A differentiable fiber bundle exists for any family of clutching functions.

Lemma: Let $F$ be a smooth manifold and let $\{U_a\}_{a\in A}$ be a covering of a manifold $B$, and let $\{g_{ab}\}_{a,b\in A}$ be a family of clutching functions. That is, $g_{ab}:U_a\cap U_b\to \text{Diff}(F)$ satisfy the cocycle condition $g_{ab}g_{bc}=g_{ac}$ on the intersection $U_a\cap U_b\cap U_c$.

Then $\{g_{ab}\}_{a,b\in A}$ are the clutch functions for some differentiable fiber bundle $P:E\to B$ with fiber $F$.

Idea: Consider the topological disjoint union $\tilde E=\cup_a U_a\times F$. Define an equivalence relation $\sim$ in $\tilde E$ by the formula $(a_1,b,p)\sim(a_2,c,q)\iff b=c \land p=g_{a_1a_2}(b)(q)$. Let $E$ be the quotient space $\tilde E/\sim$. Each $U_a\times F=\pi(U_a\times F)$ is a subspace of $E$. $\pi$ here is the natural projection onto the factor.

Introduce a smooth structure on $E$ by taking for charts the maps $x_{mn}=x_m\times x_n$ where $x_n$ and $x_m$ are charts on $B$ and $F$. Then $U_a\times F$ are open submanifolds.

Given such a structure, if we define $P:E\to B$ by $P(\pi(a,b,p))=b$, we arrive at a smooth fiber bundle with clutching functions $\{g_{ab}\}_{a,b\in A}$. Indeed, $P$ is well defined and for all $a\in A$. We need to consider trivializations $f_a:P^{-1}(U_a)=\{\pi(\alpha,b,p):\alpha\in A,b\in U_a\cap U_{\alpha},p\in F\}\to U_a\times F$, such that $f_a\circ f_b^{-1}(b,p)=(b,g_{ab}(b)(p))$.

Problems: How should we define the trivializations? Will the endowment of $E$ with the smooth structure go through if I were to work out the details?

Intuitively, we know that we want $f_a$ to send all points in the equivalence class of $(a,b,p)$ in $E$ to $(b,p)$ in $U_a \times F$. This is made possible because of the universal property of quotient spaces. In particular, we can define the trivialization using a smooth map $$\tilde f_a: \pi^{-1}([U_a \times F])\subset \tilde E \to U_a \times F$$ such that $\tilde f_a (a,b,p)=\tilde f_a(a',c,q)$ whenever we have $\pi(a,b,p)=\pi (a',c,q)$, i.e. when
1. $b=c$
2. $p=g_{aa'}(c)(q)$.
The natural choice is to define $\tilde f_a$ via $(a',c,q)\mapsto (c,g_{aa'}(c)(q))$, which is clearly smooth in $c$ and $q$. Then if $\pi(a',c,q)=\pi (a,b,p)$, we have $$\tilde f_a(a',c,q)=(c,g_{aa'}(c)(q))\overset{(1)}{=}(b,g_{aa'}(b)(q)) \overset{(2)}{=}(b,p)=\tilde f_a(a,b,p),$$ as desired. This induces the trivialization $f_a$ sending $\pi(a,b,p)$ to $(b,p)$. Unpacking the definition immediately implies that $$f_a \circ f_{a'}^{-1}(b,q)= f_a\big(\pi(a',b,q)\big)= (b,g_{aa'}(b)(q)).$$