Vertical subbundle

Let $\pi:E\to M$ be a fibered manifold . The differential of $\pi$ defines a smooth vector bundle map between the tangent bundles

The vertical subbundle of $E$ is the kernel of this map:

Equivalently, $VE$ is the smooth subbundle whose fiber at $e\in E$ is the vertical tangent space $V_eE\subset T_eE$.

Since $\pi$ is a submersion, $d\pi$ has constant rank, and therefore $\ker(d\pi)$ has constant rank $\dim(E)-\dim(M)$; this implies $VE$ is a smooth vector subbundle of $TE$. A vertical vector field is exactly a smooth section of $VE$.

Examples

1. Product projection: for $\pi=\mathrm{pr}_1:M\times F\to M$, the vertical subbundle is $\{0\}\times TF\subset T(M\times F)\cong TM\times TF$.
2. Tangent bundle: for $\tau:TM\to M$, the vertical subbundle $V(TM)=\ker(d\tau)\subset T(TM)$ is the distribution tangent to the fibers $T_xM$.
3. Fibers as leaves: for any fibered manifold, $VE$ is tangent to each fiber $E_x$; in fact $VE|_{E_x}=TE_x$.