About the curvature of a Cartan connection Let $M$ be a manifold, $L$ a Lie group and $L_0 \subset L$ a closed subgroup such that $\text{dim } M = \text{dim }L/L_0 = n$. Let $P$ be a $L_0$-bundle over $M$ and let $\omega$ be a Cartan connection on P, which is a $\mathfrak{l}$-valued 1-form on $P$ such that:
(a) $\omega(X^*) = X$ for all $X \in \mathfrak{l}_0$, (here $X^*$ is the fundamental vector field defined by $X$, which is the vector field defined by $X^*_p := \frac{d}{dt}\big|_{t = 0} p \cdot\exp(tX)$).
(b) $(R_g)^* \omega = \text{Ad}(g^{-1})\omega$, for all $g \in L_0$, (where $R_g$ denotes the right action of $L_0$ on $P$).
(c) If $X \in TP$ is non-zero, then $\omega(X) \neq 0$, (i.e., $\omega$ defines an absolute parallelism on $P$, since $P$ and $L$ have the same dimension).
Suppose that we have that the Lie algebra $\mathfrak{l}$ has a decomposition of the type $\mathfrak{l} =\mathfrak{g}_{-1} + \mathfrak{g}_0 + \mathfrak{g}_{1}$ with $\mathfrak{l}_0 = \mathfrak{g}_0 + \mathfrak{g}_{1}$, and that we write the Cartan connection as $\omega = \omega_{-1} + \omega_0 + \omega_1$.
The curvature of $\omega$ is defined as the following 2-form on $P, $ $\Omega := d\omega + \frac{1}{2}[\omega,\omega]$.
I am stuck trying to understand why the curvature only depends on the term $\omega_{-1}$. More concretely, I don't know how to prove the following assertion:
There are $\mathfrak{l}$-valued functions $K_{ij}$ such that
\begin{equation}\Omega = \frac{1}{2}\sum_{i,j}K_{ij}\omega^i \wedge \omega^j \hspace{1cm} (1)
\end{equation}
Where the $\omega^k$ are the coordinates of $\omega_{-1}$ with respect to some basis of $\mathfrak{g}_{-1}$.
I am following the book of S. Kobayashi, "Transformation Groups in Differential Geometry", and it is written that this property follows from the following 3 facts:
(i) $\omega_{-1}$ restricted to each fiber vanishes.
(ii) $\omega_0 + \omega_1$ restricted to each fiber defines an absolute parallelism.
(iii) The curvature $\Omega$ restricted to each fiber vanishes.
I understand why these 3 facts are true. The first two are due to (a) and (a) and (c) respectively, and the third one follows from (i) and the fact that the form $\omega_0 + \omega_1$ satisfies the Maurer-Cartan equation $d\omega + \frac{1}{2}[\omega,\omega]=0$. Now, we should have that $\Omega$ only depends on $\omega_{-1}$ and can be written as in (1), but I don't know how to deduce it completely. Here is what I have done.
To prove the assertion take $X, Y \in \mathfrak{X}(P)$. By the bilinearity of $\Omega$, it suffices to prove it for the cases: $X$ and $Y$ both vertical, $X$ and $Y$ both horizontal, and $X$ vertical, $Y$ horizontal (the case $X$ horizontal, $Y$ vertical might be similar). For the first two we have that $\Omega(X,Y) = 0$ and $\Omega(X,Y) = d\omega_{-1}(X,Y)$ respectively, which only depend on $\omega_{-1}$ and can be written in the form of (1). However, for the last case I can only deduce that
$$\Omega(A^*,Y) = d\omega_{-1}(A^*,Y) + \frac{1}{2}[A^*, \omega_{-1}(Y)],$$
where $A \in \mathfrak{l}_0$ is an element satisfying $X = A^*$ (since $X$ is vertical).
This expression only depends on $\omega_{-1}$, and the first term can be written as in (1). However I don't know how to manage the second term. Can it be written in the form of (1) too? Am I missing something?
Any comment or hint would be highly appreciated. Thank you in advance!
Just in case, I'm adding the page of the book below.

 A: I was able to solve the problem, so I am writing this answer just to close the post.
It might be useful for someone else in the future.
I don't know if it can be proved directly from the facts (i)~(iii) as S. Kobayashi suggests in his book, but I could prove it by using a previous (non-trivial) lemma, which I found in the book Differential Geometry. Cartan’s Generalization of Klein’s Erlangen Program, by R. W. Sharpe [S]
The key is the following
Lemma [Lem. 5.3.9, S] Let $\psi: P \rightarrow L_0$ be a smooth map and define $f: P \rightarrow P$ by $f(p) := p \cdot \psi(p)$. Then, the following holds:
$$
(f^*\omega)_p = \text{Ad}(\psi(p)^{-1})\omega_p + (\psi^*\omega_{L_0})_p \quad (\forall p \in P),\\
(f^*\Omega)_p = \text{Ad}(\psi(p))\Omega_p \quad (\forall p \in P).$$
where $\omega_{L_0}$ stands for the Maurer--Cartan form of $L_0$.
This allows us to prove the following fact.
Fact [Cor. 5.3.10, S] The curvature $\Omega(X,Y)$ of a Cartan form $\omega$ on $P$ vanishes whenever $X$ or $Y$ is vertical.
Now, since $\Omega$ is a two-form and from (c), we can express $\Omega$ as in (1) since $\omega_0(Z) = \omega_1(Z)$ for any $Z$ horizontal.
Let's prove the Fact.
Take $p \in P$, $X, Y \in T_pP$ and suppose that $Y$ is vertical. Take an arbitrary $\psi: P \rightarrow L_0$ satisfying the conditions below and define $f: P \rightarrow P$ by $f(p) = p \cdot \psi(p) = R_{\psi(p)}p$.
\begin{equation*}
\begin{gathered}
\psi(p) = e \in L_0, \enspace (d\psi)_p(Y) = -\omega_p(Y) \in \mathfrak{l}_0.
\end{gathered}
\end{equation*}
Now, by the lemma
\begin{equation*}
\begin{gathered}
(f^*\omega)_p(Y) = \omega_p((df)_p(Y)) = \text{Ad}(\psi(p)^{-1})\omega_p(Y) + (\psi^*\omega_{L_0})_p(Y) \\
= \omega_p(Y) + (\omega_{L_0})_{\psi(p)}((d\psi)_p(Y)) = \omega_p(Y) - \omega_p(Y) = 0,
\end{gathered}
\end{equation*}
and by condition (c) of the definition of a Cartan form, we deduce $(df)_p(Y) = 0$.
On the other hand, $$\Omega_p(X,Y)= \text{Ad}(\psi(p))\Omega_p(X,Y) = (f^*\Omega)_p(X,Y) = \Omega_p((df)_p(X),(df)_p(Y)) = \Omega_p((df)_p(X), 0)$$
Hence, $\Omega_p(X,Y) = 0$.
