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Well I have a doubt about a rank $k$ vector bundle. My definition of vector bundle is: A rank $k$ vector bundle is a triple $(\pi, E, M)$ where $E$ and $M$ are smooth manifolds and $\pi:E\rightarrow M$ is a smooth submersion, which satisfies:

(i) for every $p\in M$ the fiber $F_p=\pi^{-1}(p)$ is a $k$-vector space.

(ii) given $p\in M$ there is an open set $U\subseteq M$ containing $p$ and a diffeomorphism $\phi:\pi^{-1}(U)\rightarrow U\times \mathbb R^k$ such that $\textrm{pr}_1\circ \phi=\pi$.

(iii) For every $q\in U$ the map $\phi_q:E_q\rightarrow \mathbb R^k$, $e\mapsto (\textrm{pr}_2\circ \phi)(e)$, is a linear isomorphism..

My doubt is: If I have $\pi^{-1}(U)=E$ and $U=M$ in $(ii)$ and I consider $\phi^{-1}:M\times \mathbb R^k\rightarrow E$ can I say $\phi^{-1}(p, \cdot)=\phi_p^{-1}$?

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up vote 3 down vote accepted

You didn't state it in your definition, but $\phi_p\colon E_p\to\{p\}\times U$ is the restriction of $\phi$ to the fiber $E_p=\pi^{-1}(p)\subseteq\pi^{-1}(U).$ Hence the identity you asked about is certainly satisfied, since in general the inverse of a restriction of any function is the restriction of the inverse.

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Unfortunately I must follow this definition because it is from a hand out =( but I can finally justify the equality $\phi^{-1}(p, \cdot)=\phi_p^{-1}(\cdot)$: Using $(iii)$ I know $$\phi_p=\textrm{pr}_2\circ \phi,$$ and since $\phi$ is a bijection I have, $$\phi^{-1}=\phi_p^{-1}\circ \textrm{pr}_2,$$ hence $$\phi^{-1}(p, \cdot)=\phi^{-1}_p\circ \textrm{pr}_2(p, \cdot)=\phi_p^{-1}(\cdot).$$ Thanks anyway @Joe. –  PtF Apr 24 '13 at 0:07
    
Even though you didn't say explicitly that $\phi_p$ is a restriction, it does follow from things on your handout, since you give $\phi_p$ in terms of $\phi$. In fact, without observing this fact, some things you write are problematic. You write $\phi_p=\text{pr}_2\circ\phi$. But the map on the left has domain $E_p$ and the map on the right has domain $\pi^{-1}(U).$ So in fact these maps cannot be equal. If you think that objection is overly pedantic, you are surely right. –  Joe Hannon Apr 24 '13 at 2:08

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