Is $(g \circ f)^{-1} (x)$ equal to $(f^{-1}\circ g^{-1})(x)$ or not?
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1NOOOOOOOOOOOOOOOO – user 1 Jan 05 '15 at 16:28
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3Have you tried some simple examples? – Robert Israel Jan 05 '15 at 16:28
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1Think they would both have to be invertible. – Trajan Jan 05 '15 at 16:28
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Try $f(x)=2x, g(x)= 2x$. – copper.hat Jan 05 '15 at 16:28
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6Perhaps you mean $(f^{-1} \circ g^{-1})(x)$? – Robert Israel Jan 05 '15 at 16:29
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You have asked if, in general, the inverse of the composition of two functions can equal the composition of those same two functions. This is obviously not possible. Perhaps you've made a typo? – 123 Jan 05 '15 at 16:30
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@RobertIsrael Yes. It is what you have written. You have solved my question. – dj1 Jan 05 '15 at 16:32
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@mathtastic: If $f(x)=g(x) = x$ it is possible. :-) – copper.hat Jan 05 '15 at 16:32
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I edited the question to reflect @dj1's comment above. – copper.hat Jan 05 '15 at 16:33
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@ copper.hat - Notice I specifically said 'in general.' I wrote my comment that way to specifically address the rare instances where his proposition might me true. Had he specified that f(x)=g(x) then yes, that would be the correct response. – 123 Jan 05 '15 at 16:43
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Yes. $\DeclareMathOperator\id{id}$(At least if $f$ and $g$ are invertible.) In general in a group we have $(ab)^{-1}=b^{-1}a^{-1}$. Applying this to $(\langle f,g\rangle,\circ)$ yields $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$.
One can also verify directly that $(g\circ f)\circ(f^{-1}\circ g^{-1})=\id=(f^{-1}\circ g^{-1})\circ(g\circ f)$.
Bart Michels
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$\DeclareMathOperator\id{id}$ 1) For two (bijective) functions $f:A\to B$ and $g:B\to C$, there is no such group as your "$\langle f,g\rangle$". 2) $(g\circ f)\circ(f^{-1}\circ g^{-1})=\id=(f^{-1}\circ g^{-1})\circ(g\circ f)$ makes no sense either. Instead, write $(g\circ f)\circ(f^{-1}\circ g^{-1})=\id_C$ and $(f^{-1}\circ g^{-1})\circ(g\circ f)=id_A$. – Anne Bauval Apr 06 '24 at 13:52