Let \(Y\) be path-connected, locally path-connected, and simply-connected, and let \(G_1\) and \(G_2\) be subgroups of \(\mathrm{Homeo}(Y)\) defining covering space actions on \(Y\). Show that the orbit spaces \(Y/G_1\) and \(Y/G_2\) are homeomorphic if \(G_1\) and \(G_2\) are conjugate subgroups of \(\mathrm{Homeo}(Y)\).
Solution
Suppose \[ G_2=hG_1h^{-1} \] for some homeomorphism \[ h:Y\to Y. \] Let \[ q_i:Y\to Y/G_i \] be the quotient maps.
1. The map \(h\) sends \(G_1\)-orbits bijectively to \(G_2\)-orbits.
For \(y\in Y\), \[ h(G_1y)=\{hg(y):g\in G_1\} =\{(hgh^{-1})h(y):g\in G_1\} =G_2h(y). \] Thus points in the same \(G_1\)-orbit have images in the same \(G_2\)-orbit, and applying \(h^{-1}\) gives the converse.
2. Hence the formula\[ \bar h([y]_{G_1})=[h(y)]_{G_2} \]defines a bijection\[ \bar h:Y/G_1\to Y/G_2. \]
Well-definedness and injectivity follow from 1, and surjectivity follows from surjectivity of \(h\).
3. The map \(\bar h\) is a homeomorphism.
The relation \[ \bar h\circ q_1=q_2\circ h \] shows that \(\bar h\) is continuous because \(q_1\) is a quotient map. Applying the same argument to \(h^{-1}\) gives continuity of the inverse map induced on orbit spaces. Therefore \(\bar h\) is a homeomorphism.