Q24 of 70 Page 11

Using the definition, prove that the function f : A → B is invertible if and only if f is both one-one and onto.

f : A → B is one-one if the images of distinct elements of A under f are distinct, i.e.,for every a, b ∈ A, f(a) = f(b)⇒ a = b


we suppose that f : A → B is not one-one function.


Let f(a) = x and f(b) = x


⇒ f-1(x) = a and f-1(x) = b


⇒ inverse function cannot be defined as we have two images ‘a’ and ‘b’ for one pre-image ‘x’.


So, f can be invertible if it is one-one.


Now, suppose that f : A → B is not onto function.


Let B = {x,y,z} and range of f = {x,y}


We can observe that ‘z’ does not have any pre-image in A.


But f-1 has z as a pre-image which does not have any image in A.


So, f can be invertible if it is onto.


Hence, f is invertible if and only if it is both one-one and onto.


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