R defined by f2(x) = 4x1
Solutions:
a. f_{1} = {(a,1), (b,2), (c,3), (d,(4)}
f_{1} is One–One since
1 ≠2
f_{1}(a) ≠f_{1}(b)
a≠ b
f_{1} is onto as every element of {a,b,c,d} has an image in {1,2,3,4}.
f_{1} is a bijection.
b. f_{2} : R→ R where f_{2}(x) = 4x1
f_{2} is One–One since
f_{2}(x_{1}) = f_{2}(x_{2})
On canceling like terms
x1 = x2
f_{2} is onto since for every element x∈R there exists an element y∈ R such that
f(x) = y.
f_{2} is a bijection.

Find the inverse functions (if they exist) of the following.
a. f: R→ R f  x  = 2x^{2}1
f: R {3} →R  {1} defined by f(x) =
Solutions:
a. f: R→ R f (x) = 2x^{2}1
f is not One–One as
f(1) = 2  1
= 1
f(1) = 2  1
= 1
So f(1) = f(1)
But 1 ≠1
f^{1} does not exist.
b. f: R {3} →R  {1} defined by f(x) =
f is One–One
f is onto
f1 exists.
let y = f(x) =
y =
y(x3) = (x + 3)
yx3y = x+3
yx  x = 3 + 3
x(y1)= 3(y+1)
x =
x=f1(y) =
Or f1(x) =