Question

Given the bijective mapping \(\Phi: \mathrm{A} \rightarrow \mathrm{B}\), prove that \(\Phi^{-1}: \mathrm{B} \rightarrow \mathrm{A}\) is also a bijection.

Short answer

To prove that \(\Phi^{-1}: \mathrm{B} \rightarrow \mathrm{A}\) is a bijection, we will show that it's injective and surjective. To prove injectivity, suppose \(\Phi^{-1}(b_1) = \Phi^{-1}(b_2)\) for some \(b_1, b_2 \in \mathrm{B}\). Then, \(b_1 = \Phi(\Phi^{-1}(b_1)) = \Phi(\Phi^{-1}(b_2)) = b_2\), hence, \(\Phi^{-1}\) is injective. To prove surjectivity, for any \(a \in \mathrm{A}\), there exists a unique \(b = \Phi(a) \in \mathrm{B}\) such that \(\Phi^{-1}(\Phi(a)) = a\). Thus, \(\Phi^{-1}\) is surjective. Since \(\Phi^{-1}\) is both injective and surjective, it is a bijection.

Step-by-step solution

Recall definitions of bijection, injection, and surjection

We need to recall the definitions of bijections, injections, and surjections to solve the problem. A function is: - Injective (one-to-one) if different elements in the domain get mapped to different elements in the codomain, i.e., if \(\Phi(a_1) = \Phi(a_2)\), then \(a_1 = a_2\). - Surjective (onto) if every element of the codomain is an image of some element from the domain. - Bijective if it's both injective and surjective. Given that \(\Phi: \mathrm{A} \rightarrow \mathrm{B}\) is a bijection, it is injective and surjective. We will use these properties to show the injectivity and surjectivity of its inverse function \(\Phi^{-1}: \mathrm{B} \rightarrow \mathrm{A}\).

Prove \(\Phi^{-1}\) is injective

To prove that \(\Phi^{-1}\) is injective, we need to show that if \(\Phi^{-1}(b_1) = \Phi^{-1}(b_2)\), then \(b_1 = b_2\). Suppose that \(\Phi^{-1}(b_1) = \Phi^{-1}(b_2)\) for some \(b_1, b_2 \in \mathrm{B}\). Since \(\Phi\) is a bijection, we know that: 1. \(\Phi(a_1) = b_1\) for some \(a_1 \in \mathrm{A}\) 2. \(\Phi(a_2) = b_2\) for some \(a_2 \in \mathrm{A}\) Now, because \(\Phi^{-1}(b_1) = \Phi^{-1}(b_2)\), let's look at the action of \(\Phi\) on this equation: \[\Phi(\Phi^{-1}(b_1)) = \Phi(\Phi^{-1}(b_2))\] Since \(\Phi\) and \(\Phi^{-1}\) are inverse functions, they cancel each other: \[b_1 = b_2\] Therefore, \(\Phi^{-1}\) is injective, as different elements in its domain get mapped to different elements in its codomain.

Prove \(\Phi^{-1}\) is surjective

To prove that \(\Phi^{-1}\) is surjective, we need to show that for any \(a \in \mathrm{A}\), there exists \(b \in \mathrm{B}\) such that \(\Phi^{-1}(b) = a\). Since we are given that \(\Phi\) is a bijection, for every \(a \in \mathrm{A}\), we know there exists a unique element \(b \in \mathrm{B}\) such that \(\Phi(a) = b\). Hence, for each \(a \in \mathrm{A}\), consider the action of the inverse function: \[\Phi^{-1}(\Phi(a))\] Since \(\Phi\) and \(\Phi^{-1}\) are inverse functions, they cancel each other: \[\Phi^{-1}(\Phi(a)) = a\] We've shown that for any \(a \in \mathrm{A}\), there exists a \(b = \Phi(a)\), such that \(\Phi^{-1}(b) = a\). Therefore, \(\Phi^{-1}\) is surjective.

Conclude that \(\Phi^{-1}\) is a bijection

Since we have proven that the inverse function \(\Phi^{-1}: \mathrm{B} \rightarrow \mathrm{A}\) is both injective (one-to-one) and surjective (onto), we can conclude that it is also a bijection.

Key concepts

Inverse Function

An inverse function is a function that, essentially, reverses another function. If you have a function \( \Phi: A \rightarrow B \), its inverse function \( \Phi^{-1}: B \rightarrow A \) undoes what \( \Phi \) does.
Think of it like putting on and taking off a jacket. If putting on the jacket is \( \Phi \), then taking it off is \( \Phi^{-1} \). When you compose a function and its inverse, you should end up right back where you started:

• \( \Phi(\Phi^{-1}(b)) = b \) for all \( b \in B \)
• \( \Phi^{-1}(\Phi(a)) = a \) for all \( a \in A \)

This characteristic is what qualifies \( \Phi^{-1} \) as an inverse to \( \Phi \).
For a function to have an inverse, it must be both injective and surjective, which means it's essentially a bijection. This ensures that each output from the original function can be "undone" by the inverse, allowing us to uniquely identify every input.

Injective Function

An injective function, also known as a one-to-one function, means that no two different elements in the domain map to the same element in the codomain.
Imagine distributing unique gifts: if two different people (inputs) receive the same gift (output), it's not injective. However, if every gift is unique to a person, it is injective. Here's how the property is defined:

• If \( \Phi(a_1) = \Phi(a_2) \) implies that \( a_1 = a_2 \), then \( \Phi \) is injective.

For our problem: because the function \( \Phi \) from A to B is bijective, it's inherently injective.
This property is crucial for the inverse \( \Phi^{-1} \) to be well-defined, meaning that every output in B corresponds to one and only one input in A.

Surjective Function

A surjective function, also known as an onto function, ensures that every element in the codomain is mapped from some element in the domain.
Think of it like throwing a party where every seat is filled: if any seat is empty (any element in the codomain with no pre-image), the function isn't surjective. Instead, every element needs a matching input:

• For every \( b \in B \), there exists an \( a \in A \) such that \( \Phi(a) = b \).

In our exercise, since \( \Phi \) is given as bijective, it's automatically surjective.
This surjectivity ensures that the inverse function \( \Phi^{-1} \) maps every element in B back to some element in A, maintaining the relationship properly from B to A.

Proof by Contradiction

Proof by contradiction is a powerful strategy in mathematics where you start by assuming the opposite of what you want to prove.
Suppose you are proving a statement is true. You begin by assuming that the statement is false, and through logical reasoning, you try to show that this assumption leads to a contradiction.

• To prove \( \Phi^{-1} \) is injective, we can assume the opposite: that \( \Phi^{-1} \) is not injective.
• If we reach a logical impossibility, we've shown that \( \Phi^{-1} \) must be injective.

This method works because the rules of logic dictate that if an assumption leads to a contradiction, then that assumption must be false.
It's like assuming there's no fire, but then discovering smoke; the contradiction (smoke with no fire) suggests that the initial assumption was flawed.
By employing this method, you can rigorously establish truths within mathematics, ensuring logical consistency and thorough proof structures.