Can vector bundle be thought of as below? Suppose we have following data.


*

*Two smooth manifolds $E,M$ and a surjective submersion $\pi:E\rightarrow M$.

*A vector space structure on $\pi^{-1}(x)$ for each $x\in M$.

*A collection of isomorphisms $\{\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n : x\in M\}$.


I call $\pi:E\rightarrow M$ to be 


*

*a trivial bundle if, there exists a diffeomorphism $\Phi: \pi^{-1}(M)\rightarrow M\times \mathbb{R}^n$ such that  $\Phi_{\pi^{-1}(x)}$ is same as $\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n$ in some sense, I am asking if there I can glue these isomorphisms of vector spaces $\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n$ to get a diffeomorphism  $\Phi: \pi^{-1}(M)\rightarrow M\times \mathbb{R}^n$.


I do not want to restrict my interest to gluing all isomorphisms $\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n$. I want to consider the case when I can glue isomorphisms locally. This is not new thought or anything. One sees Locally compact more often than compact, Locally connected more than connected and so on. So, I want to consider the case where I can glue isomorphisms locally in the following sense.


*

*Given $x\in M$, there exists an open set $U$ of $M $containg $x$ such that, I can glue the isomorphisms $\{\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times\mathbb{R}^n:x\in U\}$ i.e., given $x\in M$, there exists a diffeomorphism $\Phi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$
such that $\Phi_{\pi^{-1}(y)}$ is same as $\psi_y:\pi^{-1}(y)\rightarrow \{y\}\times\mathbb{R}^n$ for each $y\in U$. 


I call $\pi:E\rightarrow M$ to be 


*

*a locally trivial bundle, if given $x\in M$ there exists an open set $U$ containing  $x$ and  a diffeomorphism $\Phi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$
such that $\Phi_{\pi^{-1}(y)}$ is same as $\psi_y:\pi^{-1}(y)\rightarrow \{y\}\times\mathbb{R}^n$ for each $y\in U$. 


A vector bundle is a locally trivial bundle mentioned above.
Question :

Did I miss anything in the definition?

I used notion of vector bundles very often and may be assuming something and not saying here. Please let me know if I am assuming something and not saying here
 A: Yes, you missed all the non-trivial vector bundles! Your proposed definition has two severe problems:


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*The notion of a vector bundle should generalize the notion of a vector space. In particular, when $M$ is a point, a vector bundle over $M$ should be just a vector space. An arbitrary vector space $V$ doesn't come with a preferred basis (i.e, an isomorphism with $\mathbb{R}^n$) while you propose that a vector bundle over a point is not just a vector space but a vector space together with a choice of a basis. A choice of a basis is not part of a data of a vector space and a choice of isomorphisms $\psi_x$ shouldn't be part of a data of a vector bundle.

*You write "there exists a diffeomorphism $\Phi$" but this is misleading because you require that $\Phi|_x = \psi_x$. Given $(\psi_x)_{x \in M}$ this requirement completely specifies the map $\Phi$ and the only issue is whether the set theoretic map $\Phi$ is a diffeomorphism or not. If it is a diffeomorphism, your data specifies what is called in the usual terminology a trivial vector bundle with a specific choice of a trivialization (just like in $(1)$ you didn't get a vector space but a vector space together with a choice of a specific isomorphism to $\mathbb{R}^n$).

