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Chapter 6: Problem 4

The identity function on a set \(S,\) denoted by \(I_{S},\) is defined as follows: \(I_{S:} S \rightarrow S\) by \(I_{S}(x)=x\) for each \(x \in S .\) Let \(f: A \rightarrow B\) " (a) For each \(x \in A\), determine \(\left(f \circ I_{A}\right)(x)\) and use this to prove that \(f\). \(I_{A}=f\) (b) Prove that \(I_{B} \circ f=f\).

Short Answer

Expert verified
(a) To prove \(f \circ I_A = f\), we find the composition and obtain \(\left(f \circ I_{A}\right)(x)=f(x)\) for all \(x \in A\). This confirms that \(f \circ I_A = f\). (b) To prove \(I_B \circ f = f\), we find the composition and obtain \(\left(I_{B} \circ f\right)(x)=f(x)\) for all \(x \in A\). This confirms that \(I_B \circ f = f\).

Step by step solution

01

(a) Proving \(f \circ I_A = f\)

To find \(\left(f \circ I_A\right)(x)\), we will first apply the identity function \(I_A\) to \(x \in A\), and then use the output of \(I_A\) as input for \(f\). First, we write the definition of the identity function on A: \[I_{A}(x)=x \quad \forall x \in A\] Now, the function composition (f 鈭� I_{A})(x) can be found by applying f on I_{A}(x): \[\left(f \circ I_{A}\right)(x)=f(I_{A}(x))\] We substitute the definition of \(I_A(x)\) in this expression: \[\left(f \circ I_{A}\right)(x)=f(x)\] Since we are given that \(f : A \rightarrow B\), the function \(f(x)\) is defined for any \(x \in A\). Thus, \(\left(f \circ I_{A}\right)(x)=f(x)\) for every \(x \in A\). This proves that \(f \circ I_A = f\).
02

(b) Proving \(I_B \circ f = f\)

To find \(\left(I_B \circ f\right)(x)\), we will first apply the function \(f\) to \(x \in A\), and then use the output of \(f\) as input for the identity function \(I_B\). First, we write the definition of the identity function on B: \[I_{B}(y)=y \quad \forall y \in B\] Now, the function composition (I_{B} 鈭� f)(x) can be found by applying I_{B} on f(x): \[\left(I_{B} \circ f\right)(x)=I_{B}(f(x))\] We substitute the definition of \(I_B(y)\) in this expression: \[\left(I_{B} \circ f\right)(x)=f(x)\] Since we are given that \(f : A \rightarrow B\), the function \(f(x)\) is defined for any \(x \in A\). Thus, \(\left(I_B \circ f\right)(x)=f(x)\) for every \(x \in A\). This proves that \(I_B \circ f = f\).

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Key Concepts

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

Understanding Function Composition
Function composition is a fundamental concept in mathematics that combines two functions to create a new function. When we compose two functions, we apply one function to the result of another. For example, if we have functions \( f : A \rightarrow B \) and \( g : B \rightarrow C \), their composition is denoted as \( g \circ f \) and defined as:
  • \( (g \circ f)(x) = g(f(x)) \) for every \( x \in A \)
This means that for each element \( x \) in the domain \( A \), we first apply \( f \), getting an output in \( B \), and then apply \( g \) to this output to get a result in \( C \). The identity function itself is significant in composition because it behaves like a neutral element. Whatever is composed with the identity function remains unchanged, as shown through the equations \( f \circ I_A = f \) and \( I_B \circ f = f \) discussed in the solution.
Navigating Proof Techniques
Proof techniques are essential tools for verifying the truth of mathematical statements. In this exercise, we use a direct proof technique. A direct proof starts with given assumptions and logically deduces the statement that needs to be proven.When proving that \( f \circ I_A = f \), we follow a straightforward line of reasoning:
  • Apply the identity function \( I_A \) to any element \( x \in A \), resulting in the element itself: \( I_A(x) = x \).
  • Compose \( f \) with \( I_A \), that is, apply \( f \) to the output of \( I_A(x) \), which is \( f(I_A(x)) = f(x) \).
This step-by-step approach clearly leads to the conclusion that \( f \circ I_A = f \).Similarly, the proof for \( I_B \circ f = f \) follows the same logical steps. We first apply \( f \), then apply the identity function \( I_B \) to the result, ensuring that the proof is both clear and conclusive.
Utilizing Mathematical Reasoning
Mathematical reasoning is the process of using logic to connect truths, solve problems, and understand concepts. In this example, reasoning through the identity function helps us grasp why function composition works as described.The identity function \( I_S \) is defined such that \( I_S(x) = x \) for all \( x \in S \). Recognizing this property is key to understanding why, when composed with another function, the overall output remains unchanged. This reasoning is critical, as it shows that identity functions act as the flow-through or unchanged element within a composite function.When we apply mathematical reasoning:
  • We identify the definitions and properties of the functions involved, notably the identity function.
  • We evaluate how these properties interact in our composition, ensuring no steps skip logical or crucial connections.
This logical progression enables us to see why \( f \circ I_A = f \) and \( I_B \circ f = f \) hold true, emphasizing the consistent role of identity functions in maintaining the stability and integrity of function operations.

