Solving Inverse Function Homework: Bijection & Uniqueness

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.


Homework Equations





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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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