Problem 2-1 (Naive definition of smoothness)

Question

Define $f: \mathbb{R} ightarrow \mathbb{R}$ by

$$
f(x)= \begin{cases}1, & x \geq 0, \ 0, & x<0 .\end{cases}
$$

Show that for every $x \in \mathbb{R}$, 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 $\varphi\left(U \cap f^{-1}(V)ight)$ to $\psi(V)$, but $f$ is not smooth in the sense we have defined in this chapter.

Elu Thingol · Accepted Answer

By Proposition 2.4, $f$ cannot be smooth since it is not continuous.\newline

Obviously, the point $x=0$ is going to be the problematic one. Since $f^{-1}(V)$ is not required to be open, we can use this freedom to separate $0$ from its neighbourhood on the left. In other words, for each point $x \in \mathbb{R}$, we set two global charts equipped with the $\mathrm{Id}$ map:
\begin{equation*}
    U := \mathbb{R}, \quad 
    V:= \left\{\begin{array}{ll} \left(-\infty,\frac{1}{2}\right), &x < 0\[5pt] \left[\frac{1}{2},+\infty\right), &x\geq 0\end{array} \right.
    \implies f^{-1}(V):= \left\{\begin{array}{ll} \left(-\infty,0\right), &x < 0\[5pt] \left[0,+\infty\right), &x\geq 0\end{array}\right.
\end{equation*}

Then
\begin{equation*}
    \operatorname{Id} \circ f \circ\, \operatorname{Id}^{-1}: \operatorname{Id}(U \cap f^{-1}(V)) \to \operatorname{Id}(V) =
    \left\{\begin{array}{ll} 
    f:(-\infty,0) \to  \left(-\infty,\frac{1}{2}\right), &x < 0\[5pt] 
    f:[0,+\infty) \to  \left[\frac{1}{2},+\infty\right), &x\geq 0
    \end{array} \right.
\end{equation*}

Since the restrictions $f |_{(-\infty,0)}$ and $f|_{[0,+\infty)}$ are both constant functions, they satisfy the naive smoothness condition.

Problem 2-1 (Naive definition of smoothness)

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Problem 2-1 (Naive definition of smoothness)

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