Let $M$ be a topological manifold and suppose there is a covering space action of a group $G$ on $M$. This means that each $x\in M$ has a neighbrohood $U$ such that $g_1(U)\cap g_2(U)$ nonempty implies $g_1=g_2$, for any $g_1,g_2\in G$. Then is it true that the quotient space $M/G$ is Hausdorff? (I am trying to show that $M/G$ is a manifold. The hypothesis that the action of $G$ is a covering space action gives that the projection $M\to M/G$ is a covering map, so $M/G$ is locally Euclidean. But I can't see why $M/G$ should be Hausdorff.)
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3This is false and examples were discussed on several occasions on MSE. The correct condition to require is properness of the action. – Moishe Kohan Apr 25 '21 at 13:09
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2See the example given here. – Moishe Kohan Apr 25 '21 at 14:04
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@MoisheKohan I see. Thanks! – blancket Apr 25 '21 at 14:12