Reading Nakahara's "Geometry, Topology and Physics", in Chapter 10 "Connections on fiber bundles" it is said that given a principle bundle $P$ with base space a manifold $M$ and fiber a group $G$, $P(M, G)$, you can split the tangent space at $q \in P$ as
$$ T_qP = V_qP\ \oplus\ H_qP $$
Such that the vertical tangent space at $q$, $V_qP$, is defined as the tangent space in fiber directions (i.e., tangent to the fiber $G$) and therefore,
$$ V_qP = \{v \in T_qP\ /\ \pi_*v = 0 \} $$
Where $\pi: P \rightarrow M$ so $\pi_*$ is the push-forward map $\pi_*: T_qP \rightarrow T_{\pi(q)}M$
I want to prove that the elements in vertical space have to satisfy $\pi_*v = 0$
My attempt goes as follows:
$V_qP \subseteq T_qG$, so locally in a neighbourhood of $q$ you can write $P \simeq M\times G$ so $q \simeq (\pi(q), g)$, with $g \in G$
Hence locally $\pi: M\times G \rightarrow M\times 0$ and, therefore, $\pi_*: T_qM\times T_qG \rightarrow T_{\pi(q)}M\times 0$
All in all, $T_qG$ is being projected to zero which means $\pi_*v = 0\ \forall v \in V_qP $
Is this the right way?