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I have a brain lag. I'm overthinking probably. I have the following equation: $$ f^{-1}(f(x)+f(y))=(g\circ h)^{-1}((g\circ h)(x)+(g\circ h)(y)), $$ where, say, $f,g,h$ are automorphisms on $\mathbb{R}$. Is it enough to compare the relevant parts of both sides of equation to deduce that: $$ f\equiv g\circ h? $$ What it comes from?

EDIT: Ok, the naming is confusing. I mean $f, g, h\colon\mathbb{R}\to\mathbb{R}$ are bijections.

user3725657
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  • If $ f $, $ g $ and $ h $ are automorphisms, so are $ f ^ { - 1 } $ and $ ( g \circ h ) ^ { - 1 } $. Thus $ f ^ { - 1 } ( a + b ) = f ^ { - 1 } ( a ) + f ^ { - 1 } ( b ) $, and same for $ g \circ h $. This shows that the equation must hold for any three automorphisms, and you have no further restrictions on them (just simplify both sides of the equation to see that). So you can't conclude $ f = ( g \circ h ) ^ { - 1 } $. – Mohsen Shahriari Oct 17 '18 at 06:58
  • Also, assuming you're talking about field automorphisms, there are only two distinct ones: the identity function, and additive inversion. So being an automorphism, is in fact a very strong restriction. – Mohsen Shahriari Oct 17 '18 at 07:02
  • There's a typo in the last sentece of my first comment. $ ( g \circ h ) ^ { - 1 } $ should be $ g \circ h $. – Mohsen Shahriari Oct 17 '18 at 07:06

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By manipulating this equation and making substitutions with $x$ and $y$, we may arrive at the equation $$(g\circ h\circ f^{-1})(x+y)=(g\circ h\circ f^{-1})(x)+(g\circ h\circ f^{-1})(y)$$ If you add the restriction that $g\circ h\circ f^{-1}$ is continuous (at at least one point), this reduces to Cauchy's functional equation, whose solution allows us to conclude that $$(g\circ h\circ f^{-1})(x)=ax$$ for some nonzero constant $a$, or that $$(g\circ h)(x)=af(x)$$ Which is slightly more general than the conclusion that you proposed.

Franklin Pezzuti Dyer
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