Given $f:[0,1]\to[0,1]$ and $q:[0,1]\to\mathbb{R}$, is there a $g$ such that $q(f(x))\equiv g(q(x))$? Let $f$ is a bijection and $q$ a given function, is there some function $g:\mathbb{R}\to\mathbb{R}$?  It is no need to be continuous or differentiable function.
$\matrix{ [0,1] & \to^{f} &  [0,1] \\ q\downarrow & & q\downarrow  \\ \mathbb{R} & \cdots^{g} & \mathbb{R} }$
 A: A map $g$ will exist, and be given by
$$
g(x) = q\circ f\circ q^{-1}(x)
$$
if and only if the above map is well defined. What I mean is this: there may be many elements in $q^{-1}(x)$, but as long as any choice of $s\in q^{-1}(x)$ leads to the same value of $q\circ f(s)$, then $q\circ f\circ q^{-1}$ is well defined. Another way of saying this is that $q\circ f$ must be constant on the sets $q^{-1}(x)$ for any $x\in\mathbb{R}$.
When $g$ is well defined, you can indeed show $g$ makes the diagram commute. When it is not well defined, there must exists $s,t\in q^{-1}(x)$, for some $x\in\mathbb{R}$, where $q\circ f(s)\neq q\circ f(t)$. Since, for any $g$, $g\circ q(s)=g(x)=g\circ q(t)$, the diagram cannot commute, since $g\circ q$ maps $s,t$ to the same place, while $q\circ f$ maps them to different places. 
A: In general, no. There might be $x$ and $y$ for which $q(x)=q(y)$ but $q(f(x)) \neq q(f(y))$. 
An example would be $f(x)=\sqrt{x}$ and $q(x) = \lfloor 2x \rfloor$. Now $q(0.16)=q(0.49)=0$ but $q(f(0.16))=q(0.4)=0$ and $q(f(0.49))=q(0.7)=1$.
