# Existence of smooth coordinate charts such that composition map is smooth

The following problem is problem 2.1 from John Lee's Introduction to Smooth Manifolds:

Define $f: R \to R$ by $f(x) = 1$ if $x \geq 0$ and $f(x) = 0$ if $x < 0$. Show that for every $x \in R$, there are smooth coordinate charts $(U, \varphi)$ containing $x$ and $(V, \psi)$ containing $f(x)$ such that $\psi \circ f \circ \varphi^{-1}$ is smooth as a map from $\varphi(U \cap f^{-1}(V))$ to $\psi(V)$, but $f$ is not smooth.

That $f$ is not smooth I think is pretty clear, since using the smooth atlas containing the chart $(R, Id)$, the map $f \circ Id^{-1} = f$ is clearly not smooth.

However, I have a problem with the first part of the question. I think the charts $(U, \varphi)$ and $(V, \psi)$ should be quite simple, something like open subsets $(x - 1/2, x+ 1/2)$ with linear maps, but I don't know how to make it work. Any help is appreciated.

Only $x=0$ poses a problem. Here is what you can do at $x=0$:

Take $U=\mathbb R, \phi=Id:\mathbb R\to \mathbb R$ and $V=(0,\infty), \psi=Id:(0,\infty) \to (0,\infty)$.
Then $U \cap f^{-1}(V)=[0,\infty)$ and $\varphi(U \cap f^{-1}(V))$ to $\psi(V)$ is the map $$[0,\infty) \to (0,\infty): r\mapsto 1$$ which is smooth, as required.
This solves the problem but the fact that $U \cap f^{-1}(V)$ and $\varphi(U \cap f^{-1}(V))$ are closed instead of open confirms that something fishy is going on, namely of course that $f$ is not even continuous.

A personal point of view
This problem is a quite interesting and counterintuitive caveat, confirming the excellence of Lee's book.

• I knew that the only problematic point is 0 but did not know how to "go around" it. Thank you for your answer.
– user228960
Feb 12, 2017 at 21:28
• You are welcome, dear mathstudent123. Feb 12, 2017 at 21:38