Answer :

Given : gof = IA


In order to prove that f is one-one, it is sufficient to prove that f(x)=f(y) x=y x,y A .


Let x,y A such that f(x) = f(y). Then,


f(x) = f(y)


g(f(x)) = g(f(y))


gof(x) = gof(y)


IA(x) = IA(y) [ gof = IA is given ]


x = y [by definition of Identity function, I(x) = x]


Thus, f is one-one.


Now, in order to prove that g : B A is onto, it is sufficient to prove that each element in A has pre-image in B.


Let x A.


Also, f : A B is a function f(x) B


Now,


Let f(x) = y


g(f(x)) = g(y)


gof(x) = g(y)


IA(x) = g(y) [ gof = IA is given ]


x = g(y) [by definition of Identity function, I(x) = x]


Thus, for every x A there exists y B such that g(y) = x.


g is onto.


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