One-to-One correspondence also called as Bijection, is an mathematical concepts which comes under sets. An one-to-one correspondence between two sets exists only if a mapping exists between the two sets such that each element of the first set maps to exactly one element of the second set and each element of the second set maps to exactly one element of the first set. Mapping is kind of binary relation which relates any given element to another. The below figure shows how mapping relates each element of one set with another different set of elements. |

**'A'**is evenly paired with the element of a second set, say

**'B'**. Evenly paired is nothing but the matching of each element of

**'A'**with the only one element of

**'B'**, and vice versa. Here, none of the element of 2 sets are left unpaired. As a result of this, we get an ordered pairs of

**(a, b)**, where

**'a'**is the element of the set

**'A'**and

**'b'**is the element of the set

**'B'**. In the process of pairing, no two ordered pairs are created having the same number of elements.

And, when this type of ordered pair exist, we say that one-to-one correspondence exist between

**'A'**and

**'B'**. Any two sets which have same number of the element can easily be seen in finite sets.

When a same set

**'A'**and

**'B'**relates the function of each element of

**'A'**and with that of

**'B'**which is twice larger, then this type of function is called as one-to-one function also called as injection function. One-to-One function are mainly used to find whether an one-to-one correspondence exists between infinite sets.

To have a correct pairing between the set

**'A'**and**'B'**, they should satisfy all the 4 properties below.- Each element of A must be paired with at least one element of B, and
- No element of A may be paired with more than one element of B,
- Each element of B must be paired with at least one element of A, and
- No element of B may be paired with more than one element of A.

### Solved Examples

**Question 1:**If C = {Car 1, Car 2, Car 3} and D = {Green, Orange, Black}, then write the one-to-one correspondence from C to D and D to C.

**Solution:**

Since A = {Car 1, Car 2, Car 3} and B = {Green, Orange, Black}

We have the one-to-one correspondence from C to D given by the set of ordered pairs as

C to D = {(Car 1, Green), (Car 2, Orange), (Car 3, Black)}

And, one-to-one correspondence from C to D is given by the set of ordered pairs obtained by reversing each pair of one-to-one correspondence from D to C.

D to C = {(Green, Car 1), (Orange, Car 2), (Black, Car 3)}

We have the one-to-one correspondence from C to D given by the set of ordered pairs as

C to D = {(Car 1, Green), (Car 2, Orange), (Car 3, Black)}

And, one-to-one correspondence from C to D is given by the set of ordered pairs obtained by reversing each pair of one-to-one correspondence from D to C.

D to C = {(Green, Car 1), (Orange, Car 2), (Black, Car 3)}

**Question 2:**A group of passengers board the bus which contains certain number of seats. The conductor asks everyone to be seated and then, notices that there is a bijection between the number of seats and number of passengers. Here, each passenger is paired with the seat they are seating. Now, what observation made the conductor makes him to give this conclusion.

**Solution:**

- All the passengers were seated.
- Each seats were occupied by the each individuals without sharing the seat with anyone else.
- All the passengers were seated, no seats were left empty and no seat were occupied by more than one passenger.
- Therefore, the conductor was able to conclude that each and every passenger has their own seat.