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.

  • $\begingroup$ You give each of the $U_i \times F$ the smooth structure of a product. Since the bundle is covered by such things this suffices to define the smooth structure. $\endgroup$ – user98602 Jun 27 '16 at 18:39
  • $\begingroup$ Does it not matter that you have an infinite product of these? And how does this smooth structure descend to E? $\endgroup$ – B. Pasternak Jun 27 '16 at 18:41
  • $\begingroup$ Infinite coproduct, aka disjoint union. Just do one at a time. You just need to see that the transition maps are smooth. $\endgroup$ – user98602 Jun 27 '16 at 18:42
  • $\begingroup$ Please see my edit. $\endgroup$ – B. Pasternak Jun 27 '16 at 19:00

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