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The problen can be foun in p.71 of Topology and Geometry. I state it below for convenience.

Problem: Consider the half open real line $X=[0,\infty)$. 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 $\mathbb{R}$.) Convince yourself that it is not unreasonable to believe that these structured spaces are equal, and also try to convince yourself that this is not a triviality; i.e., try to prove it.

Now, consider the following example: let $f$ to be the restriction of the identity function to an open set $U$ of $0$ (WLOG assume $U=[0,\varepsilon)$). Clearly $f\in F_{2}(U)$ and, if my understanding of what is asked in the problem is correct; i.e., in order to show that $f\in F_{1}(U)$, we need to find a smooth function $g$ on $(-\varepsilon,\varepsilon)$ such that $x=g(x^{2})$ on $U$. But that means that $1=g'(x^{2})\cdot 2x$ and the differentiability of $g$ at the origin is in trouble.

Is the problem correctly stated?

EDIT: It is pointed in an answer below that Bredon probably means "isomorphic" when he writes "equal", and that the map $\phi:(X, F_{1})\rightarrow (X,F_{2})$ given by $x\mapsto x^2$ is a possible isomorphism.

Now, consider the interval $U=(2,5)$ and the function $f:U\rightarrow \mathbb{R}$ defined by $f(x)=1/(x-2)$. Clearly, $f\in F_{2}(U)$. Let $V=\sqrt{U}=(\sqrt{2},\sqrt{5})$. Then if $\phi$ is a morphism we have $f\circ\phi\in F_{2}(V)$, i.e. there is a $C^{\infty}$ function $g$ on $-V\cup V$ such that $f\circ\phi(x)=g(x^{2})$. That is, $$g(x^{2})=\frac{1}{x^{2}-2}.$$ Then the chain rule implies that $$g'(x^{2})=\frac{-1}{(x^{2}-2)^{2}}.$$ Therefore, $g'$ does not exist at $2$. Note that $2\in V$, and thus $g$ cannot be $C^{\infty}$ on $-V\cup V$.

Does this example show that $\textbf{$\phi$}$ cannot be a morphism? Is there a way around this?

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I think that "equal" should be read as "isomorphic". An isomorphism $(X,F_1)\to (X,F_2)$ cannot be the identity map (as your observed) but it is not unreasonable to think that $x\mapsto x^2$ will work.

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@John The second structure is defined in terms of smooth functions on an open subset of $\mathbb R$ containing $U$, the first one is defined in terms of smooth functions on $U$ only. Since $U$ is not necessarily open in $\mathbb R$, there is still something to think about. – user90090 Aug 20 '13 at 22:17
I edited the question. It may be possible that the map you define is not an isomorphism after all. – John Sep 12 '13 at 14:59

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