Solving Inverse Function Homework: Bijection & Uniqueness

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SUMMARY

The discussion focuses on proving the uniqueness of the inverse function for bijections. It establishes that if a function \( f: X \to Y \) is a bijection, then any two functions \( g \) and \( h \) that are inverses of \( f \) must be identical, i.e., \( g(y) = h(y) \) for all \( y \) in \( Y \). The proof strategy involves assuming the existence of a different inverse \( g \) and deriving a contradiction, thereby confirming that the inverse is unique. The approach utilizes the properties of bijections and the definition of inverse functions.

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  • Understanding of bijections in set theory
  • Familiarity with inverse functions
  • Knowledge of proof by contradiction
  • Basic concepts of functions and mappings
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  • Study the properties of bijections in more depth
  • Learn about the formal definition of inverse functions
  • Explore proof techniques, particularly proof by contradiction
  • Review examples of bijections and their inverses in mathematical contexts
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Students studying mathematics, particularly those focusing on algebra and functions, as well as educators looking for clear explanations of bijections and inverse functions.

benjamin111
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Homework Statement


My textbook states that the inverse of a bijection is also a bijection and is unique. I understand how to show that the inverse would be a bijection and intuitively I understand that it would be unique, but I'm not sure how to show that part.


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The Attempt at a Solution



My idea is to somehow say that if the inverse function is bijective and maps S -> T such that f-1(f(x))=x, then any other function that produces the same result must be the same function, but I can't quite figure out how to make this statement mathematically...
Thanks.
 
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Well, about showing that the inverse is unique, try to prove it by using a contradiction. That is suppose that there is another function call it g that is different from f^-1 ( the inverse of f) but that is also the inverse of f, ( suppose that also g is the inverse of f) and try to derive a contradiction, in other words try to show that indeed f^-1=g.

This is the ide, the rest are details.
 
Suppose f:X->Y is a bijection. Let g and h be inverses of f. Show that g(y)=h(y) for all y in Y. To do this express y as f(x) (possible since f is a surjection), and use the definition of g and h.
 

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