One to One function is also known as Injective function.Let us consider a function Y=f(X) and if for unique element of X there is one and only one element in Y under the function f( none element of Y having two pre image in X) is said to one to one function i.e.
if Y =f(X) and if f(x1) = f(x2) then$x_{1}$ = $x_{2}$
In other words we say if each element from the range of f there is an unique element in the domain of f.

One to One Function

Let us consider some examples based on one to one function:

Example 1:
Let A = $\left \{(1,5),(2,4),(3,5),(4,6) \right \}$ and
B = $\left \{(2,3),(4,1),(1,5),(3,4) \right \}$

In above example set A is not one to one function because the elements 1 and 3 has the same image or value. And set B is an injective function since there isno repetition of the second element.

Example 2:
Is F(x) = 9x + 1 is one to one or not?
Solution:
We know that if Y= f(x) and if f(x1) = f(x2) $\Rightarrow$ x1 = x2,
then the given function is one one.

So we use this property to prove this:
Let f(a) = 9a + 1 and f(b) = 9b + 1
$\Rightarrow$ 9a + 1 = 9b + 1
$\Rightarrow$ a = b, so above function is one to one function.

Example 3:
X = { 1, 2, 3, 4} and Y = (h, i, j, k), then
(a) { (i, h), (2, i)}.
(b) { (1, i) , (3, j), (4, j)}
(c) { ( 1, i), ( 2, k), (3, k), (4, j)}
(d) { ( 1, h), ( 2, i), ( 3, j), ( 4, k)}
Which one is one to one?
Solution:
(d) is one to one function since each element of X is related to one and only one element of Y.

Example 4:
Show that f: R $\rightarrow$ R defined as f(a) = 3a3 - 4 is one to one function?
Solution:
Let f ( a1 ) = f ( a2 ) $\forall$ a1 , a2 $\in$ R
so 3a13 - 4 = 3a23 - 4
$\Rightarrow$ a13 = a23
$\Rightarrow$ a13 - a23 = 0
$\Rightarrow$ (a1 - a2) (a12 + a1a2 + a22) = 0
$\Rightarrow$ a1 = a2 and (a12 + a1a2 + a22) = 0
(a12 + a1a2 + a22) = 0 is not considered because there is no real values
of a1 and a2.
So the given function f is one one.