Let π:E→B be a smooth vector bundle. Prove π is a submersion.
Smooth Vector Bundle is a Submersion
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differential-geometry
riemannian-geometry
vector-bundles
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5please 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. – 2011-03-12
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1Also, 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. – 2011-03-12
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We need to show locally $U$ of $E$ submerged to $V$ of $B$. But note $U\cong V\times F$, this should be automatic.
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0I need help proving that... – 2011-03-12
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0Proving 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 $U\times V\to U$. – 2011-03-12
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0Is that it ?. Does that prove that $\pi$ is a submersion. – 2011-03-12
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0@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. – 2011-03-13
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0No, 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 – 2011-03-13