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:
$\pi_U\circ\Phi=\pi$ (where $\pi_U:U\times\Bbb{R}^k\to U$ is the projection);
for each $q\in U$, the restriction of $\Phi$ to $E_q$ is a vector space isomorphism from $E_q$ to $\{q\}\times\Bbb{R}^k\cong\Bbb{R}^k$.
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?
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.