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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6Why don't you show $f\circ f^{-1}=\mathrm{id}_Y$ and $f^{-1}\circ f=\mathrm{id}_X$? – yunone Oct 08 '12 at 07:24
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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 08 '12 at 07:31
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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 08 '12 at 07:59
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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 08 '12 at 08:25
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Hint: If $(x,y)\in f$ then $(y,x)\in f^{-1}$ and vice versa.
Mikasa
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@user43964: You can use what Yunone noted you above as well. I think that approach is faster. – Mikasa Oct 08 '12 at 09:01