Horizontal and vertical bundles Consider the smooth fiber bundle $$F\to E \xrightarrow{\pi}B.$$ The vertical bundle $\mathcal{V}\subset TE$ is uniquely defined to be the kernel of $d\pi$. By choosing some Riemannian metric $g$ on $E$, we can also define a horizontal bundle $\mathcal{H}\subset TE$ as the $g$-orthogonal complement of $\mathcal{V}$, i.e. $$\mathcal{H}=\mathcal{V}^{\perp}=(\ker d\pi)^{\perp}.$$ Then the tangent bundle of $E$ decomposes as $TE=\mathcal{V}\oplus \mathcal{H}$.
Is it true that $\mathcal{H}$ and $TB$ are isomorphic vector bundles? Here is my idea for trying to prove it:
Pointwise, $d\pi_p:\mathcal{H}_p\to T_{\pi(p)}B$ is an isomorphism of vector spaces since $\pi$ is a submersion (i.e. $d\pi$ is a surjective linear map) and the restriction of $d\pi_p$ to $\mathcal{H}_p$ is injective by construction: $\mathcal{H}_p\cong T_pE/\ker(d\pi_p)$. Thus, in one direction, we have the vector bundle morphism $(\pi:E\to B,\;d\pi:\mathcal{H}\to TB)$ induced by $\pi$. For the inverse morphism, let $\varphi:B\to E$ be a global section of $\pi$ and define the vector bundle morphism $(\varphi:B\to E,\;(d\pi)^{-1}:TB\to\mathcal{H})$.
Does this work? My concern is that there may not exist a global section $\varphi$ of $\pi$.
Thanks for your help.
 A: I see two issues.  First,  $\mathcal{H}$ and $TB$ are bundles over different manifolds, so they can't be isomorphic as stated.  Second, as you noted, many bundles have no global sections.  For example, the Hopf bundles $S^1\rightarrow S^3\rightarrow S^2$ does not.
What is true, however, is that $\pi^\ast(TB)$ is isomorphic to $\mathcal{H}$.
To see this, first recall that $\pi^\ast(TB) = \{(e,v_p)\in E\times TB: \pi(e) = p\}$.  Here, I am thinking of $TB =\{v_p: p\in B, v\in T_p B\}$  The projection map $\pi_1: \pi^\ast(TB)\rightarrow E$ is the projection on to the first factor.
To see this, consider the map $f:\mathcal{H}\rightarrow \pi^\ast(TB)$ given as follows.  For $v\in \mathcal{H}_e\subseteq T_e E$, we define $f(v) = (e, d \pi_e v)$.
First note that $f(v)\in \pi^\ast(TB)$ because $d\pi_e v \in T_{\pi(e)} B$.  Clearly $f$ is smooth.
The map $f$ also commutes with projections.  That is $\pi = \pi_1\circ f$.  This follows because $\pi_1(f(v)) = e  = \pi(v)$ for $v\in\mathcal{H})_e$.
The inverse is given by $f^{-1}(e,v_p) = (d\pi_e)^{-1} v_p$, where I am thinking of $d\pi_e$ as an isomorphism $\mathcal{H}_e\cong T_{\pi(e)}B$.
