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. 
> 
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>     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
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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](https://math.stackexchange.com/questions/289177/alternate-definition-of-vector-bundle) a few weeks ago but received no answers.*)



Alternate definition of vector bundle?
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 $x \mapsto x^3+x$.


