Final answer:
To show that h(x) is a bijection when g(x) and f(x) are bijections, we need to prove that h(x) is both injective and surjective. If h(x) is injective and surjective, it is a bijection.
Step-by-step explanation:
If g(x) and f(x) are bijections, we need to prove that h(x) is also a bijection.
To show that h(x) is a bijection, we need to prove two things:
- h(x) is injective (one-to-one)
- h(x) is surjective (onto)
To prove injectiveness, suppose that h(x) = h(y) for some x and y. We know that h(x) = g(f(x)) and h(y) = g(f(y)). Since g(x) and f(x) are bijections, it implies that f(x) = f(y). Since f(x) = f(y), and f(x) is injective, we can conclude that x = y. Hence, h(x) is injective.
To prove surjectiveness, let's take an arbitrary element y in the codomain of h(x). Since g(x) is surjective, there exists an element x such that g(x) = y. Similarly, since f(x) is surjective, there exists an element z such that f(z) = x. Now, consider h(z): h(z) = g(f(z)) = g(x) = y. This shows that every element in the codomain of h(x) has a pre-image in the domain, making h(x) surjective.
Since h(x) is both injective and surjective, it is a bijection.