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Let A be a finite set. If `f: AvecA`is an onto function, show that `f`is one-one also. |
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Answer» Given that `f:A->A` is an onto function. Let `A->{a_1,a_2,a_3....a_n}` As this is an onto function from `A->A`, `:. f(A) ->{f(a_1),f(a_2),f(a_2)...f(a_n)} = A` It means, `f(a_1),f(a_2)...f(a_n)` are all distinct elements of `A`. `:. f: A->A` is one-one function. |
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