[proofplan] We first prove that the displayed rule defines an [equivalence relation](/page/Equivalence%20Relation); the cocycle identity is exactly what makes transitivity work. We then define the projection to $B$ and construct local trivializations by identifying the preimage of $U_i$ with $U_i \times F$. The transition maps between these local trivializations are computed explicitly and are smooth because the maps $g_{ij}$ and the action of $G$ on $F$ are smooth. Finally, the assumed Hausdorff and second-countable atlas topology lets these compatible local models define a smooth manifold structure on the quotient, and the resulting projection is a smooth fibre bundle with fibre $F$ and structure group $G$. [/proofplan]

[step:Use the cocycle identities to prove that the gluing rule is an equivalence relation]Let $X := \bigsqcup_{i \in I} U_i \times F$ denote the disjoint union, whose elements are written as triples $(i,b,y)$ with $b \in U_i$ and $y \in F$. We first record two consequences of the hypotheses. For $b \in U_i \cap U_j$, applying the cocycle identity with $(i,j,k) = (i,j,i)$ gives \begin{align*} g_{ij}(b)g_{ji}(b) = g_{ii}(b) = e. \end{align*} Applying it with $(i,j,k) = (j,i,j)$ gives \begin{align*} g_{ji}(b)g_{ij}(b) = g_{jj}(b) = e. \end{align*} Thus $g_{ji}(b) = g_{ij}(b)^{-1}$. The relation is reflexive because, for $b \in U_i$ and $y \in F$, \begin{align*} (i,b,y) \sim (i,b,g_{ii}(b)\cdot y) = (i,b,e\cdot y) = (i,b,y). \end{align*} It is symmetric because if $(i,b,y) \sim (j,b,g_{ji}(b)\cdot y)$, then, using $g_{ij}(b)=g_{ji}(b)^{-1}$, \begin{align*} g_{ij}(b)\cdot \bigl(g_{ji}(b)\cdot y\bigr) = \bigl(g_{ij}(b)g_{ji}(b)\bigr)\cdot y = e\cdot y = y. \end{align*} Therefore $(j,b,g_{ji}(b)\cdot y) \sim (i,b,y)$. For transitivity, suppose $b \in U_i \cap U_j \cap U_k$ and \begin{align*} (i,b,y) \sim (j,b,g_{ji}(b)\cdot y), \qquad (j,b,g_{ji}(b)\cdot y) \sim (k,b,g_{kj}(b)\cdot(g_{ji}(b)\cdot y)). \end{align*} Using associativity of the action and the cocycle identity with indices $(k,j,i)$, we get \begin{align*} g_{kj}(b)\cdot(g_{ji}(b)\cdot y) = \bigl(g_{kj}(b)g_{ji}(b)\bigr)\cdot y = g_{ki}(b)\cdot y. \end{align*} Hence $(i,b,y) \sim (k,b,g_{ki}(b)\cdot y)$, so the relation is transitive. Therefore $\sim$ is an equivalence relation.[/step]

[guided]Let us check the relation carefully, because this is where the cocycle condition is used most directly. Define $X := \bigsqcup_{i \in I} U_i \times F$. An element of $X$ is a triple $(i,b,y)$, where $i \in I$, $b \in U_i$, and $y \in F$. The proposed gluing rule says that, whenever $b$ lies in both $U_i$ and $U_j$, the point $(i,b,y)$ in the $i$-th copy of $U_i \times F$ is identified with the point $(j,b,g_{ji}(b)\cdot y)$ in the $j$-th copy. Before proving symmetry, we need to know that $g_{ji}(b)$ is the inverse of $g_{ij}(b)$. This follows from the cocycle identity, not from an additional hypothesis. If $b \in U_i \cap U_j$, then using the cocycle identity with $(i,j,k)=(i,j,i)$ gives \begin{align*} g_{ij}(b)g_{ji}(b)=g_{ii}(b)=e. \end{align*} Using it with $(i,j,k)=(j,i,j)$ gives \begin{align*} g_{ji}(b)g_{ij}(b)=g_{jj}(b)=e. \end{align*} Thus $g_{ji}(b)=g_{ij}(b)^{-1}$ in the Lie group $G$. Now reflexivity follows from the identity element of the action. For $b \in U_i$ and $y \in F$, \begin{align*} (i,b,y) \sim (i,b,g_{ii}(b)\cdot y)=(i,b,e\cdot y)=(i,b,y). \end{align*} For symmetry, assume $(i,b,y)$ has been identified with $(j,b,g_{ji}(b)\cdot y)$. To return from the $j$-chart to the $i$-chart, we apply $g_{ij}(b)$ to the fibre coordinate. Since $g_{ij}(b)g_{ji}(b)=e$, the action gives \begin{align*} g_{ij}(b)\cdot \bigl(g_{ji}(b)\cdot y\bigr) = \bigl(g_{ij}(b)g_{ji}(b)\bigr)\cdot y = e\cdot y = y. \end{align*} Therefore $(j,b,g_{ji}(b)\cdot y) \sim (i,b,y)$. For transitivity, suppose we first pass from the $i$-copy to the $j$-copy and then from the $j$-copy to the $k$-copy. The fibre coordinate changes first by $g_{ji}(b)$ and then by $g_{kj}(b)$, so the resulting fibre coordinate is \begin{align*} g_{kj}(b)\cdot(g_{ji}(b)\cdot y). \end{align*} Because this is a left action, this equals \begin{align*} \bigl(g_{kj}(b)g_{ji}(b)\bigr)\cdot y. \end{align*} The cocycle identity with indices $(k,j,i)$ gives \begin{align*} g_{kj}(b)g_{ji}(b)=g_{ki}(b). \end{align*} Hence the two-stage identification sends $(i,b,y)$ to exactly the same point as the direct identification from the $i$-copy to the $k$-copy: \begin{align*} g_{kj}(b)\cdot(g_{ji}(b)\cdot y)=g_{ki}(b)\cdot y. \end{align*} Thus $(i,b,y) \sim (k,b,g_{ki}(b)\cdot y)$, proving transitivity. The relation is therefore an equivalence relation.[/guided]

