Vertical vectors on a principal bundle Let
$$\pi:P\longrightarrow M$$
be a principal $G$-bundle, and consider, for each $p\in P$, the map
$$\begin{matrix}l_p:&G&\longrightarrow&P\\&g&\longmapsto &p\cdot g\end{matrix}$$
let $x=\pi(p)$, and since $l_p(e)=p$ we get the sequence of vector spaces
$$0\longrightarrow T_e G\longrightarrow T_pP\longrightarrow T_xM\longrightarrow 0$$
and so vertical vectors are defined as the image of $d_el_p$, or (and here is where I have trouble) as the kernel of the next map, which means that the sequence must be exact.
In my notes (Fibre bundles and Chern-Weil theory by Johan Dupont) there is no proof of this fact and it's driving me crazy, I don't know how to write things so I can "see" what those linear maps are or how they work on a particular vector. Any idea or reference will be appreciated.
 A: I don't think you need an explicit description of those maps to conclude. Consider the compositon $\pi\circ l_p$. For any $g\in G$ it holds that $\pi(l_p(g))=\pi(p\cdot g)=\pi(p)=x$, so $\pi\circ l_p$ is a constant map. Hence, we have that $d(\pi\circ l_p)=0$, so $0=d\pi_p\circ(dl_p)_e$, which means that $\mathrm{im}((dl_p)_e)\subset\ker(d\pi_p)$. Can you close the argument yourself?
Geometric edit
Consider first the trivial principal G-bundle $P=M\times G$ with the trivial $G$-action. In this case, viewing the tangent space to $P=M\times G$ at $p=(m,g)$ as the $T_pM\oplus T_gG$, the exactness of the sequence is clear immediately, and the map $dl_p$ is just $v\mapsto(0,dL_g(v))$, with $L_g:G\to G$, $h\mapsto gh$. Anyways, if you just "draw" $M\times G$ as a product with the base horizontally a line and $G$ attached to $M$ at each point as a vertical line, then the tangent vectors to $G$ are actually "vertical", and I think this is where the terminology comes from, and it should serve as a sufficient interpretation/intuition. The vertical vectors are simply precise those which don't point non-trivially with respect to $M$ at all.