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Most popular questions from this chapter

In calculus, we learned that if \(f\) is real function that is continuous on the closed interval \([a, b],\) then the definite integral \(\int_{a}^{b} f(x) d x\) is a real number. In fact, one form of the Fundamental Theorem of Calculus states that $$ \int_{a}^{b} f(x) d x=F(b)-F(a) $$ where \(F\) is any antiderivative of \(f,\) that is, where \(F^{\prime}=f\) (a) Let \([a, b]\) be a closed interval of real numbers and let \(C[a, b]\) be the set of all real functions that are continuous on \([a, b]\). That is, $$ C[a, b]=\\{f:[a, b] \rightarrow \mathbb{R} \mid f \text { is continuous on }[a, b]\\} $$ i. Explain how the definite integral \(\int_{a}^{b} f(x) d x\) can be used to define a function \(I\) from \(C[a, b]\) to \(\mathbb{R}\). ii. Let \([a, b]=[0,2] .\) Calculate \(I(f),\) where \(f(x)=x^{2}+1\). iii. Let \([a, b]=[0,2] .\) Calculate \(I(g),\) where \(g(x)=\sin (\pi x)\). In calculus, we also learned how to determine the indefinite integral \(\int f(x) d x\) of a continuous function \(f\). (b) Let \(f(x)=x^{2}+1\) and \(g(x)=\cos (2 x)\). Determine \(\int f(x) d x\) and \(\int g(x) d x\) (c) Let \(f\) be a continuous function on the closed interval [0,1] and let \(T\) be the set of all real functions. Can the process of determining the indefinite integral of a continuous function be used to define a function from \(C[0,1]\) to \(T ?\) Explain. (d) Another form of the Fundamental Theorem of Calculus states that if \(f\) is continuous on the interval \([a, b]\) and if $$ g(x)=\int_{a}^{x} f(t) d t $$ for each \(x\) in \([a, b],\) then \(g^{\prime}(x)=f(x)\). That is, \(g\) is an antiderivative of \(f\). Explain how this theorem can be used to define a function from \(C[a, b]\) to \(T,\) where the output of the function is an antiderivative of the input. (Recall that \(T\) is the set of all real functions.)

Prove Part (2) of Theorem 6.34. Let \(f: S \rightarrow T\) be a function and let \(A\) and \(B\) be subsets of \(S\). Then \(f(A \cup\) \(B)=f(A) \cup f(B)\)

Prove Part (2) of Corollary 6.28. Let \(A\) and \(B\) be nonempty sets and let \(f: A \rightarrow B\) be a bijection. Then for every \(y\) in \(B,\left(f \circ f^{-1}\right)(y)=y\).

Let \(f: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}\) be defined by \(f(m, n)=m+3 n\) (a) Calculate \(f(-3,4)\) and \(f(-2,-7)\). (b) Determine the set of all the preimages of 4 by using set builder notation to describe the set of all \((m, n) \in \mathbb{Z} \times \mathbb{Z}\) such that \(f(m, n)=4\).

(a) Let \(f: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}\) be defined by \(f(m, n)=2 m+n .\) Is the function \(f\) an injection? Is the function \(f\) a surjection? Justify your conclusions. (b) Let \(g: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}\) be defined by \(g(m, n)=6 m+3 n\). Is the function \(g\) an injection? Is the function \(g\) a surjection? Justify your conclusions.
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