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

Toolbox:
• A function $f: A \rightarrow B$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
• A function f is invertible if and only if f is one-one and f is onto. also $f^{-1}$ is a function g such that $fog=I=gof$

Given:If f is invertible then f must be one -one and onto ie then exist $g=f^{-1}$ such that $fog=I=gof$

Step1: Injective or One-One function:

$f^{-1}of(x)=x$

$f^{-1}(f(x))=x$

Also $f^{-1}(f(y))=y$

$=>f(x)=f(y)=>x=y$

f is one-one

Step 2: Surjective or On-to function:

and for every element y belonging h image f there exists $f(x)=y$

f is onto

If f is one one and onto by definition

f is invertible. ie then exists $g =f^{-1}$

Such that $gof=fog=I$

answered Mar 4, 2013 by
edited Mar 27, 2013 by meena.p