Let $\pi:E\to M$ be a principal $G$-bundle. Suppose that $M$ is covered by open sets $U_i\subset M$ and that we can trivialize the bundle over each $U_i$ by choosing appropriate sections $\sigma_i:U_i\to \pi^{-1}(U_i)$.
If $x\in U_i\cap U_j$ then there has to be some $\Omega_{ij}(x)\in G$ with the property that $\sigma_j(x)=\sigma_i(x)\cdot \Omega_{ij}(x)$. It is simple to see that these maps obey $\Omega_{ii}=1$, $\Omega_{ij}^{-1}=\Omega_{ji}$ and $\Omega_{i\ell}=\Omega_{ij}\Omega_{j\ell}$ whenever the relevant open sets overlap.
I want to prove the following: Let $\{U_i\}$ be an open cover of $M$ and $\Omega_{ij}:U_i\cap U_j\to G$ maps obeying $\Omega_{ii}=1$, $\Omega_{ij}=\Omega_{ji}^{-1}$ and $\Omega_{ij}\Omega_{j\ell}\Omega_{\ell i}=1$. Then there exists one principal $G$-bundle $E$ which is trivializable over each $U_i$ with $\Omega_{ij}$ as transition maps.
I have tried the following. For every $i$ define $E_i= U_i\times G\times \{i\}$. Now consider the disjoint union $\tilde{E}=\bigcup_i E_i$ and define one equivalence relation on it by saying that $$(x,g,i)\sim (x,\Omega_{ji}(x)g,j),\quad \forall x\in U_i\cap U_j.$$
That this is an equivalence relation follows immediately from the properties of $\Omega_{ij}$. My idea is now to define $E = \tilde{E}/\sim$ and define a right $G$-action on $E$ which turns it into a principal $G$-bundle which can be trivialized over each $U_i$ giving rise to $\Omega_{ij}$ as the transition maps.
I define the projection as $\pi\left([(x,g,i)]\right)=x$, which is well-defined because all elements of $[(x,g,i)]$ have the same base point $x\in M$. I also define the $G$-action as $$[(x,g,i)]\cdot g'=[(x,gg',i)].$$
This is well-defined because another representative $(x,\Omega_{ji}(x)g,j)$ gets mapped to $(x,\Omega_{ji}(x)(gg'),j)$ which is in the class of $(x,gg',i)$.
Now comes the proof of the trivializations. For each $U_i$ we define $\phi_i:U_i\times G\to E$ by means of $\phi_i(x,g)=[(x,g,i)]$. This is clearly a principal bundle map since $$\phi_i(x,gg')=[(x,gg',i)]=[(x,g,i)]\cdot g'=\phi_i(x,g)\cdot g'.$$
So it seems everything is correct. Is this proof really correct?