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.

  • $\begingroup$ Oh yes, that is silly... $\endgroup$ – CL. Jan 2 at 2:58

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