TITLE: Alternate definition of vector bundle?
QUESTION [7 upvotes]: Recall the usual definition of a $k$-dimensional vector bundle (everything is assumed to be continuous/smooth/etc depending on the category):

A $k$-dimensional vector bundle is a triple $(E,B,\pi)$, where 
  $\pi\colon E \to B$, satisfying the following:
    a) The map $\pi$ is onto (I don't know if everyone requires this, but I will). 
For each $p\in B$:
    b) The fiber $E_p=\pi^{-1}(p)$ is a (real) vector space. 
    c) There's a neighborhood $U\ni p$ and a diffeomorphism $\phi\colon \pi^{-1}(U) \to U \times \mathbb{R}^k$ such that $P_1\circ \phi = \pi$, where $P_1$ is projection onto the first factor. 
    d)  for each $q\in U$, the restriction $\phi\colon E_q \to {q}\times \mathbb{R}^k$ (where $\phi$ is the diffeomorphism from c)) is a linear isomorphism.

My Question
Can d) be replaced with

    d') the restriction $\phi\colon E_p \to {p}\times \mathbb{R}^k$ is a linear isomorphism. 

In the original version, the restriction of $\phi$ to every fiber over $U$ must be a linear isomorphism. In the alternate version, this is only required for the single fiber $E_p$.
(Note: This was posted at Math.SE a few weeks ago but received no answers.)

REPLY [4 votes]: No. Take $E=[0,1] \times \mathbb{R} \to [0,1]=B$ with the usual vector space structure at all fibers except for the fiber of $1/2$ where the vector space structure is twisted by any non-linear diffeomorphism of $\mathbb{R}$, for example $x \mapsto x^3+x$.