Prove that $s: B \rightarrow E$ is a section ( vector bundles)

Question

I'm very very unfamiliar with vector bundles, so maybe this question is quite trivial.

Let $\pi : E \rightarrow B$ be a vector bundle and $s: B \rightarrow E$ a map sending each $p \in B$ to the $0$ vector of $\pi^{-1}(p)$. I need to show that $s$ is a section that is $\pi \circ s=1_B$

I know that $\pi$ sends a fibre $\pi^{-1}(p)$ to $p$. But i can't see clearly what $\pi$ does to elements of a fibre.

I tried to use the local triviality condition: Since there's a isomorphism $t: \pi^{-1}(p) \rightarrow p \times \mathbb{R}^n$ it's clear that $t\circ s (p)=(q,0)$ but again I'm stuck.

Answer 1

Let $U$ be a small open set of $B$. On this set $\pi^{-1}(U)=R^n\times U$. Then, locally, $s:U\rightarrow\pi^{-1}(U)$ is the map for $p\in U$, $p\mapsto (\vec{0},p)$.

On the other hand, $\pi:R^n\times U\rightarrow U$ is the map $\pi(\vec{v},p)=p$. Therefore, $\pi(\vec{0},p)=p$.

Combining these, $\pi\circ s(p)=\pi(\vec{0},p)=p$.