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This is from Bredon's Topology and Geometry, page 71. This comes right after the very definition of differentiable manifold, so I think no use of tangent space or 'differential' is permitted. (Bredon gives two defenitions, one the usual chart and atlas definition, the other using the functional structure. He then explains that the two are equivalent.)

  1. Let $X$ be the graph of the real valued function $\theta(x) = |x|$ of a real variable $x$. Define a functional structure on $X$ by taking $f \in F(U) \iff f$ is the restriction to $U$ of a $C^\infty$ function on some open set $V$ in the plane with $U = V \cap X$. Show that $X$ with this structure is not diffeomorphic to the real line with usual $C^\infty$ structure.

I thinks that if there were any diffeomorphism, something bad happens at $(0,0)$, but I just can't figure out... Please enlighten me.

Dilemian
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  • I'm not at all used to the definiton of manifolds as functionally structured spaces, but I think that it might work to assume that an isomorphism $\varphi$ of structures exist and then to show that $\pi_2\mid_U \circ \varphi^{-1}$ is not contained in $C^\infty(\varphi(U))$, where $\pi_2$ is the projection onto the second coordinate in $\mathbb{R}^2$ and $U = V \cap X$ is a neighborhood around the origin in $X$ (the idea with $\pi_2$ comes from the similar question found here). – user331406 Jul 17 '16 at 13:09

2 Answers2

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Note: This is an extension of my comment above. As I said there, I'm not very used to the definiton of manifolds as functionally structured spaces, but I think that the following works. The idea comes from the similar question found here. Any comments on possible mistakes are welcome.

Answer: According to Bredon's definition 2.4, a diffeomorphism is defined as an isomorphism of structures, hence it suffices to show (according to definition 2.3) that there are open sets $U \subseteq X$ and $V \subseteq \mathbb{R}$ such that no isomorphism between $(U, F_X(U))$ and $(V, C^\infty(V))$ exists.

Assume there exists such an isomorphism $\varphi: X \rightarrow \mathbb{R}$, and let $U \subseteq X$ and $V \subseteq \mathbb{R}$ be two neighborhoods around the origin (for example $U = X$ and $V = \mathbb{R}$). Then the induced maps on the structures are given as

\begin{equation} \phi: C^\infty(V) \rightarrow F_X(\varphi^{-1}(V)),\quad f \mapsto f \circ \varphi \end{equation}

and

\begin{equation} \tilde \phi: F_X(U) \rightarrow C^\infty(\varphi(U)),\quad f \mapsto f \circ \varphi^{-1} \end{equation}

Now consider the map $\pi_2(x,y) = y$, i.e. the projection onto the second coordinate in $\mathbb{R}^2$. This is a smooth map around the origin in the plane, hence $f := \pi_2\mid_U \in F_X(U)$. Now, since $\varphi$ is an isomorphism, we need to have a counterpart of $f$ in $C^\infty(\varphi(U))$. But the map

\begin{equation} \tilde\phi(f) = \pi_2\mid_U \circ \varphi^{-1} \end{equation}

is not smooth around the origin, hence $\tilde\phi(f) \notin C^\infty(\varphi(U))$, a contradiciton. So there exists no isomorphism of structures between $(X, F_X)$ and $(\mathbb{R}, C^\infty)$, what means that they are not diffeomorphic to each other.

Community
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user331406
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  • I think you can just take $Z = \mathbb{R}^2$ and $V = \mathbb{R}$, introducing these neighborhoods seems rather arbitrary (and not useful). Other than that I think the answer is OK. – Najib Idrissi Jul 19 '16 at 10:34
  • @NajibIdrissi: Yes you are right, thank you for your comment. I just took these arbitrary neighborhoods because I thought that it doesn't matter what neighborhoods around the origin we choose, but providing a specific example sounds better. I will make an edit. – user331406 Jul 19 '16 at 17:19
  • This is a great answer. Thank you. – Dilemian Jul 23 '16 at 19:21
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This is not a complete answer, more like a hint. I think this can be shown as follows. Given a point $x\in X$ let $J_x\subset C^\infty (X,\mathbb R)$ be the ideal of functions vanishing at $x$. Then consider the quotient $J_x/(J_x)^2$, where $(J_x)^2$ is generated by all products of two functions which both vanish at $x$. If $\phi:X\to\mathbb R$ would be a diffemorphism, then $\phi^*$ would induce linear isomorphism $J_{\phi(x)}/(J_{\phi(x)})^2\to J_x/(J_x)^2$ for each $x\in X$. But on $\mathbb R$, each of this quitents is one-dimensional (an isomorphism being given by mapping the class of $f$ to the derivative in the point in question). In contrast, for $0\in X$, the quotient space should be two-dimensional, corresponding to derivatives coming from the left and from the right. (Of course, this secretely uses cotangent spaces, but I think it should be elemenatry enough ...)

Andreas Cap
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