Compatibility conditions for Maurer-Cartan forms on a homogeneous space I am reading Maurer-Cartan forms on a homogeneous space 
and am  unable to show that $\theta_V=Ad(h_{UV}^{-1})\theta_U+(h_{UV})^*\omega_H$.
Notation : We are considering $G \to G/H$ as a homogeneous H-space, where $G$ is a Lie group, and $H$ is a closed subgroup. $\omega $ is the Maurer-cartan form. For intersecting open sets $U$ and $V$, we have sections $s_U : U \to G$ and $s_V : V \to G/H$. Further $\theta_U=s_U^*\omega$ and $\theta_V=s_V^*\omega$; on the overlap $U \cap V$, we have $h_{UV}=s_V \circ s_U^{-1}$.
Here is my attempt :  Let $X\in T_x (G/H)$. We need to show that 
$\theta_V(X)=Ad(h_{UV}^{-1})\theta_U(X)+(h_{UV})^*\omega_H(X)$.
Let $c(t)$ be a curve starting at $x \in G/H$ i.e. $c(0)=x$ with tangent vector $c'(0)=X$. Then 
$\theta_V(X)= (s_V^*(\omega))(X)=\omega((s_V)_*(X))=\omega((h_{UV}\circ s_U)_*(X))=\omega(\frac{d}{dt}|_{t=0}(h_{UV}(c(t))\cdot s_U(c(t)))$. 
But am unable to further simplify this to obtain R.H.S. Kindly help. Thanks a lot !
 A: I think there is a mixed up and $h_{UV}(x)=(s_{U}(x))^{-1} \cdot s_V(x) \in H$. Continuing where you left: 
$$\theta_V(X)= (s_V^*(\omega))(X)=\omega((s_V)_*(X))=\omega(( s_U\cdot h_{UV})_*(X))=\omega(\frac{d}{dt}|_{t=0}(s_U(c(t))\cdot h_{UV}(c(t)))$$
Now we use the chain rule to write
$$\omega(\frac{d}{dt}|_{t=0}(s_U(c(t))\cdot  h_{UV}(c(t)))=\omega\left(\left(dL_{S_{U}}\right)_{h_{UV}(x)}((h_U)_*(X))+\left(dR_{h_U}\right)_{s_{U}(x)}((s_{U})_*(X))\right)$$
Now we will use the defention of the maurer cartan form. It is given as the pushforward of a vector in $T_gG$ along the left-translation:
$$\omega(v) = (L_{g^{-1}})_* v,\quad v\in T_gG.$$
In your case we get
$$\left(L_{(s_{U}(x)h_{UV}(x))^{-1}}\right)_*\left(\left(dL_{s_{U}}\right)_{h_{UV}(x)}((h_{UV})_*(X))+\left(dR_{h_{UV}}\right)_{s_{U}(x)}((s_{U})_*(X))\right)$$
Useing the properties of left translation we can write the above as
$$(L_{(h_{UV}(x)^{-1}})_*(L_{(s_{U}(x)^{-1}})_*\left(\left(dL_{s_{U}}\right)_{h_{UV}(x)}((h_{UV})_*(X))+\left(dR_{h_{UV}}\right)_{s_{U}(x)}((s_{U})_*(X))\right).$$
Simplfing further and rembering that $(h_{UV}(x))^{-1}$ is an elment of the closed lie  subgroup $H$, 
$$(L_{(h_{UV}(x)^{-1}})_*\left(((h_{UV})_*(X))+(L_{(s_{U}(x))^{-1}})_*\left(dR_{h_{UV}}\right)_{s_{U}(x)}((s_{U})_*(X))\right)=\omega_{H}\left((h_{UV})_*(X)\right)+(L_{(h_{UV}(x)^{-1}})_*(L_{(s_{U}(x))^{-1}})_*\left(dR_{h_{UV}}\right)_{s_{U}(x)}((s_{U})_*(X))$$
The right action commute with the left action so we almost get the desired result 
$$h_{UV}^*(\omega_H)\left(X\right)+(L_{(h_{UV}(x))^{-1}})_*\left(R_{h_{UV}}\right)_*\omega((s_{U})_*(X))$$
Finally by the defention of the adjoing operator we get:
$$\theta_V(X)=Ad(h_{UV}^{-1})\theta_U(X)+(h_{UV})^*\omega_H(X)$$
