Do you mean $f:\mathbb{R}\to\mathbb{R}$ ??

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He might be saying $f:\mathbb{R}\to\mathbb{R}^{-}$

Quote: Let $g(x) = ax^2 + bx + c$ a quadratic function with real coefficients ($a \ne 0$) such that the equation $g(g(x)) = x$ has four distinct real roots. Prove that there isn't a function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = g(x)$ for all $x$ real source

Assume to the contrary that there is a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f\big(f(x)\big)=g(x)$ for all $x\in\mathbb{R}$. Let's start by observing that a fixed point of $g(x)$ is a fixed point of $g\big(g(x)\big)$ as well: $g(x_0)=x_0$ $\implies$ $g\big(g(x_0)\big)=x_0$, where $x_0$ might be a complex number depending on negativity of $\triangle=(b-1)^2-4ac$. Also from the fact that $g\big(g(x)\big)=x$ is a fourth-degree equation and has four distinct real roots, we can say it does not have any complex fixed point. So, all roots of the equation $ax^2+bx+c=x$ are real numbers and distinct, i.e., $g(x)=x$ has two distinct real roots. So, for convenience assume that $g(x)=x$ only if $x\in\{x_1,x_2\}$ and $g\big(g(x)\big)=x$ only if $x\in\{x_1,x_2,x_3,x_4\}$, where $x_1,x_2,x_3,x_4$ are four distinct real numbers. Claim. $g(x_3)=x_4$ and $g(x_4)=x_3$. Assume $g(x_3)=a$ for some $a\in\mathbb{R}$. $\implies$ $x_3=g\big(g(x_3)\big)=g(a)$ $\implies$ $g\big(g(a)\big)=g(x_3)=a$ $\implies$ $a\in\{x_1,x_2,x_3,x_4\}$. Case 1. $a=x_1$. $x_1=g(x_1)=g(a)=x_3$, contradiction. Case 2. $a=x_2$. $x_2=g(x_2)=g(a)=x_3$, contradiction. Case 3. $a=x_3$. $x_3=g(g(x_3))=g(a)=g(x_3)$ $\implies$ $x_3\in\{x_1,x_2\}$, contradiction. So the only possibility is $a=x_4$ which gives $g(x_3)=x_4$. Analogously, we get $g(x_4)=x_3$. Recall our assumption: $f\big(f(x)\big)=g(x)$ for all $x\in\mathbb{R}$. $\implies$ $f\big(g(x)\big)=f\Big(f\big(f(x)\big)\Big)=g\big(f(x)\big)$ or $f\big(g(x)\big)=g\big(f(x)\big)$ for all $x\in\mathbb{R}$. $a\in\{x_1,x_2\}$ $\implies$ $f(a)=f\big(g(a)\big)=g\big(f(a)\big)$ $\implies$ $f(a)\in\{x_1,x_2\}$. $a\in\{x_1,x_2,x_3,x_4\}$ $\implies$ $g\Big(g\big(f(a)\big)\Big)=g\Big(f\big(g(a)\big)\Big)=f\Big(g\big(g(a)\big)\Big)=f(a)$ $\implies$ $f(a)\in\{x_1,x_2,x_3,x_4\}$. Case 1. $f(x_3)=x_1$. $\implies$ $f(x_1)=f\big(f(x_3)\big)=g(x_3)=x_4$, a contradiction since $f(x_1)\in\{x_1,x_2\}$. Case 2. $f(x_3)=x_2$. $\implies$ $f(x_2)=f\big(f(x_3)\big)=g(x_3)=x_4$, a contradiction since $f(x_2)\in\{x_1,x_2\}$. Case 3. $f(x_3)=x_3$. $\implies$ $x_3=f(x_3)=f\big(f(x_3)\big)=g(x_3)=x_4$, a contradiction as $x_3\not=x_4$. Case 4. $f(x_3)=x_4$. $\implies$ $f(x_4)=f\big(f(x_3)\big)=g(x_3)=x_4$ $\implies$ $x_3=g(x_4)=f\big(f(x_4)\big)=f(x_4)=x_4$, a contradiction as $x_3\not=x_4$. So, in all cases above we reached a contradiction. Hence, there is no $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f\big(f(x)\big)=g(x)$ for all $x\in\mathbb{R}$. $\blacksquare$