Let X, Y and Z be any non-empty sets and let f and g be one-one functions of X onto Y andY onto Z respectively so that f and g are both invertible. Then, show that(a) g ◦ f is one-one(b) g ◦ f is onto(c) (g ◦ f )−1 = f −1 ◦g−1

(a) g ◦ f is one-one:

Assume x1, x2 ∈ X and g(f(x1)) = g(f(x2)). Since g is one-one, this implies that f(x1) = f(x2). And since f is also one-one, this implies that x1 = x2. Therefore, g ◦ f is one-one.

(b) g ◦ f is onto:

Let z ∈ Z. Since g is onto, there exists y ∈ Y such that g(y) = z. Since f is onto, there exists x ∈ X such that f(x) = y. Therefore, g(f(x)) = z, which means that for every z ∈ Z, there exists x ∈ X such that g(f(x)) = z. Therefore, g ◦ f is onto.

(c) (g ◦ f )−1 = f −1 ◦g−1:

Let z ∈ Z. Since g ◦ f is onto, there exists x ∈ X such that g(f(x)) = z. Therefore, (g ◦ f )−1(z) = x. Now, g(f(x)) = z implies that f(x) = g−1(z) and x = f−1(g−1(z)). Therefore, (g ◦ f )−1(z) = f−1(g−1(z)), which means that (g ◦ f )−1 = f−1 ◦g−1.