I have a question about the manifold, especially when the manifold is as well a vector space of finite dimensional $k$. Actually, let $(v_1, \dots, v_k)$ be a basis of F as a vector space. I would like to say that : $ f : \mathbb{R}^k \longrightarrow F$ which send $(x_1, \dots, x_k)$ to $x_1v_1 + \dots + x_kv_k$ is a $\mathcal{C}^\infty$ diffeomorphism. But I don't figure out how to prove it (and if it's right).
Actually, this question is directly related to the fact that given a vector bundle of rank $k$ : $p : E \rightarrow B$, we say the bundle trivializes if and only if it's isomorphic (as a vector bundle) to the trivial bundle $B \times \mathbb{R}^k$. On the other side, in some cases, we have to only prove that it's the case if and only if it's isomorphic to the trivial bundle $B \times F$, where $F$ is a vector space of finite dimensional $k$. So, I suppose that the two definitions are the same.
And I wanted to prove it, that's why I try to figure out if the application given at the beginning if a diffeomorphism, but I don't succeed to do it.
Someone could help me ?
Thank you !
Let's take $F = \Bbb R^4$ with the standard basis (as a vector space). Then your map $f$ is the identity, and you'd like to claim that this is a diffeomorphism from the standard $\Bbb R^4$ to $F$.
But whether this is a diffeomorphism depends on the differentiable structure on $F$. Donaldson and Freedman proved that there is an "exotic $\Bbb R^4$" (see https://en.wikipedia.org/wiki/Exotic_R4) for which this map is not a diffeomorphism (indeed, for which there's no diffeomorphism to the standard $\Bbb R^4$).
In short: the thing you're trying to prove is false.
You might also want to look at Manifold with different differential structure but diffeomorphic.