is an $n$-dimensional smooth manifold, $G$ is a Lie group. Then we have the reduced map $\varphi: G/K\times Y\rightarrow X$. Suppose $dx, dy$, and $d\mu$ are smooth measures on $X,Y$, and $G/K$, respectively, with $dx$ and $d\mu$ to be $G$-invariant. Our goal is to convert the integration over $X$ to the integration over $Y$. Suppose the conditions (a), (b), and (c) hold. We hope the map $\varphi: G/K\times Y'\rightarrow X'$ satisfies the conditions as given in Proposition \[P:jifenbianhuan\].

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\[P:localdiffeo\] Suppose conditions (a), (b) and (c) hold. Then $\varphi: G/K\times
Y'\rightarrow X'$ is a local diffeomorphism.

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0.2cm Let $e$ be the unit element in $G$. For $([e],y)\in G/K\times Y'$, $d\varphi_{([e],y)}(0,v)=v, \forall v\in T_y Y'$, so $T_y
Y'\subset Im(d\varphi_{([e],y)})$. Furthermore, $\varphi|_{G/K\times\{y\}}:G/K\times\{y\}\rightarrow O_y\cong
G/G_y$ is a local diffeomorphism, so $T_y O_y\subset
Im(d\varphi_{([e],y)})$. Thus $d\varphi_{([e],y)}$ is surjective. But $\mathrm{dim}(G/K\times Y')=\mathrm{dim}X'$, so $d\varphi_{([e],y)}$ is in fact an isomorphism. For general $([g],y)\in G/K\times Y'$, notice that $\varphi\circ
l_g=\sigma_g\circ\varphi$, where $l_g([h],y)=([gh],y)$, so $d\varphi_{([g],y)}\circ
(dl_g)_{([e],y)}=(d\sigma_g)_{([e],y)}\circ d\varphi_{([e],y)}$, and $d\varphi_{([e],y)}$ is isomorphic implies $d\varphi_{([g],y)}$ is isomorphic. Thus $\varphi$ is everywhere regular, and hence is a local diffeomorphism.

To make Proposition \[P:jifenbianhuan\] available, we endow the following covering condition.

0.3cm (*covering condition*) The map $\varphi: G/K\times Y'\rightarrow X'$ is 