Suppose we have $f:X\to Y$, where $f$ is a bijection. How would we show that $(f^{-1})^{-1}=f$
Prove that the inverse of an inverse is the function itself. $(f^{-1})^{-1}=f$
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calculus
real-analysis
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2Why don't you show $f\circ f^{-1}=\mathrm{id}_Y$ and $f^{-1}\circ f=\mathrm{id}_X$? – 2012-10-08
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0I 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 – 2012-10-08
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0Yes, 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 $f=(f^{-1})^{-1}$. – 2012-10-08
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0Thank 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. – 2012-10-08