After reading some introductory texts i've got a question on smooth fibre bundles. I understand that we require both the base space and the typical fibre to be smooth manifolds. But then i am not completely sure why the total space becomes a smooth manifold itself.
Consider a fibre bundle $(E, B, \pi, F, G)$. Then there exist smooth trivializations $\varphi_i:E\rightarrow U_i\times F$ where $U_i$ is open in B. Does the total space E then become a smooth manifold because we can compose these trivializations with those of the base space B and the typical fibre $F$ getting a trivialization $\psi_i:E\rightarrow\mathbb{R}^k\times\mathbb{R}^n\cong\mathbb{R}^{k+n}$ where $k$ is the dimension of B and $n$ is the dimension of $F$?