The relevant definitions are listed below. They can be found in Chapter VI, pages 297-298 of Bredon's Introduction to Compact Transformation Groups; and Section 2, Chapter II of Bredon's Topology and Geometry.
Relevant definitions:
On a topological space $X$ a functional structure $F_{X}$ is an assingment defined on the collection of open sets $U\subset X$ taking $U\mapsto F_{X}(U)$ such that:
$F_{X}(U)$ is a subalgebra of the algebra of continuous real valued functions on $U$ and contains all constant functions.
If $V$ is open, $V\subset U$ and $f\in F_{X}(U)$ then $f|_{V}\in F_{X}(V)$.
If $\{U_{i}\}$ is a collection of open sets and $f|_{U_{i}}\in F_{X}(U_{i})$ for all $i$ then $f\in F_{X}(\bigcup_{i}U_{i})$.
The pair $(X,F_{X})$ is called a functionally structured space. E.g. $(\mathbb{R}^{n},C^{\infty})$, where $\mathbb{R^{n}}$ has the usual topology and $C^{\infty}(U)=\text{all $C^{\infty}$ functions on $U$.}$
If $(X,F_{X})$ is a functionally structured space and $U\subset X$ is open then for $V\subset U$ open we define $F_{U}(V)=F_{X}(V)$ so that $(U,F_{U})$ is a functionally structured space.
A morphism of functionally structured spaces $(X, F_{X})$ and $(Y,F_{Y})$ is a continuous map $\varphi:X\rightarrow Y$ such that for any open set $V\subset Y$ and $f\in F_{Y}(V)$ we have $f\circ \varphi\in F_{X}(\varphi^{-1}(V))$. A morphism $\varphi$ is an isomorphism if $\varphi^{-1}$ exists as a morphism. We then say that $(X,F_{X})\simeq (Y,F_{Y})$.
Problem set-up: (This is Problem 5 page 71 in Bredon's Topology and Geometry)
Consider the functionally structured spaces $(X,F_{1})$ and $(X,F_{2})$ defined as follows: Let $X$ be the half open real line $[0,\infty)$. Define a functional structure $F_{1}$ by taking $f\in F_{1}(U) \Leftrightarrow f(x)=g(x^{2})$ for some $C^{\infty}$ function $g$ on $\{x| x\in U\text{ or} -\negmedspace x\in U\}$. Define another functional structure $F_{2}$ by taking $f\in F_{2}(U)\Leftrightarrow f$ is the restriction to $U$ of some $C^{\infty}$ function on an open subset of $\mathbb{R}$. (Note that $U$ is open in $[0,\infty)$ but not necessarily in $\mathbb{R}$.)
Question: As functionally structured spaces, are $\textbf{$(x,F_{1})$}$ and $\textbf{$(x,F_{2})$}$ isomorphic?
Remark 1: Consider the way $F_{1}$ is defined: "$f\in F_{1}(U) \Leftrightarrow f(x)=g(x^{2})$ for some $C^{\infty}$ function $g$ on $\{x| x\in U\text{ or} -\negmedspace x\in U\}$". I understand it is implied that $g$ is defined not only on $\{x| x\in U\text{ or} -\negmedspace x\in U\}$, but also on $\{x^{2}: x\in U\}$ (and perhaps somewhere else), BUT it is only required to be $C^{\infty}$ on $\{x| x\in U\text{ or} -\negmedspace x\in U\}$.
Remark 2: It seems that Bredon wants to define the functional structure $F_{1}$ such that the map $\phi:(X,F_{1})\rightarrow (X,F_{2})$ given by $x\mapsto x^2$ is an isomorphism of functionally structured spaces.
Remark 3: When trying to show that $\phi$ above is a morphism I run into trouble. Consider the following example: Let $U=(2,5)$ and let $f:U\rightarrow \mathbb{R}$ be defined by $f(x)=1/(x-2)$. Clearly $f\in F_{2}(U)$, and if $\phi$ is a morphism then we must have $f\circ\phi\in F_{1}(\phi^{-1}(U)$, where $V:=\phi^{-1}(U)=\sqrt{U}=(\sqrt{2},\sqrt{5})$. That is, there is a $C^{\infty}$ function $g$ on $-V\cup V$ such that for $x\in V$ we have $f\circ\phi(x)=f(x^{2})=g(x^{2})$. In particular $g$ is $C^{\infty}$ at $2\in V$. However, the chain rule yields $$g'(x^{2})=\frac{-1}{(x^{2}-2)^{2}},$$ which prevents the smoothness of $g$ at $2\in V$. Thus either I am missing something or $\phi$ is not a morphism.
