Redundancy in the definition of vector bundles?

Question

In John Lee's classic Introduction to Smooth Manifolds, the following definition of vector bundle is given.

Definition. 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 $\pi:E\to M$ satisfying the following conditions:

(i) 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.

(ii) For each $p\in M$, there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $\Phi:\pi^{-1}(U)\to U\times\Bbb{R}^k$ (called a local trivialization of $E$ over $U$*), satisfying the following conditions:

1. $\pi_U\circ\Phi=\pi$ (where $\pi_U:U\times\Bbb{R}^k\to U$ is the projection);

2. for each $q\in U$, the restriction of $\Phi$ to $E_q$ is a vector space isomorphism from $E_q$ to $\{q\}\times\Bbb{R}^k\cong\Bbb{R}^k$.

But if we skip conditions (i) and 2, can't we just define the vector space structure on $E_p$ by using its set-theoric bijection with $\{p\}\times\Bbb{R}^k$?

In other words:

Question: Let $E$ and $M$ be topological spaces and $\pi:E\to M$ a continuous map such that for each $p\in M$ there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $\Phi:\pi^{-1}(U)\to U\times\Bbb{R}^k$ such that $\pi_U\circ\Phi=\pi$.

Is $E$ is vector bundle?

Answer 1

No Look at $[0,1]\times \mathbb{R}$, now identify $\{0\} \times \mathbb{R}$ and $\{1\} \times \mathbb{R}$ via the map $$f:\{0\} \times \mathbb{R}\rightarrow \{1\} \times \mathbb{R}$$ $$f(0,x)=(1,x^3)$$ a non linear map. The base space is then $S^1$. Basically in a vector bundle the maps between the fibres must be linear.

Answer 2

As already pointed out, your definition is incomplete in that it lacks compatibility conditions among the different trivializations. The original definition guarantees this compatibility using a god-given vector space structure on the fibers, but it's true that instead you could take your new definition and add the following condition:

Given $U$ and $V$ admitting a trivialization, the map $$\Phi_U \circ \Phi_V^{-1}: (U \cap V) \times \mathbb{R}^k \to (U \cap V) \times \mathbb{R}^k$$ is given by $$(x,w) \mapsto (x,g_{UV}(x)(w))$$ where $g_{UV}: U \cap V \to GL_n(\mathbb{R})$ is smooth.

P.S.: I am not completely sure on how to prove that the original definition implies the smoothness of the $g_{UV}$... I hope that somebody will point it out. It should be the only point missing to show the equivalence of the two definitions.

Edit: As pointed out by Karl Kronenfeld in the comment, if we define smooth functions

$$ f_i: U\cap V \to (U \cap V) \times \mathbb{R}^k, \ x \mapsto (x, e_i),$$ $$ g_j: (U\cap V)\times \mathbb{R}^k \to \mathbb{R}, \ (x,v) \mapsto v_j,$$

then we obtain $g_{UV}^{ij}$ as the composition $g_j \circ \Phi_U \circ \Phi_V^{-1}\circ f_i$.