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When I was doing my exercise about functions, I came across a question that asks to prove that composite function f^-1 g^-1 (x) is equivalent to the composite function (gf)^-1 (x) given random one-one functions for f(x) and g(x).

I answered the question to show they are equivalent by using the example given, and after double checking by using another set of functions for f(x) and g(x), I realized that the rule f^-1 g^-1 (x) = (gf)^-1 (x) applies to all one-one functions.

But I don't know why this particular arrangement (gf)^-1 (x) equals to f^-1 g^-1 (x) rather than (fg)^-1 (x) being equal to it. I just want to ask if there is a certain algebraical method to determine which compound function is equivalent to the other compound function?

nabu1227
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  • The compositions only make sense in one order if the sets involved are different. If $f \colon A \to B$ and $g \colon B \to C$, then $f\circ g$ isn't defined (in general, it would be defined if $C\subset A$). – Daniel Fischer May 26 '17 at 16:05
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    If you do $f$ first and then $g$, then to unwind this you'll gave to unwind $g$ first and then unwind $f.$ – zhw. May 26 '17 at 16:05
  • Function composition is associative but not commutative, $f^{-1}\circ f=f\circ f^{-1}=\text{Id}$, and $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$. These are the only rules. –  May 26 '17 at 16:10
  • I see, thank you all! But @zhw. do you mean have rather than gave? – nabu1227 May 27 '17 at 01:47
  • Yes, lol, have $,$ – zhw. May 27 '17 at 02:33

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