Normal fundamental domain

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I am studying quotient manifolds and there is something in the notes I took in class that I don't understand.

First I will write some definitions, because I know they may be different if you follow a different book.

Let $G$ be a group and $M$ a manifold. We write $G\setminus M$ for the quotient manifold.

A fundamental domain $\bar F$ is the closure of the set $F$ that contains a point of every equivalence class.

I understand this.

Let $A\subset M$ and $\bar F$ fundamental domain, $K(A)=\{ g\in G \colon g(A)\cap \bar F \neq \emptyset \}$

My first question: What is $K$? Does it have a name?

A fundamental domain is said to be normal if $\forall m \in M$ there exists a neighbourhood $U$ of $m$ such that $K(U)$ is a finite set.

Then I have this note:

In a normal fundamental domain, every point $m\in M$ has a neighbourhood such that $K(m)=K(U)$.

My second question: How could I prove this? I don't find this to be trivial.

Thank you and sorry for my poor English.

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My teacher told me the answer, so I will answer my own question.

We have that $K(m)=\{g\in G: g(m)\in \bar F\}$ and $K(V)=g\in G: g(V)\cap \bar F \neq \emptyset\}$.

If $m\in V$ it is trivial that $K(m)\subseteq K(V)$.

$\bar F$ is a normal fundamental domain, so $K(V)$ is a finite set. Therefore $K(V)\setminus K(m)=\{g_1, \dots , g_p \}$, where $g_\alpha \notin K(m)$ ($\alpha=1,\dots, p$), so $g_\alpha (m) \notin \bar F$.

$\bar F $ is closed, so $g_\alpha (m) \notin \bar F$ has a neighbourhood $V_\alpha$ such that $V_\alpha \cap \bar F= \emptyset$.

So we consider $U_\alpha= g_\alpha ^{-1} (V_\alpha)$ and $$U=\bigcap _{\alpha=1} ^p U_\alpha$$We have that $g_\alpha (U) \subset g_\alpha (U_\alpha)$ and $g_\alpha (U_\alpha) \cap \bar F =\emptyset$, so $g_\alpha \notin K(U)$.

Therefore, $K(U)=K(m)$.