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Do there exist two functions $f$ and $g$ from the reals to itself satisfying $f\circ g (x)=x^2 , g\circ f (x)=x^3$ for any $x\in\mathbb{R}$?

From the given equations I could get the following information:

  1. $f$ is injective.

  2. $g$ is surjective and an even function.

  3. $f(x^3)=f(x)^2$ for every real number $x$.

  4. $g(x^2)=g(x)^3$ for every real number $x$.

How these information help us to decide whether such functions exist or not?

Thank you.

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    $\begingroup$ Hint: Prove that if $S$ and $T$ are two sets, and if $f : S \to T$ and $g : T \to S$ are two maps, then the maps $f \circ g : S \to S$ and $g \circ f : T \to T$ have the same number of fixed points. (The number can be infinite, though -- but not in this case.) $\endgroup$ – darij grinberg Oct 27 '18 at 23:20
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    $\begingroup$ @darij grinberg Thank you for your help i will try it. Here the fixed points of the first are 0 ,1 while those of the second are -1, 0, 1. So such functions do not exist by the fact you mentioned. How the theorem is restated for infinite fixed points? $\endgroup$ – Fermat Oct 27 '18 at 23:30
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    $\begingroup$ The general statement is that there is a bijection from the set of fixed points of $f \circ g$ to the set of fixed points of $g \circ f$. (Can you find this bijection? It's very simple.) $\endgroup$ – darij grinberg Oct 27 '18 at 23:33
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    $\begingroup$ If $a$ is a fixed point of $f\circ g$, then $g(a)$ is a fixed point for $g\circ f$. The other direction is similar. $\endgroup$ – Fermat Oct 27 '18 at 23:52
  • $\begingroup$ @Fermat , should $f, g$ be injective? $\endgroup$ – Doyun Nam Oct 28 '18 at 0:25
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No, they don't.

I have a strong feeling this is a duplicate, but can't find the original, so let's repeat it anyway.

Say, $f(0) = a$. Then $f\circ g\circ f(0) = f(0^3)=a$, but at the same time it equals $f(0)^2=a^2$. So $a$ is either $0$ or $1$.

The same reasoning applies to $f(1)$ and $f(-1)$, with the same result. So at least some of these three values must coincide, which contradicts with $f$ being injective.

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