One to one correspondence

One to One Correspondence

One-to-One functions define that each element of one set called Set (A) is mapped with a unique element of another set called Set (B).
Or

It can  be defined as each element of Set A has a unique element on Set B.

In short , let us  take  ‘f’ as a function whose domain is set A. The function is said to be injective if for all x and y in A,

When there is  f(x)=f(y), then x=y

And equally , if x ≠ y, then f(x) ≠ f(y)

Now, Onto functions are basically functions whose range equates to its codomain..

Thus,

It is stated as, if f(x) = f(y)  implies x=y, then f is one-to-one mapped and also if the codomain equates to the range of the function, f is one to one correspondence.

Formally , if “f” is a function which is one to one correspondence, with domain A and range B, then the inverse of function f is given by;
f-1(y) = x ; if and only if f(x) = y

A function f : X → Y is said to be one to one correspondence, if the images of unique elements of X under f are unique, i.e., for every x1 , x2 ∈ X, f(x1 ) = f(x2 ) implies x1  = x2 and also range = codomain. Otherwise, we call it a non invertible function or not bijective function.

Therefore we can say, every element of the codomain of one to one correspondence is the image of only one element of its domain.

One to one correspondence example

Let f : A ----> B be a function. Codomain = {7,9,10,8,4} 

The function f is  say is  one to one, if it takes different elements of A into different elements of B.

So then , we say f is one to one

If  f is one-one, if no element in B is associated with more than one element in A. 

Also, range is equal to codomain given the function. Thus, this is one to one correspondence or an invertible function.

Solved Problems

Problem 1 : 

Let f : A ----> B. A, B and f are defined as. Codomain is the set { 5,6,7,8} 

A  =  {1, 2, 3}

B  =  {5, 6, 7, 8}

f  =  {(1, 5), (2, 8), (3, 6)}

Prove that  whether f is a function. If so, then is it one to one correspondence?

Solution :

Write the elements of f (ordered pairs) using arrow diagram as shown below

In the above arrow diagram, all the elements of A have images in B and every element of A has a unique element .

That is, no element of A has more than one image or unique value..

So, f is a function. 

Every element of A has a different value in B.

That is, no two or more elements of A have the same value  in B.

Therefore, f is one to one function. 

BUT

Codomain is given as {5,6,7,8} while Range is {5,6,8}. Thus codomain is not equal to range.

Therefore, f is not one to one correspondence. 

Problem 2 : 

Let f : X ----> Y. X, Y and f are defined as 

X  =  {a, b, c, d}

Y  =  {d, e, f}

f  =  {(a, e), (b, f), (c, e), (d, d)}

Is f one to one correspondence? Explain.  

In the above arrow diagram, all the elements of X have images in Y and every element of X has a unique element..

That is, no element of X has more than one image.

The elements "a" and "c" in X have the same image "e" in Y. 

So, f is not one to one or injective. 

Thus, f is not an invertible function or not one to one correspondence. 

Problem 3 : 

Let f : A ----> B. A, B and f are defined as 

A  =  {1, 2, 3, 4}

B  =  {5, 6, 7, 8}

f  =  {(1, 8), (2, 6), (3, 5), (4, 7)}

Codomain= {5,6,7,8}

Is f invertible or one to one correspondence ? Explain. 

Solution :

Please write the elements of f (ordered pairs) using an arrow diagram as shown below. 

In the above  diagram, all the elements of A have images in B and every element of A has a unique image.

That is, no element of A has more than one image.

So, f is a function. 

Every element of A has a different element  in B.

That is, no two or more elements of A have the same image in B.

Therefore, f is one to one or injective function.

Also, the codomain is equal to the range which is {5,6,7,8}

Therefore the above mapping is both one to one and onto which makes it one to one correspondence.

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