Problem 5 - A Book of Abstract Algebra

Question

Each of the following functions \(f\) is bijective. Describe its inverse. \(A=\\{a, b, c, d\\}, B=\\{1,2,3,4\\}\), and \(f: A \rightarrow B\) is given by $$ f=\left(\begin{array}{llll} a & b & c & d \\ 3 & 1 & 2 & 4 \end{array}\right) $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}\) given by: \(1 \to b\), \(2 \to c\), \(3 \to a\), \(4 \to d\).

Step 1: Understand the Given Function

The function \(f: A \rightarrow B\) is a mapping, where \(A\) is the domain consisting of elements \(\{a, b, c, d\}\), and \(B\) is the codomain consisting of elements \(\{1, 2, 3, 4\}\). The mapping provided is: \(f(a) = 3\), \(f(b) = 1\), \(f(c) = 2\), and \(f(d) = 4\).

Step 2: Identify Characteristics of a Bijective Function

A bijective function is both injective (one-to-one) and surjective (onto). This means each element in the domain maps to a unique element in the codomain, and every element in the codomain is covered by the function.

Step 3: Find the Inverse Function

To find the inverse of a function, we swap the roles of the domain and codomain. Using the mapping provided, the inverse function \(f^{-1}\) can be written as:- \(f^{-1}(1) = b\)- \(f^{-1}(2) = c\)- \(f^{-1}(3) = a\)- \(f^{-1}(4) = d\)

Step 4: Write the Inverse Function

The inverse function \(f^{-1}: B \rightarrow A\) can be expressed as the array:\[\begin{array}{cccc}1 & 2 & 3 & 4 \b & c & a & d \\end{array}\]

Key Concepts

Bijective Functions

Bijective functions are special because they create a perfect pair-up between every element of the domain and the codomain. Think of it like a dance where everyone has a partner, and no one is left without one. In mathematics, this means:

• Each element in the domain is paired with a unique partner in the codomain.
• No two elements in the domain map to the same element in the codomain.
• Every element in the codomain has a partner from the domain.

When a function is bijective, it’s possible to reverse it, meaning you can find an inverse function. The given problem shows a one-to-one correspondence, confirming the bijectiveness because for each element in set A, there is a unique and corresponding element in set B.

Injective Functions

An injective function is also known as one-to-one. This means every element in the domain maps to a unique element in the codomain. If you think about a row of lockers, each locker (domain) has its own unique combination (codomain), and no two lockers share the same combination.

• If function f is injective, then for any two different elements in the domain, their images under f will also be different.
• A key test for injectivity is that if f(x₁) = f(x₂), then x₁ = x₂, meaning no duplicates allowed.

In our problem, each letter from set A maps to a unique number in set B, which confirms the injective nature of the function.

Surjective Functions

A surjective function, sometimes called 'onto', requires every element in the codomain to be mapped to by at least one element from the domain. It's like ensuring every seat at a table is filled, and no one is left standing.

• This means that the image of the domain, under the function, is the entire codomain.
• There cannot be an element in the codomain without a pre-image from the domain.

In the problem, every number from 1 to 4 is mapped from one of the elements in set A, showing a surjective relationship.

Mapping and Functions

Mappings and functions in mathematics describe relationships between sets, specifically the linking of elements from one set to another. A mapping tells us how each element of the first set (domain) is paired with an element of the second set (codomain).

• Function: A rule or policy which pairs elements from two sets.
• Mapping: Often used interchangeably with function, it stresses on the act of linking elements.

In the exercise, function f describes a specific mapping: from every element in set A, we move uniquely to set B. Understanding this mapping helps in thinking about how the function can be reversed to find its inverse, effectively swishing the role of sets A and B.