Can any atlas of the manifold act as "local trivialization"? I'm learning Manifolds theory and on one concept on a vector bundle I couldn't really understand for weeks.
In Loring Tu's Introduction to smooth manifolds, in order to find the local trivialization, we find the collection of $U$ covering the manifold $M$ such that we get a fiber-preserving diffeomorphism $\phi:\pi ^{-1}(U)\to U\times \mathbb{R}^r$ with each fiber being vector space with dimension $r$.
This notion of local trivialization is introduced in the section of vector bundle (not necessarily a tangent bundle) and it seems that every manifold admits local trivialization (not global trivialization). So for the case of manifolds, can we regard this $U$ in local trivialization as an atlas? Even more basically, vector bundle is the generalization of tangent bundle, is it correct?
 A: Tu defines a surjective smooth map $p : E  \to M$ to be locally trivial of rank $r$ if the two conditions (i) and (ii) on the top of p. 134 are satisfied.
Any such collection $\{(U,f )\}$, with $\{U\}$ an open cover of $M$, is called a local trivialization for $E$, and $\{U\}$ is called a trivializing open cover of $M$ for $E$.
You may also say that $\{(U,f )\}$ is a bundle atlas in that case.
If I understand correctly, you ask whether a trivializing open cover $\{U\}$ of $M$ is an atlas for $M$. The answer is no.

*

*We are not given homeomorphisms $\phi_U : U \to V$, where $V$ is an open subset of $\mathbb R^n$. Only a collection $\{(U,\phi_U) \}$ can be a (smooth) atlas.


*You could ask whether for each bundle atlas $\{(U,f )\}$ there exists an associated  smooth atlas $\{(U,\phi_U) \}$ for $M$ having the same open sets $U$. This is not the case. The trivial bundle $M \times \mathbb R^k$ has a trivializing open cover with one set (which is the whole manifold $M$). But in general there is no smooth atlas on $M$ with a single chart.
However, what you can do is this: Given a trivializing open cover $\mathcal U$ of $M$, then any open cover $\mathcal V$ which refines $\mathcal U$ is again  a trivializing open cover for $M$. Now take $\mathcal U'$ to be set of all open $U'$ which are contained in some $U \in \mathcal U$ on which there exists a chart $\phi_{U'} : U' \to V'$ belonging to the smooth structure of $M$.
