neighborhood $U_i$ of $x_i$, $i=1,\cdots,d$, such that $U_i\cap U_j=\emptyset$ for $i\neq j$, and $\varphi_i=\varphi|_{U_i}:U_i\rightarrow\varphi(U_i)$ is a diffeomorphism. Let $V=\bigcap_{i=1}^d\varphi(U_i)$, and let $V_i=\varphi_i^{-1}(V)$, then $\varphi|_{V_i}$ is also a diffeomorphism onto $V$. We conclude that $\varphi^{-1}(V)=\bigcup_{i=1}^d V_i$. In fact, $\forall
z\in\varphi^{-1}(V)$, let $z_i=\varphi_i^{-1}(\varphi(z))$, then $z_i\in\varphi^{-1}(\varphi(z))$ and $z_i\neq z_j$ for $i\neq j$. But $z\in\varphi^{-1}(\varphi(z))$ and $|\varphi^{-1}(\varphi(z))|=d$, this force $z=z_{i_0}$ for some $i_0$. Hence $z\in\bigcup_{i=1}^d V_i$. Therefore $\varphi^{-1}(V)=\bigcup_{i=1}^d V_i$. The Lemma is proved.

0.3cm

\[C:covering\] Suppose conditions (a), (b) and (c) hold. If furthermore $\exists
d\in \mathbb{N}$ such that $\forall y\in Y'$,

0.3cm (1) the isotropic subgroup 0.1cm $G_y=K$, 0.6cm (2) $|O_y\cap Y'|=d$,

0.3cm then $\varphi: G/K\times Y'\rightarrow X'$ is a $d$-sheeted covering map.

0.3cm

By Proposition \[P:localdiffeo\], $\varphi$ is a local diffeomorphism. So by the above Proposition, we need only to show that for each $x\in X'$, $\varphi^{-1}(x)$ has $d$ points.

0.3cm For $x\in Y'$, suppose $O_x\cap
Y'=\{y_1,\cdots,y_d\}$. Then there exists $g_i\in G$ such that $\sigma_{g_i}(y_i)=x$ for each $i\in\{1,\cdots,d\}$. Then $([g_i],y_i)\in\varphi^{-1}(x)$. On the other hand, if $([g],y)\in\varphi^{-1}(x)$, then $y=y_{i_0}$ for some $i_0\in\{1,\cdots,d\}$. Now we have $\sigma_{gg_{i_0}^{-1}}(x)=\sigma_{g}(y_{i_0})=x$, that is $gg_{i_0}^{-1}\in G_x=K$, so $[g]=[g_{i_0}]$ and $([g],y)=([g_{i_0}],y_{i_0})$. Thus $\varphi^{-1}(x)=\{([g_1],y_1),\cdots,([g_d],y_d)\}$.

0.3cm In general for $x\in X'$, suppose $\sigma_h(x)\in Y'$ for some $h\in G$, then the relation $\varphi^{-1}(\sigma_{h}(x))=l_h(\varphi^{-1}(x))$ 