Question

prove that only one one onto function possess an Inverse function​

Answer 1

we have to prove that , A function is invertible function if and only if it is bijective ( means one to one and onto).

proof : suppose $f:A\rightarrow B$ is an invertible function. and $f^{-1}$ is inverse of it.

first we have to show that f is injective (one to one ).

suppose $a,b\in A$ are such that f(a) = f(b)

then we can apply $f^{-1}$ to both sides to get, $f^{-1}of(a)=f^{-1}of(b)$.

now by the definition of inverse, both the above compositions are the identity on A. so, we can $f^{-1}of(a)=f^{-1}of(b)\implies a=b$

but we know, if $a,b\in A$ are such that f(a) = f(b) $\implies$a =b, then f is one-to-one function or injective function.

now, to show surjective function. let $b\in B$ and let $a=f^{-1}(b)$ then,

$f(a)=fof^{-1}(b)=(fof^{-1})(b)=b$ [ as f is inverse so, $fof^{-1}(x)=x$

but we know, from definition of surjective, if $b\in B$ in such a way that $a=f^{-1}(b)$ and $f(a)=b$

then, f is surjective or onto function.

hence, given function, f is also surjective or onto function.

now it is proved that a function is invertible function if and only if it is one to one and onto.