A (maybe) trivial question on trivial vector bundles: alternative definition of trivial vector bundle

I am studying Loring W. Tu's Differential geometry, Connections, Curvature and Characteristic Classes and I am having a doubt (the same doubt I had when studying the same topic in the author's An Introduction to Manifolds).

The following definition of a vector budle is given

Definition 7.1. A $$C^{\infty}$$ surjection $$\pi : E \to M$$ is a $$C^{\infty}$$ vector bundle of rank $$r$$ if

1. For every $$p \in M$$, the set $$E_p:=\pi^{−1}(p)$$ is a real vector space of dimension $$r$$;
2. every point $$p \in M$$ has an open neighborhood $$U$$ such that there is a fiber-preserving diffeomorphism $$\phi_U:\pi^{−1}(U) \to U\times \mathbb{R}^r$$ that restricts to a linear isomorphism $$E_p \to {p}\times \mathbb{R}^r$$ on each fiber.

The following definition for bundle map is then given:

Definition 7.5. Let $$\pi_{E}: E \to M$$ and $$π_F: F \to N$$ be $$C^{\infty}$$ vector bundles. A $$C^{\infty}$$ bundle map from $$E$$ to $$F$$ is a pair of $$C^{\infty}$$ maps $$(\phi: E \to F, \underline{\phi}: M \to N)$$ such that

2. $${\phi}$$ restricts to a linear map $$\phi_p: E_{p} \to F_{\underline{\phi(p)}}$$ of fibers for each $$p \in M$$.

A bundle map over $$M$$ is when $$\underline{\phi}$$ is the identity map. The author then says that

If there is a bundle map $$\psi: F \to E$$ over $$M$$ such that $$\psi \circ \phi = \mathbb{1}_E$$ and $$\phi \circ \psi= \mathbb{1}_F$$, then $$\phi$$ is called a bundle isomorphism over $$M$$, and the vector bundles $$E$$ and $$F$$ are said to be isomorphic over $$M$$.

Finally, here is the definition of trivial bundle:

Definition 7.6. A vector bundle $$\phi: E \to M$$ is said to be trivial if it is isomorphic to a product bundle $$M \times \mathbb{R}^r \to M$$ over $$M$$.

Here are my questions:

1. Can we say, equivalently, that a trivial bundle is a bundle in the sense of Definition 7.1 where there exists a open $$U = M$$, i.e. when there is a fiber-preserving diffeomorphism with the same properties $$\phi_M: \pi^{-1}(M) \to M \times \mathbb{R}^r$$ ? To me $$\psi \circ \phi = \mathbb{1}_E$$ and $$\phi \circ \psi= \mathbb{1}_F$$ is equivalent to say that $$\phi$$ is a diffeomorphism, i.e. a bijective $$C^{\infty}$$ map with $$C^{\infty}$$ inverse.
2. Can we say that that any manifold with a single chart has trivial tangent bundle?

Apologies in advance if my question is obvious or if, on the contrary, I am missing some macroscopic obstruction to my idea. I was wondering why the definition of trivial bundle has been given after the one of bundle isomorphism, while right after definition 7.1 the author defines trivializing open subset ($$U$$), trivialization ($$\phi_U$$ is a trivialization for $$\phi^{-1}(U)$$) and trivializing open cover.

thanks

Question 2 is certainly true. A single chart for $$M$$ is a diffeomorphism $$M \xrightarrow{\psi} V \subset \mathbb R^m$$, where $$m = \text{dimension}(M)$$ and $$V$$ is open. By definition of the tangent bundle of $$M$$, the map $$D\psi : TM \to TV$$ is a bundle isomorphism, and $$TV$$ is the restriction of $$T\mathbb R^m$$ which is trivial, hence $$TV$$ is trivial.
Regarding Question 1 and your closing paragraph, those trivialization operations that you are concerned about are only defined over individual elements of an atlas for $$M$$. So there is indeed a "macroscopic obstruction", but it is a simple one: Not every manifold has an atlas with a single chart, so triviality of individual charts is not yet sufficient to define what it means for $$\pi_E : E \to M$$ as a whole to be trivial.
You could easily proceed directly to a definition of triviality for the whole of $$\pi_E : E \to M$$, without first defining the concept of bundle isomorphism.
However, after later proceeding to a definition of bundle isomorphism, you would then have to prove a theorem (or lemma) saying that triviality of $$\pi_E : E \to M$$ is equivalent to the statement that $$\pi_E : E \to M$$ is isomorphic to a trivial bundle; and then your readers would object by saying "Wait, isn't that a trivial statement?" (pun intended).
• Yes, that's basically correct. I would add that if one chooses to define vector bundle isomorphism first, in that case your definition of triviality basically translates, directly, into the statement that a vector bundle $E \mapsto M$ is trivial if and only if it is isomorphic to the product bundle $M \times R^r \mapsto M$. Jan 29 at 23:58