f: A -> B

A relation from A to B is a function if every element of set A has one and only one image in set B.

∴ For a function

- Every element of set A will have an image.
- Every element of set A will only one image in set B

Let us take an example

Let A = {a, b, c, d}

and B = {Aman, Ankit, Baljinder, Chandu, Eklavya}

## Check if the following are functions?

-a-

For a function

- Every element of set A will have an image.
- Every element of set A will only one image in set B

Since d doesn’t have a image,

**
it is not a function
**

-ea-

-a-

For a function

- Every element of set A will have an image.
- Every element of set A will only one image in set B

Since a has two images,

it is not a function

-ea-

-a-

For a function

- Every element of set A will have an image.
- Every element of set A will only one image in set B

Since every element has an image,

and every element has only one image.

Hence, it is a function

So, the function is

**
f
**
= {(a, Aman), (b, Baljinder), (c, Chandu), (d, Eklavya)}

**
Domain
**
= Set of first elements = {a, b, c, d}

**
Range
**
= Set of second elements = {Aman, Baljinder, Chandu, Eklavya}

**
Codomain
**
= Second set = {Aman, Ankit, Baljinder, Chandu, Eklavya}

-ea-

Note: In a function, domain will always be equal to first set.

-a-

For a function

- Every element of set A will have an image.
- Every element of set A will only one image in set B

Since every element has an image,

and every element has only one image.

Hence, it is a function

So, the function is

**
f
**
= {(a, Ankit), (b, Baljinder), (c, Chandu), (d, Eklavya)}

**
Domain
**
= Set of first elements = {a, b, c, d}

**
Range
**
= Set of second elements = {Ankit, Baljinder, Chandu, Eklavya}

**
Codomain
**
= Second set = {Aman, Ankit, Baljinder, Chandu, Eklavya}

-ea-

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