[proofplan] We send a principal bundle trivialized over the cover to its transition functions and verify that the transition functions form a nonabelian Čech $1$-cocycle. We then show that changing local sections, or replacing the bundle by an isomorphic one, changes the cocycle exactly by the displayed coboundary formula. Conversely, we reconstruct a principal bundle from a cocycle by gluing the local products $U_i \times G$ with the prescribed transition functions. Finally, we check that coboundary equivalent cocycles give isomorphic reconstructed bundles and that every isomorphism between reconstructed bundles yields such a coboundary. [/proofplan]

[step:Extract a smooth cocycle from local trivializations] Let $\mathcal U := (U_i)_{i \in I}$ denote the fixed open cover of $M$. Define $Z^1(\mathcal U,G)$ to be the set of families $g=(g_{ij})_{i,j \in I}$ of smooth maps \begin{align*} g_{ij}: U_i \cap U_j \to G \end{align*} such that, for every $x \in U_i \cap U_j \cap U_k$, \begin{align*} g_{ij}(x)g_{jk}(x) = g_{ik}(x). \end{align*} These are the smooth nonabelian Cech $1$-cocycles on the cover $\mathcal U$ with values in $G$. Let $\pi: P \to M$ be a smooth principal right $G$-bundle such that $\pi^{-1}(U_i) \to U_i$ is smoothly trivial for every $i \in I$. Choose, for each $i \in I$, a smooth local section \begin{align*} \sigma_i: U_i \to P \end{align*} of $\pi$, meaning $\pi \circ \sigma_i = \operatorname{id}_{U_i}$. On $U_i \cap U_j$, the points $\sigma_i(x)$ and $\sigma_j(x)$ lie in the same $G$-orbit in the fiber $P_x = \pi^{-1}(\{x\})$. Since the right $G$-action on each fiber is free and transitive, there is a unique element $g_{ij}(x) \in G$ such that \begin{align*} \sigma_j(x) = \sigma_i(x)g_{ij}(x). \end{align*} This defines a map \begin{align*} g_{ij}: U_i \cap U_j \to G. \end{align*} The map $g_{ij}$ is smooth because, in the local trivialization determined by $\sigma_i$, the section $\sigma_j$ has second coordinate $g_{ij}$. More explicitly, the trivialization \begin{align*} \Phi_i: U_i \times G \to \pi^{-1}(U_i), \quad (x,h) \mapsto \sigma_i(x)h \end{align*} is a smooth principal bundle isomorphism, and \begin{align*} \Phi_i^{-1}(\sigma_j(x)) = (x,g_{ij}(x)). \end{align*} Thus $g_{ij}$ is smooth. On a triple overlap $U_i \cap U_j \cap U_k$, we compute using associativity of the right action: \begin{align*} \sigma_k(x) = \sigma_j(x)g_{jk}(x) = \sigma_i(x)g_{ij}(x)g_{jk}(x). \end{align*} Also, by definition of $g_{ik}$, \begin{align*} \sigma_k(x) = \sigma_i(x)g_{ik}(x). \end{align*} Freeness of the right $G$-action on the fiber $P_x$ gives \begin{align*} g_{ij}(x)g_{jk}(x) = g_{ik}(x). \end{align*} Hence $(g_{ij}) \in Z^1(\mathcal U,G)$. [/step]

[step:Identify the effect of changing local sections]Suppose that $\sigma_i': U_i \to P$ is another family of smooth local sections. For each $i \in I$, freeness and transitivity of the right action on fibers give a unique smooth map \begin{align*} a_i: U_i \to G \end{align*} such that \begin{align*} \sigma_i'(x) = \sigma_i(x)a_i(x). \end{align*} Let $g'_{ij}: U_i \cap U_j \to G$ be the transition functions determined by the sections $\sigma_i'$, so that \begin{align*} \sigma_j'(x) = \sigma_i'(x)g'_{ij}(x). \end{align*} On $U_i \cap U_j$, we have \begin{align*} \sigma_i(x)g_{ij}(x)a_j(x) = \sigma_j(x)a_j(x) = \sigma_j'(x) = \sigma_i'(x)g'_{ij}(x) = \sigma_i(x)a_i(x)g'_{ij}(x). \end{align*} Freeness of the right action gives \begin{align*} g_{ij}(x)a_j(x) = a_i(x)g'_{ij}(x). \end{align*} Multiplying on the left by $a_i(x)^{-1}$ gives \begin{align*} g'_{ij}(x) = a_i(x)^{-1}g_{ij}(x)a_j(x). \end{align*} Thus changing local sections changes the cocycle by a coboundary.[/step]

