The Lie bracket of two horizontal vector fields is vertical? $P(M, G, \pi)$ is a principal bundle, $u\in P$, vectors $X, Y\in H_u P$ where $H_u P$ is the horizontal subspace at $u$.
A theorem that says $[X, Y] \in V_u P$ where $V_u P$ is the vertical subspace. [Reference: Nakahara p. 387, and p. 388 (10.34) for in tuition, where the book is proving the Cartan's structure equation].
The argument given in p. 388 suggests that the integral curves of the projections of two horizontal vectors $V=\pi_* X$ and $W=\pi_* Y$ must form a coordinate basis as $[V, W]=0$. How could this be true? We could have started with non-coordinate basis vectors in $T_p M$ and lifted them into $P$. Furthermore, why could $\pi_*$ and the commutator swap order in (10.34)?

Consider $\pi: P \to M$, your principal bundle projection, which is a submersion. Also, you have that
$$
D_u \pi: H_uP \to T_{x}M
$$
is a vector space isomorphism. As a result, you can conclude that $X \sim \epsilon V$ and $Y \sim \delta W$ are $\pi$-related (often called $f$ related in literature). This is a step out of OPs textbook, it is not obvious. This implies that $\pi$ can be "swapped" with the commutator, in this case restricted to horizontal vector fields, and gives us
$$
\pi_*([X,Y])=[\pi_*X,\pi_*Y]=[\epsilon V, \delta W]=\epsilon \delta [V,W] \to 0.
$$
We can do this since the leftmost term does not depend on $\epsilon$ or $\delta$. Hence the commutator has no horizontal part and is therefore entirely vertical.
