# Prove that $X/G$ is Hausdorff.

Let $G$ be a finite group acting on a Hausdorff topological space $X$. Prove that $X/G$ is Hausdorff. Deduce that the projective space $P^n$ is Hausdorff for all $n$.

My Try:

Consider the quotient projection $p:X\rightarrow X/G$. Let $Gx\neq Gy$. Then, $x\neq y$. There exists neighborhoods $U$ of $x$ and $V$ of $y$, such that $U\cap V=\emptyset$. $p$ is an open map. So, I was going to prove that $p(U)\cap p(V)=\emptyset$. But, in order to prove it, I need to have $p^{-1}(p(U))=U$ and $p^{-1}(p(V))=V$. But, I could not prove it. My question is, is it true? Then how may I prove it? Moreover, I am confused with $P^n$ here. What is it? Any help is appreciated.

• You need to use that the fibres of $p$ are finite. – Daniel Fischer Nov 1 '15 at 20:26
• The result is false unless the group acts freely. Look at the first counterexample. – Najib Idrissi Nov 1 '15 at 20:26
• @NajibIdrissi I have a problem with the first counterexample there. When one talks of the action of a group $G$ on a topological space $X$, isn't it understood that the maps $x \mapsto gx$ are continuous (and hence homeomorphisms) for all $g\in G$? Otherwise, the group just acts on the underlying set, doesn't it? – Daniel Fischer Nov 1 '15 at 20:38
• You can use the argument of the answer to your previous question. – Daniel Fischer Nov 1 '15 at 21:12
• $P^n$ is presumably real $n$-dimensional projective space $S^n/\{\pm 1\}$. – Rob Arthan Nov 1 '15 at 21:35

Hint: given $x \neq y$, try to find neighbourhoods $U$ of $x$ and $V$ of $y$ such that $GU \cap GV = \emptyset$.