[guided]The point of this step is to verify that the cocycle does not depend on the arbitrary choice of local sections except through the prescribed [equivalence relation](/page/Equivalence%20Relation). Since $\sigma_i$ and $\sigma_i'$ are two sections of the same principal bundle over $U_i$, they meet the same fiber over each $x \in U_i$. The right $G$-action on that fiber is free and transitive, so there is a unique element $a_i(x) \in G$ with \begin{align*} \sigma_i'(x) = \sigma_i(x)a_i(x). \end{align*} This defines a smooth map $a_i: U_i \to G$ because it is the second coordinate of $\sigma_i'$ in the smooth trivialization associated to $\sigma_i$. Now compare the two ways of expressing $\sigma_j'(x)$ over an overlap $U_i \cap U_j$. First use the old transition function and then the change of section on $U_j$: \begin{align*} \sigma_j'(x) = \sigma_j(x)a_j(x) = \sigma_i(x)g_{ij}(x)a_j(x). \end{align*} Second use the new section on $U_i$ and the new transition function: \begin{align*} \sigma_j'(x) = \sigma_i'(x)g'_{ij}(x) = \sigma_i(x)a_i(x)g'_{ij}(x). \end{align*} Both expressions are points in the same fiber written as $\sigma_i(x)$ acted on by an element of $G$. Freeness of the right action therefore forces the two group elements to agree: \begin{align*} g_{ij}(x)a_j(x) = a_i(x)g'_{ij}(x). \end{align*} Solving this equation in the group $G$ gives \begin{align*} g'_{ij}(x) = a_i(x)^{-1}g_{ij}(x)a_j(x). \end{align*} This is exactly the nonabelian Čech coboundary formula.[/guided]

[step:Show that bundle isomorphisms preserve the cocycle class] Let $\pi: P \to M$ and $\pi': P' \to M$ be smooth principal right $G$-bundles trivialized over $\mathcal U$, and let \begin{align*} F: P \to P' \end{align*} be a principal bundle isomorphism over $M$. Choose smooth local sections $\sigma_i: U_i \to P$ and $\sigma_i': U_i \to P'$, with transition functions $g_{ij}$ and $g'_{ij}$ respectively. For each $i \in I$, define the smooth map \begin{align*} b_i: U_i \to G \end{align*} by the condition \begin{align*} F(\sigma_i(x)) = \sigma_i'(x)b_i(x). \end{align*} On $U_i \cap U_j$, equivariance of $F$ gives \begin{align*} F(\sigma_j(x)) = F(\sigma_i(x)g_{ij}(x)) = F(\sigma_i(x))g_{ij}(x) = \sigma_i'(x)b_i(x)g_{ij}(x). \end{align*} Let $P'_x := (\pi')^{-1}(\{x\})$ denote the fiber of $P'$ over $x \in M$. On the other hand, \begin{align*} F(\sigma_j(x)) = \sigma_j'(x)b_j(x) = \sigma_i'(x)g'_{ij}(x)b_j(x). \end{align*} Freeness of the right action on $P'_x$ gives \begin{align*} b_i(x)g_{ij}(x) = g'_{ij}(x)b_j(x). \end{align*} Define $a_i: U_i \to G$ by $a_i(x) = b_i(x)^{-1}$. Then \begin{align*} g'_{ij}(x) = a_i(x)^{-1}g_{ij}(x)a_j(x). \end{align*} Hence isomorphic bundles determine coboundary equivalent cocycles. [/step]

