I want two smooth functions $f_1$ and $f_2$ such that $f_1$ has critical points but $f_2\circ f_1$ does not.
At first I considered $f_1:\mathbb{R}^2 \to \mathbb{R}$ to be $f_1(x,y)=xy$, and tried to find $f_2:\mathbb{R}\to \mathbb{R}$ such that $f_2 \circ f_1(x,y)=f_2 (xy)$ would be something with constant derivative but I couldn’t get any thing.
Then I tried to assume that $f_2 \circ f_1(x,y)=x+y$ and then trying to go backward to find $f_1$ and $f_2$ but no results too.
Any help would be appreciated.
If $f_1$ and $f_2$ are both differentiable functions, it is not possible beucase of the chain rule $(f_2 \circ f_1)' (x) = (f_2' \circ f_1(x)) f_1'(x)$. So if $f_1'(x)=0$ we have $(f_2 \circ f_1)' (x)=0$.
If we allow $f_2$ to be not differentiable, here is a hint to find an example in one variable. Consider $f_1(x)=x^2 sign(x)$, which has a critical point at $x=0$. Can you find $f_2$ such that $f_2 \circ f_1 = Id$ ?