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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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