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This question is about the content of page 134 of Tu's An Introduction to Manifolds.

A $C^{\infty}$ vector bundle or rank r is a triple $(E,M,\pi)$ consisting of manifolds $E$ and $M$ and a surjective smooth map $\pi:E\rightarrow M$ that is locally trivial of rank $r$. More precisely,

(i) each fiber $\pi^{-1}(p)$ has the structure of a vector space of dimension $r$,

(ii) for each $p\in M$, there are an open neighborhood $U$ of $p$ and a fiber-preserving diffeomorphism $\phi:\pi^{-1}(U)\rightarrow U\times\mathbb{R}^n$ such that for every $q\in U$ the restriction $$\phi\mid_{\pi^{-1}(q)}:\pi^{-1}(q)\rightarrow \{q\}\times\mathbb{R}^r$$ is a vector space isomorphism.

My question is: Given any regular submanifold $S\subset M$, is the triple $(\pi^{-1}S,S,\pi\mid_{\pi^{-1}S})$ also a $C^{\infty}$ vector bundle over S? I tried to resolve this issue by checking the conditions (i) and (ii).

Condition (i) is obviously satisfied. To prove (ii), I chose an adapted chart $(U,\phi)$ with $p\in U$ (which gives a chart for $S$) and tried showing that the map $$\phi\mid_{\pi^{-1}(U\cap S)}:\pi^{-1}(U\cap S)\rightarrow U\cap S\times\mathbb{R}^r$$ is a fiber-preserving diffeomorphism. But is this map necessarily a diffeomorphism? This is a restriction of a diffeomorphism, and even though it possesses an inverse (using $\phi^{-1}$) which looks like a smooth inverse for $\phi$, we don't know what manifold structures the sets $\pi^{-1}(U\cap S),U\cap S\times\mathbb{R}^r$ have yet, and I'm hesistated to say for sure that those maps are smooth maps, let alone inverse maps to each other.

In summary:

  • Do the sets $\pi^{-1}(U\cap S),U\cap S\times\mathbb{R}^r$ have manifold strutures? Canonical ones?
  • Is this true in general? In what sense? : Restriction of a vector bundle is a vector bundle.

Thanks in advance.

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First, note that $\pi: E \to M$ is a submersion, so in particular is transverse to any submanifold $S \subset M$. So $\pi^{-1}(S)$ has a canonical manifold structure. Similarly for $\pi^{-1}(U \cap S)$. Any smooth map, restricted to a submanifold, is automatically smooth, so $\phi$ is smooth, as is its inverse.

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  • $\begingroup$ This is a great answer. Thank you. $\endgroup$ – Dilemian Jul 23 '16 at 19:21

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