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Am I right that a vector bundle $(E,M)$ of rank 0 means that sections of $E$ are functions $f:M \to M$?

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Why do you think you are right? Have you tried to prove it? –  Jason DeVito Dec 2 '12 at 18:45
    
@JasonDeVito Well, rank 0 means $E$ and $M$ are diffeomorphic. So a section of $E$ is a map from $M$ to $E \equiv M$. –  maximumtag Dec 2 '12 at 18:47
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A vector bundle comes equipped with a map from $E(\cong M)$ to $M$. Sections have to respect this map somehow. How does that enter in? –  Jason DeVito Dec 2 '12 at 18:48
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A section of $E$ is a map from $M$ to $E$ that ...... –  Neal Dec 2 '12 at 18:49
    
Sorry I don't get how I can use that $\Pi \circ S = id_M $ for a section $S$. All I know is that $S\circ d:M \to M$ where $d$ is the said diffeomorphism. –  maximumtag Dec 2 '12 at 19:10

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If $E$ is a rank $0$ vector bundle over $M$, identify $E \cong M\times\{0\}$. The projection map $\pi:E\to M$ is projection onto the first factor of the product. Every section of $E$ is a map $s:M\to E$ such that $\pi\circ s = id_M$.

With these three facts, you can completely characterize every single section of $E$ by asking what sort of function $M\to M\times\{0\}$ composed with projection to $M$ gives the identity on $M$.

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Why did you identify $E$ with $M \times \{0\}$. What's the use of the tag 0? –  maximumtag Dec 4 '12 at 13:41
    
The empty basis is a global trivialization of $E$, so we can identify $E$ with the product bundle $M\times\{0\}$. –  Neal Dec 4 '12 at 14:10

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