I am self studying smooth manifolds and I have encountered the notion of a vector bundle. My current intuition for these objects is that they somehow generalize what we encountered when studying tangent bundles, which is each point in the base space is in some way associated with a vector space. However, looking at the formal definition of a vector bundle I do not understand the role of a local trivialization and why we need it. For reference, Lee's Introduction to Smooth Manifolds defines vector bundles as follows:
Let $M$ be a topological space. A (real) vector bundle of rank $k$ over $M$ is a topological space $E$ together with a surjective continuous map $\pi: E \rightarrow M$ satisfying the following conditions:
- For each $p \in M$, the fiber $E_p = \pi^{-1}(p)$ over $p$ is endowed with the structure of a $k$-dimensional real vector space.
- For each $p \in M$, there exist a neighorhood $U$ of $p$ in $M$ and a homeomorphism $\phi: \pi^{-1}(U) \rightarrow U \times \mathbb{R}^k$ (called a local trivialization of $E$ over $U$) satisfying the following conditions:
- $\pi_U \circ \phi = \pi$ (where $\pi_U: U \times \mathbb{R}^k \rightarrow U$ is the projection);
- for each $q \in U$, the restriction of $\phi$ to $E_q$ is a vector space isomorphism from $E_q$ to $\{q\} \times \mathbb{R}^k \cong \mathbb{R}^k$.
I do not have a good feel for what a local trivialization is. More specifically, my confusion is regarding the second item in the definition. What is this conditions saying exactly and why do we need it? I must be missing something, but what extra information does it give us that first item in the defintion does not?
I took a look at the following question:
Redundancy in the definition of vector bundles?
but I still do not understand what the second item in the definition (and specifically local trivializations) is saying on an intuitive level.