Let f : X → Y and g : Y → X. If g ◦ f is an identity function on X and f ◦ g is an identityfunction on Y , then, show that(a) f is one-one(b) f is onto(c) g = f −1
Question
Let f : X → Y and g : Y → X. If g ◦ f is an identity function on X and f ◦ g is an identityfunction on Y , then, show that(a) f is one-one(b) f is onto(c) g = f −1
Solution
(a) f is one-one (Injective)
Proof: Assume that there are two elements x1, x2 in X such that f(x1) = f(x2). Since g ◦ f is the identity function on X, we have g(f(x1)) = x1 and g(f(x2)) = x2. But since f(x1) = f(x2), we have g(f(x1)) = g(f(x2)). Therefore, x1 = x2. This shows that f is one-one.
(b) f is onto (Surjective)
Proof: Let y be any element in Y. Since f ◦ g is the identity function on Y, we have f(g(y)) = y. This shows that for every element y in Y, there is an element g(y) in X such that f maps g(y) to y. Therefore, f is onto.
(c) g = f^(-1)
Proof: Since g ◦ f is the identity function on X and f ◦ g is the identity function on Y, it follows that g is the inverse of f and f is the inverse of g. Therefore, g = f^(-1).
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