# Question about subgroups on the group of homeomorphisms of a topological space

Let Y be path-connected, locally path-connected, and simply connected. Let $$G_1$$ and $$G_2$$ be subgroups of Homeo(Y) defining covering space actions of Y. (this means that each point y has a neighborhood U such that for $$g_1, g_2 \in G_1, g_1(U) \cap g_2(U) \neq \emptyset\ \text {implies}\ g_1 = g_2$$) Show that the orbit spaces $$Y/G_1$$ and $$Y/G_2$$ are homeomorphic iff $$G_1$$ and $$G_2$$ are conjugate subgroups of Homeo(Y).

Left to right is easy since we have that $$G_1$$ is isomorphic to $$\pi _1(Y/G_1)$$ because Y is simply connected. Same with $$G_2$$. (proposition 1.40 in Allen Hatcher's algebraic topology) It's right to left that I'm having trouble with. I was thinking of just building a map from $$Y/G_1$$ to $$Y/G_2$$ by directly sending one orbit to another, but I wasn't sure how to show it was continuous. Is there a way to use the quotient topology to do this? Or should I be looking at this problem another way entirely?

Say $$G_2 = h G_1 h^{-1}$$. Then the map $$f: Y \to Y/G_2$$ that is given by the action of $$h$$ followed by the quotient map, i.e., $$y \mapsto \overline{h \cdot y}$$
• is continuous, since $$h$$ acts via a homeomorphism and the quotient map is continuous, and
• collapses orbits of $$G_1$$: if $$y' = g_1 \cdot y$$ for some $$g_1 \in G_1$$, then $$f(y') = \overline{hg_1 \cdot y} = \overline{(h g_1^{-1} h^{-1}) \cdot hg_1 \cdot y} = \overline{h \cdot y} = f(y).$$
Therefore, by the universal property of quotient maps, there exists a continuous map $$\bar{f}: Y/G_1 \to Y/G_2$$ extending $$f$$.
By exchanging the roles of $$G_1$$ and $$G_2$$, you can construct another continuous map $$Y/G_2 \to Y/G_1$$ which you can check is inverse to $$\bar{f}$$. Hence $$Y/G_1 \cong Y/G_2$$.