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Suppose we have $f:X\to Y$, where $f$ is a bijection. How would we show that $(f^{-1})^{-1}=f$

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Why don't you show $f\circ f^{-1}=\mathrm{id}_Y$ and $f^{-1}\circ f=\mathrm{id}_X$? – yunone Oct 8 '12 at 7:24
I was wondering about that, that is what I did and it should suffice to prove this correct? I just was not sure that I was proving the same thing – user43964 Oct 8 '12 at 7:31
Yes, it should work. A function $g$ is the inverse of $f^{-1}$ if it satisfies $g\circ f^{-1}=\mathrm{id}_Y$, etc. Since the inverse is unique, $g=f$, so $f=(f^{-1})^{-1}$. – yunone Oct 8 '12 at 7:59
Thank you I completely understood it now. I was just missing that much I proved it and happy with the end result as well. Thanks for the tips. – user43964 Oct 8 '12 at 8:25

Hint: If $(x,y)\in f$ then $(y,x)\in f^{-1}$ and vice versa.

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Oh I see you want me to use the inverse of inverse relation! Thanks – user43964 Oct 8 '12 at 7:39
@user43964: You can use what Yunone noted you above as well. I think that approach is faster. – Babak S. Oct 8 '12 at 9:01

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