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Chapter 5: Problem 96

Prove that if a function has an inverse function, then the inverse function is unique.

Short Answer

Expert verified
The exercise concludes by proving the uniqueness of an inverse function. It is shown that if two functions can act as inverses for a given function, their composite with the original functions bring the original elements back. Thus, they must be the same function implying that inverse function is unique.

Step by step solution

01

Definitions and Assumptions

Let's assume that function \( f: A \rightarrow B \) has an inverse function. We will show that this inverse function is unique. For this, consider two functions \( g: B \rightarrow A \) and \( h: B \rightarrow A \) both claimed to be inverse of \( f \). The primary condition for a pair of functions to be inverse is that for each element in A (with f) and B (with g or h), the composition of the functions will result to the original elements.
02

Establishing an Equality

By the definition of inverse, for \( f(g(b)) = b \) for every \( b \) in B, \( f(h(b)) \) should also be \( b \). Now, consider the composite functions \( g \) after \( f \) and \( h \) after \( f \). According to the definition of inverse function, \( g(f(a)) = a \) for every \( a \) in A, and similarly for \( h(f(a)) \). This mean, \( g(f(a)) = h(f(a)) \) for every \( a \) in A.
03

Concluding Uniqueness

Since we established that \( g(f(a)) = h(f(a)) \) for all \( a \) in A, and \( f(g(b)) = f(h(b)) \) for all \( b \) in B. Thus, the composite functions with \( f \) for both \( g \) and \( h \) yields the same results on domain A and B. Thus, the two functions \( g \) and \( h \) have the same outputs for the same inputs. Therefore, it can be concluded that the function \( g \) and \( h \) are actually the same function, implying that the inverse function is unique.

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