Let $E$ and $M$ be smooth manifolds (of finite dimension, Hausdorff and second countable). Let $\pi:E\longrightarrow M$ be a surjective submersion such that:


1. $E\_p:=\pi^{-1}(p)$ is a real vector space isomorphic to $\mathbb R^n$, $\forall p \in M$.
2. The pointwise sum, multiplication by scalar and zero section are smooth:
$$ +:E\times\_M E \longrightarrow E $$
$$ \cdot: \mathbb R \times E \longrightarrow E $$
$$ 0:M \longrightarrow E$$


Then, is $E$ a vector bundle with this structure?


I have thought the following:


Let $p\in M$, then exists $\sigma\_i:U \longrightarrow E $, local sections such that $\{\sigma\_i(p)\}$ is a basis of $E\_p$. Define:
$$ V:=\{q\in U \; | \; \{\sigma\_i(q)\} \text{ is a basis of } E\_q \}=\{q\in U \; | \; \sigma\_1(q)\wedge \dots \wedge \sigma\_n(q)\not=0(q) \} $$
$$ \Xi:V \times \mathbb R^n \longrightarrow \pi^{-1}(V) $$
$$ (q,\lambda\_1,...,\lambda\_n) \mapsto \lambda\_1\sigma\_1(q)+\dots+\lambda\_n\sigma\_n(q) $$


Is $V$ open?


$\Xi$ would be the candidate for a vector bundle chart around $p$: It's an isomorphism on each fiber. If $V$ is open, then $\Xi$ will be smooth (by 2.). Is $\Xi^{-1}$ smooth?



Local triviality condition in vector bundles
If you see $\Xi$ as a map from $U\times\mathbb R^n$ to $\pi^{-1}(U)$, then it is smooth, and by the inverse mapping theorem it is locally invertible around $(p,0)$ (because of the submersion condition and the condition on the $\sigma\_i(p)$). It means that for all $q$ close enough to $p$, the restriction $d\Xi\_{(q,0)}:\mathbb R^n\to E\_q$ is invertible, so $(\sigma\_i(q))\_i$ is a basis. So you can take $V$ to be a neighbourhood of $p$, or you can use the same reasoning to show that the $V$ you describe is open. Then $\Xi:V\times\mathbb R^n\to\pi^{-1}(V)$ is a diffeomorphism, because it is a smooth bijection, and the inverse is locally smooth by the inverse mapping theorem.


Note that this is not enough to make $E$ a vector bundle by the usual definition: technically, one needs to check that the change of coordinates are actually vector space isomorphisms. However, in your setting, this is obviously the case since the $\Xi$'s come from the unified vector space structure on $E$.


