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

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

Answers (1)

Say, f: A \rightarrow B be many-one functions.

If, f(a) = p \text{ and } f(b) = p\\

Then f\textsuperscript{-1}(p) = a \: \: and \: \: f\textsuperscript{-1}(p) = b\\

In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.

However, to be one-one, f must be invertible.

Say, f: A \rightarrow B is not onto function.

B = \{ p, q, r \} and \{ p, q \}   is the range of f.

There is no pre-image for the image r, which will have no image in set A.

And, f must be onto to be invertible.

Thus, to be both one-one and onto f must be invertible

If f is a bijective function, then f = X \rightarrow Y is invertible.
 

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