Let $X$ be a connected, locally path connected, locally compact metric space and $G$ be a group of homeomorphisms of $X$. Let $\bar{G}$ be the closure of $G$ in Homeo$(X)$ endowed with the compact open topology. Let $\pi : X \rightarrow X/G$ and $\phi : X/G \rightarrow X/\bar{G}$ be the natural projections.

Define $\psi : X/\bar{G} \rightarrow X/G$ as follows - given $z\in X/\bar{G}$ choose a point $y \in X/G$ which satisfies $\phi (y) = z$ and set $\psi(z) =y$.

My question is, why is $\psi$ well-defined? I can't see why it is independent of the choice of $y$. Isn't it possible for two elements of $X$ to be in different $G$ - orbits but the same $\bar{G}$ - orbit?

Thank you!

  • 1
    $\begingroup$ It doesn't have to be well-defined. Consider $\Bbb Q/\Bbb Z$ acting on the circle by rotations. Then $S^1/\overline G$ is just a point, but $S^1/G$ is uncountable, and you have a choice of $\psi$ for every point in $S^1/G$. Maybe you could give more context about where this arises so we could suss out why it doesn't matter that $\psi$ isn't well defined. $\endgroup$ – user98602 Oct 7 '15 at 14:02
  • $\begingroup$ Thank you. I read it in a paper by M A Armstrong called "calculating the fundamental group of an orbit space". I guess we only need to be able to define a continuous map from $X/\bar{G} $ to $X/G$. It needn't be independent of the choice of $y$ then. $\endgroup$ – R_D Oct 7 '15 at 14:58

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