Math, asked by hasan167550, 1 year ago

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

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Answered by abhi178
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) \impliesa =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.

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