Let $\pi:P\longrightarrow M$ be a $G$-principal bundle. For $p\in P$, $V_p$ denotes the space of tangent vertical vectors, that is, $V_p:=T_pP_{\pi(p)}$. The space of vertical vector fields will be denoted by $\mathfrak{X}^v(P)$ whereas $\mathfrak{g}$ stands for the Lie algebra of $G$ (i.e., the space of left invariant vector fields on $G$).
I'm trying to show::
Theorem.
$(i)$ There is an isomorphism of vector space $T_eG\simeq V_p$ for every $p\in P$;
$(ii)$ There is an isomorphism of Lie algebras $\mathfrak{g}\simeq \mathfrak{X}^v(P)$.
Attempt:
$(i)$ Define the map: \begin{align*} \displaystyle \Phi:V_p\longrightarrow T_eG, \end{align*} choosing a local trivialization $(U, \phi)$ with $\pi(p)\in U$ and setting: \begin{align*} \displaystyle \Phi(v):=d(L_{\phi_{\pi(p)}(p)}^{-1})_{\phi_{\pi(p)}(p)}(d(\phi_{\pi(p)})_p(v)). \end{align*} Above $\phi_{\pi(p)}:P_{\pi(p)}\longrightarrow G$ is the diffeomorphism induced by $\phi$, that is, $$\phi_{x}:=\textrm{pr}_2\circ \phi|_{P_x},\ \forall x\in M,$$ where $P_x:=\pi^{-1}(x)$. The inverse of $\Phi$ is the map: \begin{align*} \displaystyle \Psi:T_eG\longrightarrow V_p, \end{align*} defined by: \begin{align*} \displaystyle \Psi(v):=d(\phi_{\pi(p)}^{-1})_{\phi_{\pi(p)}(p)}(d(L_{\phi_{\pi(p)}(p)})_e(v)). \end{align*} Obs. 1 It is not difficult to show $\Phi$ and $\Psi$ don't depend on the choice of local trivializations.
$(ii)$ As to $(ii)$ I'd like to use $(i)$ to define the isomorphism. Define: $$\begin{align*} \displaystyle \Psi:\mathfrak{g}\longrightarrow \mathfrak{X}^v(P), X\longmapsto \tilde{X} \end{align*}$$ choosing a local trivialization $(U, \phi)$ with $\pi(p)\in U$ and setting: $$\begin{align*} \displaystyle \tilde{X}_p:=d(\phi_{\pi(p)}^{-1})_{\phi_{\pi(p)}(p)}(X_{\phi_{\pi(p)}(p)}). \end{align*}$$ Notice this map came from the map $\Psi$ of $(i)$ as follows: applying $X_e$ in the previous $\Psi$ and using the fact $X$ is left invariant we find this new $\Psi$.
Can anyone help me defining the inverse of $\Psi$? I thought I should use the composition $$\mathfrak{X}^v(P)\longrightarrow V_p\stackrel{\simeq}{\longrightarrow} T_eG\stackrel{\simeq}{\longrightarrow} \mathfrak{g},$$ the problem is how to choose the first map $\mathfrak{X}^v(P)\longrightarrow V_p$ for I don't have a distinguished point on $P$.
Thanks.
Remark. A vertical vector field on $P$ is a vector field $X\in\mathfrak{X}(P)$ such that $X_p\in V_p:=T_pP_{\pi(p)}$ for every $p\in P$.
I'd like to share my conclusions. The isomorphism $$T_eG\simeq V_p\quad (\textrm{or}\quad \mathfrak{g}\simeq V_p)$$ is correct and can also be done using the exponential map. As to my second statement $\mathfrak{g}\simeq \mathfrak{X}^v(P)$ is not right. As @Andreas Cap has observed, what really holds is $$\mathfrak{X}^v(P)\simeq C^\infty(P; \mathfrak{g}).$$ Intuitively the argument is as follows:
We have an isomorphism $\mathfrak{g}\simeq V_p$ for every $p\in P$ and since $$P\times \mathfrak{g}=\bigcup_{p\in P} \{p\}\times \mathfrak{g}\quad \textrm{and}\quad V(P)=\bigcup_{p\in P} \{p\}\times V_p,$$ one gets an isomorphism of vector bundles $$P\times \mathfrak{g}\simeq V(P),$$ where $V(P)=\{(p, v): p\in P, v\in V_p\}$ is the vertical bundle. From this it follows: $$\mathfrak{X}^v(P)=\Gamma(V(P))\simeq \Gamma(P\times \mathfrak{g})\simeq C^\infty(P; \mathfrak{g}).$$