Let $G$ be a topological group acting on the topological space $X$ and let $\phi\colon X\xrightarrow{\simeq} Y$ be a homeomorphism between the spaces $X$ and $Y$. Can I induce a $G$-action on $Y$ by $\phi$ via $gy:=\phi(gx)$, where $x\in X$ is the unique element satisfying $\phi(x) = y$?
Quick verfications show
- $g(hy) = g\phi(hx) = \phi((gh)x) = (gh)y$, and
- $ey = \phi(ex)= \phi(x)= y$,
for all $g,h\in G$ and $y\in Y$, where $e\in G$ is the neutral element. To check the continuity let $\rho\colon G\times X\to X$ denote the $G$-action on $X$ and $\tilde\rho\colon G\times Y\to Y$ the induced action. Then $\tilde\rho(g,y) = \phi(\rho(g,\phi^{-1}(y)))$ for $(g,y)\in G\times Y$, i.e. $\tilde\rho = \phi\circ\rho\circ(\mathrm{Id}_G\times\phi^{-1})$ Now let $V\subseteq Y$ be open and compute $\tilde\rho^{-1}(V) = (\mathrm{Id}_G\times\phi)\circ\rho^{-1}\circ\phi^{-1}(V)$, which is indeed open in $G\times Y$ since $\phi$ is a homeomorphism.
This induced $G$-action on $Y$ would simultaneously make $\phi$ into a $G$-equivariant homeomorphism. Is this argument correct? Thank you in advance!