Question

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 Iₜ​, is the function Iₜ​:T→T defined by Iₜ​(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. Corollary 6.28. Let A and B be nonempty sets and let f:A→B be a bijection. Then 1. For every x in A,(f⁻¹∘f)(x)=x. 2. For every y in B,(f∘f⁻¹)(y)=y.

Answer 1

Corollary 6.28 establishes that a bijection f between two sets A and B results in compositions f⁻¹∘f and f∘f⁻¹ being the identity functions on A and B, respectively, signifying that f⁻¹∘f is the same as IA, and that f∘f⁻¹ is the same as IB.

Step-by-step explanation:

The concept of the identity function, IṬ, plays a crucial role when understanding Corollary 6.28 in the context of bijections between two sets, A and B. By definition, an identity function on a set A, which we'll denote by IA, is the function IA:A→A such that for all elements a in A, IA(a) = a. Similarly, an identity function on set B, which we'll denote by IB, is defined for all b in B such that IB(b) = b.

Considering f to be a bijection, Corollary 6.28 states that (f⁻¹∘f)(x) = x for all x in A and (f∘f⁻¹)(y) = y for all y in B. Using the concept of the identity function, we can express this corollary by saying that the composition of f and its inverse, f⁻¹, is the identity function on set A, meaning f⁻¹∘f = IA. Likewise, the composition of f⁻¹ and f is the identity function on set B, so f∘f⁻¹ = IB. Therefore, for every element x in A, f⁻¹∘f acts as IA resulting in x, and for every element y in B, f∘f⁻¹ acts as IB, resulting in y.