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In Progress Check 6.6 on page 298 , we defined the identity function on a set. The identity function on the set \(T,\) denoted by \(I_{T},\) is the function \(I_{T}: T \rightarrow T\) defined by \(I_{T}(t)=t\) for every \(t\) in \(T\). Explain how Corollary 6.28 can be stated using the concept of equality of functions and the identity functions on the sets \(A\) and \(B\).

Short Answer

Expert verified
"If \(f: A \rightarrow B\) and \(g: B \rightarrow A\) are bijections, then their compositions \(g \circ f\) and \(f \circ g\) are equal to the identity functions on sets \(A\) and \(B\, respectively, i.e., \(g \circ f = I_A\) and \(f \circ g = I_B\)."

Step by step solution

01

Let there be two bijections: 1. \(f: A \rightarrow B\) 2. \(g: B \rightarrow A\) We are interested in their compositions: 1. \(g \circ f: A \rightarrow A\) 2. \(f \circ g: B \rightarrow B\) #Step 2: Define the equality of functions#

Two functions \(f\) and \(g\) are equal, denoted as \(f = g\), if they have the same domain and codomain, and for all \(x\) in their common domain, \(f(x) = g(x)\). #Step 3: Define the identity function on sets A and B#
02

The identity function on sets \(A\) and \(B\) are defined as: 1. \(I_A : A \rightarrow A\) such that \(I_A(a) = a\) for all \(a \in A\) 2. \(I_B : B \rightarrow B\) such that \(I_B(b) = b\) for all \(b \in B\) #Step 4: State Corollary 6.28 using the concept of equality of functions and identity functions on sets A and B#

Using the equality of functions and the identity functions on sets \(A\) and \(B\), the statement of Corollary 6.28 can be reframed as follows: "If \(f: A \rightarrow B\) and \(g: B \rightarrow A\) are bijections, then their compositions \(g \circ f\) and \(f \circ g\) are equal to the identity functions on sets \(A\) and \(B\, respectively, i.e., \(g \circ f = I_A\) and \(f \circ g = I_B\)."

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equality of Functions
Understanding when two functions are equal is a fundamental concept in mathematics. Two functions, say \( f \) and \( g \), are considered equal if they exhibit three critical characteristics:
  • They have identical domains.
  • They share the same codomain.
  • For every element \( x \) within their common domain, the outputs are exactly the same, i.e., \( f(x) = g(x) \).
This means if you input any element from their domain into both functions, they will yield the same result.
Let's take an example to make this clearer. Consider two functions \( f \) and \( g \), both defined from \( A \) to \( B \). If for every \( a \in A \), \( f(a) \) gives \( g(a) \), then you can say that \( f = g \).
This concept is particularly important when dealing with compositions and identity functions, as these require functions to operate within defined specifics of equality.
Bijections
Bijections are a special type of function with some unique characteristics. A function \( f: A \rightarrow B \) is a bijection if it is both injective and surjective.
  • Injective: Each element in the codomain \( B \) is mapped by at most one element from domain \( A \). Essentially, no two elements in \( A \) map to the same element in \( B \).
  • Surjective: Every element in the codomain \( B \) is the image of some element from the domain \( A \). There are no leftovers in \( B \), ensuring full coverage.
Think of bijections as perfect matches or pairings between sets where every element has a unique partner, and there are no unmatched elements left in either set. In the exercise context, having bijections \( f: A \rightarrow B \) and \( g: B \rightarrow A \) implies both functions are inverses of each other.
Therefore, composing them, as in \( g \circ f \) or \( f \circ g \), should result in identity functions, reflecting their uniqueness and complete nature. This property is why bijections are known as "one-to-one" and "onto" functions.
Composition of Functions
Composition of functions is an essential operation, especially when working with multiple functions. If you have two functions \( f: A \rightarrow B \) and \( g: B \rightarrow C \), their composition \( g \circ f \) is a new function from \( A \) to \( C \).
To compose, you do the following:
  • First, apply \( f \) to an element from set \( A \).
  • Then, take the result and apply \( g \) to it to get the final output in set \( C \).
It's like chaining them together, where the output of \( f \) becomes the input for \( g \).
For the identity and bijective functions discussed before, compositions like \( g \circ f = I_A \) or \( f \circ g = I_B \) simplify because they map back to the original set perfectly, illustrating one key ability of bijections.
This means applying the composition leads perfectly back to where you started, mimicking the role of an identity function where nothing changes, showcasing a harmony between the function operations.

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Most popular questions from this chapter

Let \(C\) be the set of all real functions that are continuous on the closed interval \([0,1] .\) Define the function \(A: C \rightarrow \mathbb{R}\) as follows: For each \(f \in C,\) $$A(f)=\int_{0}^{1} f(x) d x$$ Is the function \(A\) an injection? Is it a surjection? Justify your conclusions.

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x)=-2 x+1\). Let \(A=[2,5] \quad B=[-1,3] \quad C=[-2,3] \quad D=[1,4]\) Find each of the following: (a) \(f(A)\) (b) \(f^{-1}(f(A))\) (c) \(f^{-1}(C)\) (d) \(f\left(f^{-1}(C)\right)\) (e) \(f(A \cap B)\) (f) \(f(A) \cap f(B)\) (g) \(f^{-1}(C \cap D)\) (h) \(f^{-1}(C) \cap f^{-1}(D)\)

Let \(f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}\) be the function defined by \(f(x, y)=-x^{2} y+3 y,\) for all \((x, y) \in \mathbb{R} \times \mathbb{R} .\) Is the function \(f\) an injection? Is the function \(f\) a surjection? Justify your conclusions.

The number of divisors function. Let \(d\) be the function that associates with each natural number the number of its natural number divisors. That is, \(d: \mathbb{N} \rightarrow \mathbb{N}\) where \(d(n)\) is the number of natural number divisors of \(n\). For example, \(d(6)=4\) since \(1,2,3,\) and 6 are the natural number divisors of 6 (a) Calculate \(d(k)\) for each natural number \(k\) from 1 through 12 . (b) Does there exist a natural number \(n\) such that \(d(n)=1 ?\) What is the set of preimages of the natural number \(1 ?\) (c) Does there exist a natural number \(n\) such that \(d(n)=2 ?\) If so, determine the set of all preimages of the natural number \(2 .\) (d) Is the following statement true or false? Justify your conclusion. For all \(m, n \in \mathbb{N},\) if \(m \neq n,\) then \(d(m) \neq d(n) .\) (e) Calculate \(d\left(2^{k}\right)\) for \(k=0\) and for each natural number \(k\) from 1 through 6 (f) Based on your work in Exercise (6e), make a conjecture for a formula for \(d\left(2^{n}\right)\) where \(n\) is a nonnegative integer. Then explain why your conjecture is correct. (g) Is the following statement is true or false? For each \(n \in \mathbb{N},\) there exists a natural number \(m\) such that \(d(m)=n\)

(a) Define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x)=e^{-x^{2}}\). Is the inverse of \(f\) a function? Justify your conclusion. (b) Let \(\mathbb{R}^{*}=\\{x \in \mathbb{R} \mid x \geq 0\\} .\) Define \(g: \mathbb{R}^{*} \rightarrow(0,1]\) by \(g(x)=e^{-x^{2}}\). Is the inverse of \(g\) a function? Justify your conclusion.

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