# A smooth vector bundle $p:E \rightarrow B$ is a submersion

Prove that a smooth vector bundle $$p:E \rightarrow B$$ is a submersion. With $$B$$ being a smooth $$k$$ dimensional manifold, and $$E$$ be a bundle of dimension $$n$$.

I would like to see how one prove this. Below is my attempt, I would be happy for some comments too.

My proof relies on Theorem 4.14, pg 83, of Lee's Manifold,

(Global rank theorem). Let $$M$$ and $$N$$ be smooth manifolds, and suppose $$F:M \rightarrow N$$ be a smooth map of constant rank. If $$F$$ is surjective, then $$F$$ is a smooth submersion.

It suffices to show that $$p$$ has constant rank. But with a local trivialization of $$E$$, over local chart $$U$$ of $$B$$, we know that $$p$$ has the coordinate representation, $$(x^1,\ldots, x^k, x^{k+1}, \ldots, x^{k+n}) \rightarrow (x^1, \ldots, x^k)$$ The Jacobian matrix has rank $$k$$.

Your proof is fine but kind of silly. You have computed that the differential has rank $$k=\dim B$$ at every point, which is the definition of what it means for $$p$$ to be a submersion. There's no need to go through the theorem you mentioned.