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

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  • 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:
Also $f^{-1}(f(y))=y$
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 meena.p
edited Mar 27, 2013 by meena.p

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