Using the definition,

Question:

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

Solution:

Let f: A → B be many-one function.

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

So, for inverse function we will have f-1(p) = a and f-1(p) = b

Thus, in this case inverse function is not defined as we have two images ‘a and b’ for one pre-image ‘p’.

But for f to be invertible it must be one-one.

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

Let B = {p, q, r} and range of f be {p, q}.

Here image ‘r’ has not any pre-image, which will have no image in set A.

And for f to be invertible it must be onto.

Thus, ‘f’ is invertible if and only if ‘f’ is both one-one and onto.

A function f = X → Y is invertible iff f is a bijective function.

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