and dense submanifolds of $X$ and $Y$, respectively. So $\forall y\in Y'$, $T_y X' = T_y X$, $T_y Y' = T_y Y$.

0.3cm Let $K=\{g\in G:\sigma_g(y)=y, \forall y\in Y\}$, then $K$ is a closed subgroup of $G$. For $g\in G$, we denote $[g]=gK$ in $G/K$. The map $\sigma: G\times X\rightarrow X$ reduces to a map $\varphi: G/K\times Y\rightarrow X$ by $\varphi([g],y)=\sigma_g(y)$, and then induces a map $G/K\times
Y'\rightarrow X'$ by restriction, which we also denote by $\varphi$. By the assumption above, $\varphi:G/K\times
Y'\rightarrow X'$ is surjective. For $x\in X$, Let $G_x=\{g\in
G:\sigma_g(x)=x\}$ be the isotropic subgroup associated with $x$. Then $K\subset G_y, \forall y\in Y$. Let $dx, dy$ be smooth measures on $X$ and $Y$, respectively. We suppose $dx$ is $G$-invariant. In the following we suppose that

0.3cm (*dimension condition*) $\mathrm{dim}G_y=\mathrm{dim}K, \quad \forall
y\in Y'$.

0.3cm [This]{} means that $\forall y\in Y'$, $G_y$ and $K$ have the same Lie algebras, and the only difference between $G_y$ and $K$ is that $G_y$ may have more components than $K$. Then for some $y\in Y'$, we have $$\begin{aligned}
\mathrm{dim}X
=&\mathrm{dim}T_y X\\
=&\mathrm{dim}T_y Y+\mathrm{dim}T_y
O_y\\
=&\mathrm{dim}Y+\mathrm{dim}G-\mathrm{dim}G_y\\
=&\mathrm{dim}Y+ \mathrm{dim}G-\mathrm{dim}K .\end{aligned}$$ So $\varphi$ is a map between manifolds of the same dimension, and the pull back 