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Chapter 5: Problem 97

Prove that a function has an inverse function if and only if it is one-to-one.

Short Answer

Expert verified
The statement has been proved that a function \( f \) has an inverse function if and only if it is one-to-one. The proof comprised two parts; firstly proving that if \( f \) is one-to-one it has an inverse, and secondly proving that if \( f \) has an inverse then it is one-to-one.

Step by step solution

01

Define the Function

Firstly, define that \( f: X 鈫� Y \) is a given function.
02

Prove the 'if' Part

Assume the function \( f \) is one-to-one. That means for every \( x1, x2 鈭� X \), if \( f(x1)=f(x2) \), then \( x1=x2 \). To show that \( f \) has an inverse, there must exist a function \( f^{-1} \) such that for every \( x 鈭� X, y 鈭� Y \), if \( f(x)=y \) then \( f^{-1}(y)=x \), and \( f^{-1}(f(x))=x \). Given our one-to-one constraint, such a function \( f^{-1} \) exists. Hence, we have proved the 'if' part.
03

Prove the 'only if' Part

Now, assume that \( f \) has an inverse function, \( f^{-1} \). We need to prove that \( f \) is one-to-one. Assume for some \( x1, x2 鈭� X, f(x1)=f(x2) \). Applying \( f^{-1} \) on both sides, we have \( f^{-1}(f(x1))=f^{-1}(f(x2)) \). From the property of the inverse function, we get \( x1=x2 \). Hence, the function \( f \) is one-to-one, thus proving the 'only if' part.

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Most popular questions from this chapter

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