Why the vertical lift of a vector bundle takes values on the vertical bundle?

Question

Let $(E,\pi,M)$ be a vector bundle. The vertical bundle of this vector bundle is the subbundle $VE$ of the tangent bundle $TE$ (as a vector bundle $(TE, T\pi, TM)$ over $TM$), such that: $VE = (T\pi)^{-1}(0)$. Then the $\textit{vertical lift}$ is the map $\text{vl}_E: E \times_M E \rightarrow VE$ defined fiberwise as $\text{vl}_E(u_p,v_p) = \frac{d}{dt}|_{t=0}(u_p + tv_p)$. Why this map takes values exactly on $VE$, and not generally on $TE$?

Answer 1

From the definition of $VE$ we see that what we want to show is $$T\pi({\rm vl}_E(u_p,v_p))=0.$$

We know from the chain rule that $$T\pi({\rm vl}_E(u_p,v_p))=\frac d{dt} \Big|_{t=0}(\pi(u_p + t v_p)).$$

But $\pi(u_p + t v_p)$ is the constant $p$, so we are done.