In the article Injective and Surjective Functions, the concepts of injective functions and surjective functions are discussed separately. Some functions are injective but not surjective while others are surjective but not injective. In fact, It is possible for a function to be both injective and surjective. Such function is called a bijection or one-to-one correspondence.

 

Definition

Let A and B be sets and f is a function from A to B. We say that f is a bijective function (or bijection) if f is injective and surjective.

 

Example 1

Let f : \mathbb{R} \rightarrow \mathbb{R} be defined by f (x) = 3x. Prove that f is a bijection.

Answer

To prove that f is a bijective function, we have to show that f is injective and bijective. Thus, the proof falls into two parts. We begin by proving that f is injective. The assertion will be established if we prove f (a) = f (b) ⇒ a = b. For this purpose, let f (a) = f (b). By the definition of f, 3a = 3b. Dividing both sides of the equation by 3, we have a = b, which completes the proof. It remains to prove that f is surjective. To prove this, we have to show that for every p ∈ \mathbb{R} there exists a t ∈ \mathbb{R} such that f (t) = p. Now, let p ∈ \mathbb{R}. Choose t = \frac{p}{3}. We claim that t ∈ \mathbb{R} and f (t) = p. We have shown that f is injective as well as surjective. Accordingly, f is a bijective function.

 

Example 2

Let {\mathbb{R}}^+ = \{ x \in \mathbb{R} \vert x \geqslant 0 \} and g be a function from \mathbb{R} to {\mathbb{R}}^+ defined by g(x) = x2. Prove that g is not bijective.

Answer

We claim that g is not injective (and therefore not bijective). To prove that g is not injective, we have to show that g(a) = g(b) but a ≠ b. Choose a = -3 and b = 3. It can be easily seen that g(a) = g(b) = 9 but a ≠ b. Thus, g is not injective and therefore not bijective.

 

Example 3

Let {\mathbb{R}}^+ = \{ x \in \mathbb{R} \vert x \geqslant 0 \} and h be a function from {\mathbb{R}}^+ to \mathbb{R} defined by h(x) = x2. Prove that h is not a bijection.

Answer

It is sufficient to show that h is not surjective. We have to prove that there is a p ∈ \mathbb{R} such that for every t ∈ {\mathbb{R}}^+ h(t) ≠ p. Choose (for example) p = -1. Since the square of any real number cannot be negative, it must hold that h(t) ≠ p for every t ∈ {\mathbb{R}}^+. So, h is not surjective hence not bijective.

 

Example 4

Let {\mathbb{R}}^+ = \{ x \in \mathbb{R} \vert x \geqslant 0 \} and k be a function from {\mathbb{R}}^+ to {\mathbb{R}}^+ defined by k(x) = \sqrt{x}. Prove that k is a bijection.

Answer

As in the proof in Example 1, we have to show that k is both injective and surjective. To prove that k is an injection, let k(a) = k(b). This gives \sqrt{a} = \sqrt{b}. Square both sides of the equation. This results in a = b. We have shown that k(a) = k(b) ⇒ a = b. Thus, k is injective. What is left is to prove that k is surjective, i.e. for every p ∈ {\mathbb{R}}^+ there is a t ∈ {\mathbb{R}}^+ such that k(t) = p. To do this, let p ∈ {\mathbb{R}}^+. Choose t = p2 . We claim that t ∈ {\mathbb{R}}^+ and it satisfies k(t) = p. Consequently, k is surjective. As k is injective and surjective, we conclude that k is a bijection.

 

 

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