1. ## Inverse functions

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.

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.

2. Originally Posted by zacharyk123
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.

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.
ok, so showing the bijection part is fine, you said you can do that (i hope you did not just mean you know it intuitively). so now for uniqueness.

let $f(x)$ be an invertible function, and let $g(x)$ be one of its known inverses. assume there is some other inverse out there, call it $h(x)$. then we know that $f(g(x)) = x$ and $f(h(x)) = x$

thus we have: $f(g(x)) = f(h(x))$

but $f(x)$ is a bijective function, so it is one-to-one. thus by the definition of what it means to be one-to-one, we must have $g(x) = h(x)$. thus, the two inverses are the same. we conclude that the inverse must be unique