Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.