[step:Reconstruct a principal bundle from a cocycle] Let $g = (g_{ij}) \in Z^1(\mathcal U,G)$. Let $e_G \in G$ denote the identity element of the Lie group $G$. Define the smooth disjoint union \begin{align*} Q := \bigsqcup_{i \in I} U_i \times G. \end{align*} For points $(i,x,h) \in U_i \times G$ and $(j,y,k) \in U_j \times G$, define an equivalence relation by declaring \begin{align*} (i,x,h) \sim (j,y,k) \end{align*} if and only if $x=y \in U_i \cap U_j$ and \begin{align*} h = g_{ij}(x)k. \end{align*} The relation is reflexive because the cocycle identity with $i=j=k$ gives $g_{ii}(x)=e_G$. It is symmetric because the cocycle identity with indices $i,j,i$ gives $g_{ij}(x)g_{ji}(x)=e_G$, hence $g_{ji}(x)=g_{ij}(x)^{-1}$. It is transitive because, if $h = g_{ij}(x)k$ and $k = g_{j\ell}(x)m$, then \begin{align*} h = g_{ij}(x)g_{j\ell}(x)m = g_{i\ell}(x)m \end{align*} by the cocycle identity on $U_i \cap U_j \cap U_\ell$. Let \begin{align*} P_g := Q/{\sim} \end{align*} and denote the equivalence class of $(i,x,h)$ by $[i,x,h]$. Define \begin{align*} \pi_g: P_g \to M, \quad [i,x,h] \mapsto x. \end{align*} This is well-defined because equivalent representatives have the same base point. Define a right action of $G$ on $P_g$ by \begin{align*} [i,x,h]\cdot r := [i,x,hr] \end{align*} for $r \in G$. This is well-defined because, if $h = g_{ij}(x)k$, then $hr = g_{ij}(x)kr$. For each $i \in I$, define \begin{align*} \Psi_i: U_i \times G \to \pi_g^{-1}(U_i), \quad (x,h) \mapsto [i,x,h]. \end{align*} The map $\Psi_i$ is bijective. Surjectivity follows because every class over $x \in U_i$ has a representative $[j,x,k]$, and then \begin{align*} [j,x,k] = [i,x,g_{ij}(x)k]. \end{align*} Injectivity follows because $[i,x,h]=[i,x,k]$ implies $h=g_{ii}(x)k=k$. Endow $P_g$ with the [quotient topology](/page/Quotient%20Topology) from $Q$. The subsets $\Psi_i(U_i\times G)$ cover $P_g$, and the quotient map restricts on each $U_i\times G$ to the bijection $\Psi_i$. We use the maps $\Psi_i^{-1}:\Psi_i(U_i\times G)\to U_i\times G$ as charts. On overlaps, the transition map is \begin{align*} \Psi_i^{-1}\circ \Psi_j: (U_i \cap U_j)\times G \to (U_i \cap U_j)\times G, \quad (x,h) \mapsto (x,g_{ij}(x)h), \end{align*} with inverse \begin{align*} (x,h) \mapsto (x,g_{ji}(x)h). \end{align*} Both maps are smooth because $g_{ij}$ and $g_{ji}$ are smooth and multiplication in the Lie group $G$ is smooth. Thus the product charts are smoothly compatible. Since $M$ and $G$ are smooth manifolds of the intended Hausdorff and second-countable class, and since the cover $\mathcal U$ is numerable, the standard chart-gluing criterion applies: the quotient topology is Hausdorff and second-countable, the charts above form a smooth atlas, and the sets $\Psi_i(U_i\times G)$ are open. With this structure the maps $\Psi_i$ are diffeomorphisms onto open submanifolds covering $P_g$, and the projection and right action are smooth in these charts. Therefore $\pi_g: P_g \to M$ is a smooth principal right $G$-bundle, trivialized over each $U_i$ by $\Psi_i$. [/step]

