0

If an equivalence relation is closed and the projection map into the quotient: $M \mapsto M/G$ is open , $M$ Metric, therefore Hausdorff, How do we proof that the quotient is Hausdorff? Take two points of $M/G$: $x_1$ and $x_2$ s.t $[x_1] \neq[x_2]$ then $x_1 \neq x_2$. By Hausdorffness of $M $ there are disjoint open sets $U_1$ and $U_2$ of $x_1 $ and $x_2$. By openness of $\pi, \pi(U_1)$ and $\pi(U_2)$ are disjoint nbhds of $[x_1]$ and $[x_2]$, but How do I prove that they are disjoint?

On another hand I know that the diagonal set of $M$ is closed and that if I prove that the diagonal set of $M/ G$ is closed, I can also prove Hausdorfness, but I don-t know how one proves that the diagonal is closed

Can you help me finish this?

BTW I am trying to make sense of the second point in the answer of this post If $A : G \times M \rightarrow M$ is a proper action and that $M$ is a metric space, the quotient space $M/G$ is Hausdorff

darkside
  • 605
  • 1
    There is no guarantee $\pi(U_1)$ and $\pi(U_2)$ are going to be disjoint for arbitrary choice of disjoint neighbourhoods $U_1\ni x_1$, $U_2\ni x_2$. You need to work a bit more. See this question – user10354138 Oct 24 '23 at 12:53
  • @user10354138 is the answer in the question I posted wrong then? – darkside Oct 24 '23 at 12:57
  • The answer is at least sloppy as it stands if not wrong. We need to pick $U_i$ away from $G$ (see answer in link) not just the diagonal of $M$ (although if "diagonal" in the answer by @ArhamMehta is interpreted as the diagonal of $M/G$ pulled back to $M\times M$ then it is the relation $G$). – user10354138 Oct 24 '23 at 13:18
  • @user10354138 In that problem regarding point 1 there in the answer, can't I just argue that if f is proper, thus closed, the image (which is the relation) of the whole set G×M (which is a closed set) is closed"? I don-t see the necessity of all that argument there – darkside Oct 24 '23 at 13:23
  • @user10354138 There in no need for that, that's proven here https://math.stackexchange.com/questions/4791610/if-y-is-a-metric-space-and-f-x-rightarrow-y-is-proper-f-is-closed/4791663?noredirect=1#comment10191312_4791663 – darkside Oct 24 '23 at 13:33
  • Ah yes if you assume $M$ is metric. I was thinking about the situation here so you need local compactness or compactly-generated to reduce to something nicer. – user10354138 Oct 24 '23 at 13:38
  • @user10354138 Actually I also wanted a metric space since I was analyzing that question, it's just that for my qeuestion situation I think we just need the Hausdorffness right? I updated the question just in case – darkside Oct 24 '23 at 13:56

0 Answers0