Let $G$ is a topological group i.e. operation of the group and inverse are continuous wrt topology.
I should show that if $G$ is Hausdorff then $G/N$ is Hausdorff where $N$ is a normal subgroup of $G$.
Is it valid with only normalness of $N$?
My attempt :
Let $x_1,x_2 \in G$
$x_1H \neq x_2H \Rightarrow x_1 \neq x_2 $ then since $G$ is Hausdorff there exists a neighborhood of $x_1$ say $V_1$ and there exists a neighborhood of $x_2 $ say $V_2$ such that $V_1\cap V_2 = \emptyset $. Also let $\pi$ is the projection which creates the topology of quotient. $\pi : G \to G/H$ Also I’ve shown that $\pi$ is open function but I don’t know how can I find disjoint nbds for $G/H$
Thanks in advance and pardon me for any mistake
No, this is false without closedness of $N$: take $G=(\Bbb R,+)$, $N=\Bbb Q$ then $G{/}N$ is indiscrete and uncountable, so certainly not even $T_0$, let alone Hausdorff. Open maps do not preserve Hausdorffness (as we see an example of here).