[step:Show that reconstruction inverts the transition function construction] Let $\pi: P \to M$ be a smooth principal right $G$-bundle with local sections $\sigma_i: U_i \to P$ and transition cocycle $g=(g_{ij})$. Let $P_g$ be the reconstructed bundle. Define \begin{align*} \Theta: P_g \to P \end{align*} by \begin{align*} \Theta([i,x,h]) := \sigma_i(x)h. \end{align*} This is well-defined: if $[i,x,h]=[j,x,k]$, then $h=g_{ij}(x)k$, and therefore \begin{align*} \sigma_i(x)h = \sigma_i(x)g_{ij}(x)k = \sigma_j(x)k. \end{align*} The map $\Theta$ covers $\operatorname{id}_M$ because $\pi(\sigma_i(x)h)=x$, and it is $G$-equivariant because \begin{align*} \Theta([i,x,h]\cdot r) = \Theta([i,x,hr]) = \sigma_i(x)hr = \Theta([i,x,h])r. \end{align*} On each $U_i$, the map $\Theta$ is exactly the local trivialization \begin{align*} U_i \times G \to \pi^{-1}(U_i), \quad (x,h) \mapsto \sigma_i(x)h. \end{align*} Thus $\Theta$ is a smooth principal bundle isomorphism. Hence reconstructing from the transition cocycle of $P$ recovers a bundle isomorphic to $P$. [/step]

[step:Relate coboundaries to isomorphisms of reconstructed bundles] Let $g,g' \in Z^1(\mathcal U,G)$ be cocycles satisfying \begin{align*} g'_{ij}(x) = a_i(x)^{-1}g_{ij}(x)a_j(x) \end{align*} for smooth maps $a_i: U_i \to G$. Define a map \begin{align*} F: P_g \to P_{g'} \end{align*} by the local formula \begin{align*} F([i,x,h]) := [i,x,a_i(x)^{-1}h]'. \end{align*} Here $[i,x,h]'$ denotes the equivalence class in $P_{g'}$. To check that $F$ is well-defined, suppose $[i,x,h]=[j,x,k]$ in $P_g$. Then $h=g_{ij}(x)k$. In $P_{g'}$, \begin{align*} a_i(x)^{-1}h = a_i(x)^{-1}g_{ij}(x)k = g'_{ij}(x)a_j(x)^{-1}k. \end{align*} Therefore \begin{align*} [i,x,a_i(x)^{-1}h]' = [j,x,a_j(x)^{-1}k]'. \end{align*} The local formula is smooth in the product charts, covers $\operatorname{id}_M$, and is $G$-equivariant because right multiplication commutes with the displayed formula: \begin{align*} F([i,x,h]\cdot r) = [i,x,a_i(x)^{-1}hr]' = F([i,x,h])\cdot r. \end{align*} The inverse is given locally by \begin{align*} [i,x,h]' \mapsto [i,x,a_i(x)h]. \end{align*} Thus coboundary equivalent cocycles reconstruct isomorphic principal bundles. Conversely, let \begin{align*} F: P_g \to P_{g'} \end{align*} be a principal bundle isomorphism over $M$. In the local trivialization over $U_i$, define the smooth map \begin{align*} b_i: U_i \to G \end{align*} by \begin{align*} F([i,x,e_G]) = [i,x,b_i(x)]'. \end{align*} Equivariance gives \begin{align*} F([i,x,h]) = [i,x,b_i(x)h]'. \end{align*} Since $[j,x,h]=[i,x,g_{ij}(x)h]$ in $P_g$, applying $F$ and using the equivalence relation in $P_{g'}$ gives \begin{align*} [j,x,b_j(x)h]' = [i,x,b_i(x)g_{ij}(x)h]'. \end{align*} By the defining relation in $P_{g'}$, this equality for all $h \in G$ implies \begin{align*} b_i(x)g_{ij}(x) = g'_{ij}(x)b_j(x). \end{align*} With $a_i(x)=b_i(x)^{-1}$, this becomes \begin{align*} g'_{ij}(x) = a_i(x)^{-1}g_{ij}(x)a_j(x). \end{align*} Therefore two reconstructed bundles are isomorphic if and only if their cocycles are coboundary equivalent. [/step]

[step:Conclude the classification bijection] The preceding steps define a map from isomorphism classes of smooth principal right $G$-bundles trivialized over $\mathcal U$ to coboundary equivalence classes in $Z^1(\mathcal U,G)$. The construction of $P_g$ from a cocycle gives a map in the reverse direction. The reconstruction step shows that starting with a bundle and then reconstructing from its transition cocycle gives an isomorphic bundle. The coboundary-isomorphism step shows that starting with a cocycle and passing to the reconstructed bundle remembers exactly its coboundary class. Hence the two maps are inverse bijections, proving the classification. [/step]