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Let $f:[a,b]\rightarrow[c,d]$ be a continuous and piecewise $C^{\infty}$ function sending $a$ to $c$ and $b$ to $d$ and let $g:[c,d]\rightarrow \mathbb{R}$ be a real continuous and piecewise $C^{\infty}$ function. Is the composition $g\circ f$ a continuous and piecewise $C^{\infty}$ function? The difficulty is to prove that the preimage via $f$ of a partition of $[c,d]$ can "generate" a partition of $[a,b]$. The matter consists to prove that the set of zeroes of a function $h\in C^{\infty}([a,b],\mathbb{R})$ is a finite set of isolated points united a finite number of closed intervals: $h^{-1}(0)=\{x_1,x_2,...,x_m \} \cup I_1 \cup...\cup I_s $ where $I_j$ is a closed interval in $[a,b]$.

Thanks for any useful hint/answer.

NOTE: A function $f:[a,b]\rightarrow\mathbb{R}$ is called piecewise $C^{l}$ if there is a partition $a=a_0 < a_1 < ...< a_n =b$ of the interval $[a,b]$ such that $f|_{[a_k,a_{k+1}]}\in C^l({[a_k,a_{k+1}]},\mathbb{R})$ for every $0\le k \le n-1$.

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  • $\begingroup$ How exactly you define "piecewise" here? $\endgroup$
    – freakish
    Oct 30, 2023 at 11:22
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    $\begingroup$ Consider a function $f(x) := e^{-1/x} \sin 1/x$ on $[0,1]$ - of course, we define $f(0):=0$. $\endgroup$
    – Roman Hric
    Oct 30, 2023 at 18:27
  • $\begingroup$ I would consider $f(x)=x^{1/3} \cos(1/x)$ on $[-1,1]$ and some $g$ which is not differentiable around the origin. $\endgroup$ Oct 30, 2023 at 18:27
  • $\begingroup$ @RomanHric You've beaten me by 28 seconds ^^ $\endgroup$ Oct 30, 2023 at 18:30
  • $\begingroup$ The function $$g:[-\cos(1),\cos(1)]\rightarrow\mathbb{R},g(x)=\vert x\vert$$ should work. $\endgroup$ Oct 30, 2023 at 18:38

2 Answers 2

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The property you want of the set of zeros of a smooth function does not hold in general: If you define $f\colon \mathbb R\to \mathbb R$ $$ f(x) = \left\{\begin{array}{cc} \sin(1/x)\exp(-\frac{1}{x}) & \text{ if } x>0\\ 0, & \text{ if } x\leq 0, \end{array} \right. $$ then using the fact that if $p(x)$ is a polynomial function of $x$ then $$ \lim_{x\to 0} p(1/x)\exp(-1/x) =0, $$ you can show that $f(x)$ is a smooth function on all of $\mathbb R$. Clearly $f^{-1}(0)$ contains no intervals of positive length, but $0$ is it is standard that $f$ is infinitely differentiable on all of $\mathbb R$.

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Conclusions
In order to make this topic more readable, I think it's useful to add some comments.
@RomanHric and @krm2233 have provided us two counterexamples that disprove the proposition. So, the answer to the initial question is: no, it isn't. The composition of continuous and piecewise $C^{\infty}$ functions, in general, is not a piecewise $C^{\infty}$ function.
To better explain the reason why, I provide a third counterexample: the function defined as $h(x)= \left\{ \begin{array}{rcl} e^{-\frac{1}{x^2}}\sin(\frac{\pi}{x}) & \mbox{for} & x \in [-1,1]-\{0\} \\ 0 & \mbox{for} & x = 0 \end{array}\right.$ ; of course $h\in C^{\infty}([-1,1])$.
The set of its zeroes is $h^{-1}(0)=\{0 \}\cup\{\frac{1}{n} $for $n\in\mathbb{Z}, n\neq0\}$; this set has an infinite number of isolated points, and this contradicts what claimed above, about the function $h$.

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