Let $M,F$ be smooth manifolds, $\{U_i:i\in I\}$ an open cover of $M$ and a cocycle $\{t_{ij}:U_i\cap U_j\to\mathrm{Diff}(F)\}$. In almost any book which discusses fibre bundles, one can find the theorem that says that you can construct a smooth fibre bundle with fibre $F$ from these data, but no one proves it explicitly. So I thought, let's prove it.
We take the disjoint union $\coprod_{i\in I}U_i\times F$ and the equivalence relation which relates $(p,f)\in U_i\times F$ and $(q,g)\in U_j\times F$ iff $p=q$ and $f=t_{ij}(p)g$. Then $E$ is the quotient space equipped with the quotient topology, and the map $\pi:E\to M$ sending $\overline{(p,f)}$ to $p$ is continuous. The restrictions $q|_{U_i\times F}:U_i\times F\to q(U_i\times F)=\pi^{-1}(U_i)$ are homeomorphisms, and should become the local trivialisations. It remains to show that $E$ is a smooth manifold, that $\pi$ is smooth and that these local trivialisations are smooth (and that the topology on $E$ is Hausdorff/second countable), and this is where I got stuck. Does anyone have any idea how the smooth structure on $E$ is defined? It should be defined by the smooth structure on $M$ and $F$, but I don't see how.
Edit: obviously $E$ is Hausdorff and second countable because $M$ is and $\pi$ is continuous.
Edit 2: a smooth athlas for each $U_i\times F$ is given by $\mathcal{A}_i\{((U\cap U_i)\times V,\phi|_{U\cap U_i}\times\psi)\,|\,(U,\phi)\in\mathcal{A}_M,(V,\psi)\in\mathcal{A}_F\}$, clearly. How is then the atlas for the infinite coproduct defined? Is it just $\{(\coprod_{i\in I}W_i,\prod_{i\in I}\Phi_i)\,|\,(W_i,\Phi_i)\in\mathcal{A}_i\}$? I still don't see how this descends to $E$.
As for the transition functions, these are given by $\phi_i\circ\phi_j^{-1}:U_i\cap U_j\times F\to U_i\cap U_j\times F$, and $(\phi_i\circ\phi_j^{-1})(p,f)=(p,t_{ij}(p)f)$, which are smooth.