If $X$ and $Y$ are g-equivariant homeomorphic then $X/G$ and $Y/G$ are homeomorphic

Question

Let $X$ and $Y$ be $G$-sets (That is the group $G$ acts on $X$ and $Y$). We say that the function $f: X\to Y$ is G-equivariant if $f(g.x) = g.f(x)$ for all $x\in X$ and all $g\in G$. Prove that if $X$ and $Y$ are topological spaces and $f$ is a G-equivariant homeomorphism (i.e. both G-equivariant and a homeomorphism) then $X/G$ and $Y/G$ are homeomorphic.

So far, my idea are only define a map $f':X/G\to Y/G$ such that $f'([x])=[f(x)]$, for all equivalence class $[x]\in X/G$. Then it is easily to show that $f'$ is a bijective. my trouble is to show $f$ is continous and $f$ is an open mapping.

Please help, thanks a lot!

Answer 1

The key point is that $f$ sends $G$-orbits in $X$ to $G$-orbits in $Y$; that’s what $G$-equivariance does for you.

Let $\pi_X:X\to X/G$ and $\pi_Y:Y\to Y/G$ be the canonical quotient maps. For continuity, suppose that $U$ is open in $Y/G$. Let

$$\widehat{U}=\pi_Y^{-1}[U]=\bigcup_{[y]\in U}[y]=\{y\in Y:[y]\in U\}\;;$$

$\widehat{U}$ is open in $Y$ by the definition of the quotient topology on $Y/G$. And $f$ is continuous, so $f^{-1}[\widehat{U}]$ is open in $X$; let $V=f^{-1}[\widehat{U}]$. Now check that $\pi_X[V]=f'^{-1}[U]$ and conclude that $f'^{-1}[U]$ is open in $X$.

Showing that $f'$ is open is similar.

Answer 2

Continuity of $f'$ is because we equip $X/G$ with the final topology with respect to the map $q_X:X\to X/G$ which sends an $x\in X$ to its orbit $G\cdot x\in X/G$. Say $q_Y:Y\to Y/G$ is the corresponding map for $Y$, then since $q_Y\circ f$ sends an orbit to a unique element of $Y/G$, there's a unique map $f':X/G\to Y/G$ such that $f'q_X=q_Yf$, and this $f'$ is continuous. The uniqueness of the induced map can also be used to easily show that the map induced by $f^{-1}$ is the inverse of $f'$.

Answer 3

It's worth mentioning that is a very useful result to identify two different topological space.

For example, as an application, you can prove that Hopf surface topologically the same as $S^3\times S^1$ , the most easy way to prove it is using the diffeomorphism $$\Bbb{C}^{2}\setminus \{0\} \to \Bbb{R}^1 \times S^3\\ z \mapsto (\log |z| , z/|z|)$$ and you can transport the action used to define Hopf surface to $\Bbb{R}^1 \times S^3$ which is simply translation action on $\Bbb{R}^1$, therefore $(\Bbb{R}\times S^3 )/ G \cong S^1 \times S^3$. Using the lemma in your post, you then can show the hopf surface $X\cong S^1\times S^3$.

(I guess it's hard to see this identification without using this lemma)