Determine the fibre bundle from the cocycle

Question

I hope someone can give me some clarifications about fiber bundle. My notes say that the cocycle determines the fiber bundle up to isomorphism without any explanation at all. In other words, I can reconstruct the fiber bundle $E$ and its projection $\pi$ once given $(B,F,\{U_\alpha\},\{g_{\alpha\beta}\})$, where $B$ is the base space, $F$ the fiber, $\{U_\alpha\}$ the open cover of $B$ and $\{g_{\alpha\beta}\}$ the cocycle; but I can't understand why.

Answer 1

The fiber bundle can be (re)constructed as the disjoint union $$\coprod_\alpha U_\alpha\times F$$ modulo the equivalence relation defined by $(a, f)\sim (b, h)$ for $(a, f)\in U_\alpha\times F$, $(b, h)\in U_\beta\times F$ if and only if $a = b\in U_\alpha\cap U_\beta$ and $g_{\alpha\beta}(a)h = f$. In other words, the cocycles tell you how to glue the overlaps.

Answer 2

The idea is that over each element $U_\alpha$ of the open covering, the fibre bundle is trivial, that is, is isomorphic to $U_\alpha\times F$. However, if $U_\alpha\cap U_\beta\neq\varnothing$, then we must specify how to identify the subset $(U_\alpha\cap U_\beta)\times F\subset U_\alpha\times F$ with the subset $(U_\alpha\cap U_\beta)\times F\subset U_\beta\times F$. This is done via the "transition function" $g_{\alpha\beta}$. Each of the transition functions gives isomorphisms of the fibre, so mapping $(b, f)\mapsto (b, g_{\alpha\beta}(b)(f))$ defines an isomorphism $(U_\alpha\cap U_\beta)\times F\to (U_\alpha\cap U_\beta)\times F$.

Requiring that this plays well with triple intersections $U_\alpha\cap U_\beta\cap U_\gamma$ gives the "cocycle condition". All of this data together gives gluing data which allows us to glue together the spaces $U_\alpha\times F$ in a meaningful way, defining the fibre bundle.