Proving a Function is a Bijection

Prove that a function $f : [0, 1] \to [2, 4]$ is a bijection by showing it is well-defined, injective, and surjective.

In this problem, you will prove that a function is a bijection. A function is considered bijective if it is both injective (one-to-one) and surjective (onto). This means that each element in the domain maps to a unique element in the range (injective), and every element in the range is mapped by an element from the domain (surjective). The concept of bijection is central in understanding the relationship between sets, especially when discussing cardinalities and if sets have the same size in terms of elements.

Firstly, to prove the function is well-defined, you need to ensure that for every input from the domain [0, 1], the function provides an output in the range [2, 4]. The well-defined nature of a function is a prerequisite before considering injectivity and surjectivity. Consider how functions can be constructed or given in this form and ensure the mapping respects these boundaries.

For injectivity, you will typically demonstrate that whenever two elements in the domain, say x1 and x2, map to the same element in the codomain (f(x1) = f(x2)), it must follow that x1 equals x2. This ensures that no two distinct inputs can produce the same output, respecting the one-to-one property.

To prove surjectivity, illustrate that for every element b in the codomain [2, 4], there exists an element a in the domain [0, 1] such that f(a) = b. Often this may involve solving for the input in terms of the output or showing that the function spans the entire codomain. By tackling these steps, you not only establish the bijectiveness of the function but also gain deeper insights into how functions relate different sets together.