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.
If f is a function defined as y = f(x), then inverse function of f is x = f -1(y) i.e. f-1 defined from y to x. In inverse function co-domain of f is the domain of f -1 and the domain of f is the co-domain of f -1.

One to One Function Inverse
Only one-to-one functions has its inverse since these functions has one to one correspondences i.e. each elements from the range correspond to one and only one domain element. 

Let a function f : A$\rightarrow$ B is defined, then f is said to be invertible if there exists a function g : B $\rightarrow$ A in such a way that if we operate f{g(x)} or g{f(x)} we get the starting point or value.

Example:
Show that the function f : X $\rightarrow $ Y, such that f(x)= 5x + 7,
$\forall$ x,y $\in$ N is an invertible.

Solution:
Let y$\in$ N $\rightarrow$ y = f(x) = 5x + 7 for x$\in$N

$\Rightarrow$ x = $\frac {y-7} { 5} $.

If we define h : Y $\rightarrow$ X by h(y) = $\frac {y-7} { 5} $.

Again hof(x) = h[ f(x) ] = h{ 5x + 7 } = $\frac {5(y-7)} { 5} $ + 7 = x

And foh(y) = f[ h(y) ] = f( $\frac {y-7} { 5} $) = $\frac {5(y-7)} { 5} $ + 7 = y

Hence f is invertible function and h is the inverse of f. 
Let y = x3 + 1 be a function where y is depends on x, then graph of the function

One to One Function Graph
If  we draw a horizontal line then that cuts the graph at one point, so the given function is one-one function.