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Chapter 7: Problem 11

State and prove a theorem on inverse functions analogous to the one that says that if a matrix has an inverse, that inverse is unique.

Short Answer

Expert verified
A function inverse, when it exists, is unique.

Step by step solution

01

Define the Theorem

Theorem: If a function \( f: A \rightarrow B \) has an inverse \( g: B \rightarrow A \), then the inverse \( g \) is unique.
02

Understand the Concept of Inverse Functions

A function \( f: A \rightarrow B \) is said to have an inverse \( g: B \rightarrow A \) if \( f(g(b)) = b \) for all \( b \) in \( B \) and \( g(f(a)) = a \) for all \( a \) in \( A \). This essentially means applying \( g \) to \( f \) results in the identity function on \( A \) and vice versa on \( B \).
03

Assume Existence of Another Inverse

Assume \( h: B \rightarrow A \) is another inverse of \( f \), meaning \( f(h(b)) = b \) for all \( b \) in \( B \) and \( h(f(a)) = a \) for all \( a \) in \( A \).
04

Show Function Equality

For any \( b \in B \), consider \( g(b) \). Since \( f \) and \( g \) are inverses, \( f(g(b)) = b \). Applying \( h \) to both sides gives us \( h(f(g(b))) = h(b) \). Since \( h \) and \( f \) are also inverses, \( h(f(g(b))) = g(b) \). Thus, \( g(b) = h(b) \) for all \( b \in B \).
05

Conclude Uniqueness

Since we assumed \( h \) was another inverse, and given that \( g(b) = h(b) \) for all \( b \in B \), we have shown \( h = g \). Therefore, any two inverses of \( f \) must be identical.

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Key Concepts

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

Function Composition
Function composition is a process of combining two functions to get a new function. It involves applying one function to the result of another function. For example, if you have two functions, say \( f: A \rightarrow B \) and \( g: B \rightarrow C \), the composition \( g \circ f \) is a function from \( A \) to \( C \). This is expressed as \( (g \circ f)(a) = g(f(a)) \) for all elements \( a \) in the set \( A \).
  • It helps in understanding how two related processes connect to produce a final outcome.
  • Function composition is associative, meaning \( (h \circ g) \circ f = h \circ (g \circ f) \).
  • However, it is not commutative, which means \( g \circ f eq f \circ g \) in most cases.
Understanding composition is crucial when working with inverse functions and proving their uniqueness.
Unique Inverses
Unique inverses refers to the property that if a function has an inverse, it can have only one such inverse. Assume you have a function \( f: A \rightarrow B \) which has an inverse \( g: B \rightarrow A \). The inverse \( g \) is unique, meaning there cannot be another different function that does the same job. This idea is illustrated in the steps of proving uniqueness in inverse functions.
  • Implying the notion: If \( h \) is another inverse, then for every element \( b \) in \( B \), both \( f(g(b)) = b \) and \( f(h(b)) = b \).
  • Comparing results: Applying \( h \) and \( g \) interchangeably leads to \( h(f(g(b))) = g(b) \) showing \( h = g \).
  • Conclusion: Since the functions output the same result across all elements, \( h \) must equal \( g \), proving uniqueness.
Identity Function
An identity function plays a key role in defining the concept of inverse functions. It is a special function that acts as a sort of 'do nothing' operation. Formally, for a set \( A \), the identity function \( \text{id}_A \) is defined such that \( \text{id}_A(a) = a \) for every \( a \) in \( A \).
  • On applying the identity function, every element maps to itself, keeping the set unchanged.
  • In terms of function inverses, when function \( f \) is composed with its inverse \( g \), it returns the identity function: \( f(g(b)) = \text{id}_B(b) \) and \( g(f(a)) = \text{id}_A(a) \).
  • This property confirms that applying a function followed by its inverse (or vice versa) yields no net change, representing a fundamental aspect of inverse relationships.
Understanding identity functions is crucial in defining and proving the existence and uniqueness of inverse functions.

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Most popular questions from this chapter

(a) Prove that the set of finite strings of 0 's and 1 's is countable. (b) Prove that the set of odd integers is countable. (c) Prove that the set \(\mathbb{N} \times \mathbb{N}\) is countable.

Let \(f\) be a function with domain \(A\) and codomain \(B\). Consider the relation \(K \subseteq A \times A\) defined on the domain of \(f\) by \((x, y) \in K\) if and only if \(f(x)=f(y) .\) The relation \(K\) is called the kernel of \(f\). (a) Prove that \(K\) is an equivalence relation. (b) For the specific case of \(A=\mathbb{Z},\) where \(\mathbb{Z}\) is the set of integers, let \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) be defined by \(f(x)=x^{2} .\) Describe the equivalence classes of the kernel for this specific function.

In your own words explain the statement "The sets of integers and even integers have the same cardinality."

Prove by induction that if \(n \geq 2\) and \(f_{1}, f_{2}, \ldots, f_{n}\) are invertible functions on some nonempty set \(A,\) then \(\left(f_{1} \circ f_{2} \circ \ldots \circ f_{n}\right)^{-1}=f_{n}^{-1} \circ \ldots \circ f_{2}^{-1} \circ f_{1}^{-1}\). The basis has been taken care of in Exercise 7.3 .4 .12 .

Which of the following are one-to-one, onto, or both? (a) \(f_{1}: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f_{1}(x)=x^{3}-x .\) (b) \(f_{2}: \mathbb{Z} \rightarrow \mathbb{Z}\) defined by \(f_{2}(x)=-x+2\). (c) \(f_{3}: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}\) defined by \(f_{3}(j, k)=2^{j} 3^{k}\) (d) \(f_{4}: \mathbb{P} \rightarrow \mathbb{P}\) defined by \(f_{4}(n)=\lceil n / 2],\) where \([x]\) is the ceiling of \(x\), the smallest integer greater than or equal to \(x\). (e) \(f_{5}: \mathbb{N} \rightarrow \mathrm{N}\) defined by \(f_{5}(n)=n^{2}+n\) (f) \(f_{6}: \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}\) defined by \(f_{6}(n)=(2 n, 2 n+1)\).
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