Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let π:E→B be a smooth vector bundle. Prove π is a submersion.

share|cite|improve this question
please don't phrase questions in the imperative mood. You are asking for help, not robbing me at gunpoint. Phrasing the question in imperative mood also creates the image that your are just copying word for word a question from your homework set: if that is indeed the case, please say so, and tell us what you know and what confuses you, and possibly show some work. – Willie Wong Mar 12 '11 at 22:53
Also, as user7887 noted below, local triviality of vector bundles implies $\pi$ is locally expressible as a canonical submersion, which means $\pi$ is a submersion. But usually these types of "definition-pushing" homework questions (assuming it is so) like this is tailored to the way various concepts are precisely defined in your class. So you should at least state in your question how you've seen submersion and vector bundle defined. – Willie Wong Mar 12 '11 at 22:58

We need to show locally $U$ of $E$ submerged to $V$ of $B$. But note $U\cong V\times F$, this should be automatic.

share|cite|improve this answer
I need help proving that... – user8169 Mar 12 '11 at 22:51
Proving what? Part of the definition of being a vector bundle inclused the fact that over small open sets $U\subseteq B$ the restriction $\pi:\pi^{-1}(U)\to U$ is isomorphic to the first projection $U\times V\to U$. – Mariano Suárez-Alvarez Mar 12 '11 at 22:53
Is that it ?. Does that prove that $\pi$ is a submersion. – user8169 Mar 12 '11 at 22:55
@Danny: (it is rather confusing that you are now no longer eric) do you know the definition of a submersion? If yes, just compute $d\pi$ using the local parametrisation that Mariano and user7887 wrote. – Willie Wong Mar 13 '11 at 1:44
No, I don't. I'm a little confused on the definition because I haven't done it before. I get user7887's point on showing the isomorphism – user8169 Mar 13 '11 at 4:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.