The motivation for this question comes from the study of lifts of an orbifold chart, which is simplified as the following:


Suppose that $ U $ is an open connected subset of $ \mathbb{R}^n $ and $ G $ is a finite group of diffeomorphisms of $ U $, which naturally acts on $ U $. Let $ \pi:U\rightarrow U/G $ denote the quotient map.


Assume we are given two smooth maps $ f,g:V\rightarrow U $ and a point $ x\in V $ such that $ \pi\circ f=\pi\circ g $ and $ f(x)=g(x) $. Obviously, $ f $ and $ g$ play the role of lifts of the map $ \pi\circ f=\pi\circ g $ along $ \pi $. There are examples that show that $ f $ and $ g $ cannot coincide, even on any open neighborhood of $ x $. So the question arises what is the difference between $ f $ and $ g$?


Is it possible to find a smooth map $ h:U'\rightarrow U $ defined on an open neighborhood $U'\subseteq U$ of $ f(x)=g(x) $ such that $ f|\_{V'}=h\circ g|\_{V'} $ for some open neighborhood $V'\subseteq V$ of $ x $?


An example of a situation where there is no such an $h$ would be welcome.



The difference between two lifts along an orbifold chart
Let $\mathbb Z/2$ act on $U=\mathbb R$ by reflection through the origin $x=0$. Let $V=\mathbb R$ as well.


Consider $f,g:\mathbb R\rightarrow \mathbb R$ given by


$$
f(x)=\begin{cases} e^{-1/x^2} \qquad \text{if }\qquad & x>0 \\
0 & x=0 \\
-e^{-1/x^2} & x<0
\end{cases}
$$


and $g(x)=e^{-1/x^2}$ for $x\not=0$ and $g(0)=0$.


So $g$ "bounces" at the origin, and $f$ continuous through. In the quotient both $f$ and $g$ "bounce" at the singularity. You cannot find an $h$ as you want, as $f$ is injective, but $g$ is not.


The situation is much better if $f$ and $g$ are assumed to be diffeomorphisms. Then $h$ exists, and in fact is given by applying an element of $G$ to $U$.


