socrates wrote: Let $A$ be a finite set of functions $f: \Bbb{R}\to \Bbb{R.}$ It is known that: If $f, g\in A$ then $f (g (x)) \in A.$ For all $f \in A$ there exists $g \in A$ such that $f (f (x) + y) = 2x + g (g (y) - x),$ for all $x, y\in \Bbb{R}.$ Let $i:\Bbb{R}\to \Bbb{R}$ be the identity function, ie, $i (x) = x$ for all $x\in \Bbb{R}.$ Prove that $i \in A.$ Wrong. Choose as counter-example $A=\emptyset$

Let's assume $A$ is not empty...

socrates wrote: Let $A$ be a finite set of functions $f: \Bbb{R}\to \Bbb{R.}$ It is known that: If $f, g\in A$ then $f (g (x)) \in A.$ For all $f \in A$ there exists $g \in A$ such that $f (f (x) + y) = 2x + g (g (y) - x),$ for all $x, y\in \Bbb{R}.$ Let $i:\Bbb{R}\to \Bbb{R}$ be the identity function, ie, $i (x) = x$ for all $x\in \Bbb{R}.$ Prove that $i \in A.$ socrates wrote: Let's assume $A$ is not empty... Let $P(x,y)$ be the assertion $f(f(x)+y)=2x+g(g(y)-x)$ Let $f\in A$ and its associated $g\in A$ $P(\frac{f(0)-x}2,-f(\frac{f(0)-x}2))$ $\implies$ $g(g(-f(\frac{f(0)-x}2))-\frac{f(0)-x}2)=x$ and so $g(x)$ is surjective. Let then $x\in\mathbb R$ and $y$ such that $g(y)=x$ : $P(x,y)$ $\implies$ $f(f(x)+y)=2x+g(0)$ and so $f(x)$ is surjective. So any element of $A$ is surjective. Since $A\ne\emptyset$, $\exists f\in A$ and so $f,f^{[2]},f^{[3]},f^{[4]}, ...,f^{[n]}\in A$ But $A$ is a finite set and so $\exists m>n$ such that $f^{[m]}\equiv f^{[n]}$ And since $f(x)$ is surjective, this means $f^{[m-n]}\equiv i$ Q.E.D.

Nice one