Zero sections of any smooth vector bundle is smooth?

Could any one give me hint how to show that the zero section of any smooth vector bundle is smooth?

Zero section is a map $\xi:M\rightarrow E$ defined by $$\xi(p)=0\qquad\forall p\in M.$$

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Smoothness can be checked locally on $M$ and locally $E$ is trivial.

Can you use these two facts to conclude what you want?

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No........................... – miosaki Aug 7 '12 at 23:48
@KutukKatuk If $M$ and $N$ are smooth manifolds and $p\in N$, then the inclusion $M\to M\times N$ given by the rule $m\to (m,p)$ is smooth. If you understand this and combine it with Mariano's answer, then you will have found an answer to your original question. – Amitesh Datta Aug 8 '12 at 0:32
I am a little confused; do we explicitly use the mapping to zero in the proof? It seems like local triviality would apply no matter what constant vector we mapped to. – Tarnation Oct 9 '12 at 2:26
@Tarnation: but mapping to another «constant vector» would in general not give an actual section of the bundle: most vector bundles do not have non-zero sections! Zero is different: we always have the zero section. – Mariano Suárez-Alvarez Oct 9 '12 at 4:17

Say $E\to M$ is a smooth vector bundle, and let $p\in M$. Then there is a neighborhood $U\subseteq M$ of $p$ and a smooth local trivialization $\Phi:\pi^{-1}(U)\to U\times\mathbb R^k$ of $E$ over $U$. What can you say about $(\Phi\circ\zeta)(q)$ where $q\in U$? Can you conclude that $\Phi\circ\zeta$ is smooth? Now remember that $\Phi$ is a diffeomorphism.

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