[step:Define the projection and local trivializations over the cover] Let $q: X \to E$ be the quotient map, given by $q(i,b,y) = [(i,b,y)]$. Define $\pi: E \to B$ by $\pi([(i,b,y)]) = b$. This is well-defined because the relation only identifies triples with the same base point $b$. For each $i \in I$, define $\Psi_i: U_i \times F \to \pi^{-1}(U_i)$ by $\Psi_i(b,y) = [(i,b,y)]$. The map $\Psi_i$ is surjective: if $[(j,b,y)] \in \pi^{-1}(U_i)$, then $b \in U_i \cap U_j$, and \begin{align*} (j,b,y) \sim (i,b,g_{ij}(b)\cdot y), \end{align*} so $[(j,b,y)] = \Psi_i(b,g_{ij}(b)\cdot y)$. It is injective: if \begin{align*} \Psi_i(b,y)=\Psi_i(b',y'), \end{align*} then applying the well-defined map $\pi$ to both sides gives $b=b'$. Since the equivalence relation is generated by the displayed gluing rule and has already been proved to be exactly the relation obtained from that rule, equivalence between two representatives in the same $i$-copy over the same base point forces the fibre coordinate to be changed by $g_{ii}(b)$. Hence \begin{align*} y' = g_{ii}(b)\cdot y=e\cdot y=y. \end{align*} Thus each $\Psi_i$ is a bijection. Let $\Phi_i: \pi^{-1}(U_i) \to U_i \times F$ denote its inverse. [/step]

[step:Compute the transition maps and prove that they are smooth] For $i,j \in I$, the transition map on $\pi^{-1}(U_i \cap U_j)$ is the map $\Phi_j \circ \Phi_i^{-1}: (U_i \cap U_j)\times F \to (U_i \cap U_j)\times F$. For $(b,y) \in (U_i \cap U_j)\times F$, \begin{align*} \Phi_j\bigl(\Phi_i^{-1}(b,y)\bigr) = \Phi_j([(i,b,y)]) = (b,g_{ji}(b)\cdot y). \end{align*} Hence $\Phi_j \circ \Phi_i^{-1}$ is the map $(b,y) \mapsto (b,g_{ji}(b)\cdot y)$. This map is smooth because $g_{ji}:U_i \cap U_j \to G$ is smooth and the action map $\alpha:G\times F\to F$ is smooth. Its inverse is the map $\Phi_i \circ \Phi_j^{-1}: (U_i \cap U_j)\times F \to (U_i \cap U_j)\times F$ given by $(b,y) \mapsto (b,g_{ij}(b)\cdot y)$, which is smooth by the same argument. Therefore the transition maps are diffeomorphisms. [/step]

[step:Put the smooth structure on the quotient using the local trivializations]Declare a subset $W \subset E$ to be open precisely when $\Phi_i(W \cap \pi^{-1}(U_i))$ is open in $U_i \times F$ for every $i \in I$. This defines a topology: arbitrary unions and finite intersections are tested after applying each $\Phi_i$, and the corresponding properties hold in the topological spaces $U_i \times F$. For each $i$, the set $\pi^{-1}(U_i)$ is open because, for every $j \in I$, \begin{align*} \Phi_j(\pi^{-1}(U_i) \cap \pi^{-1}(U_j))=(U_i \cap U_j)\times F, \end{align*} which is open in $U_j \times F$. The map $\Phi_i$ is a homeomorphism from $\pi^{-1}(U_i)$ onto $U_i \times F$. Its inverse $\Psi_i$ is continuous because, if $A \subset \pi^{-1}(U_i)$ is open, then $\Psi_i^{-1}(A)=\Phi_i(A)$ is open in $U_i \times F$. The map $\Phi_i$ is continuous because, for every [open set](/page/Open%20Set) $O \subset U_i \times F$ and every $j \in I$, \begin{align*} \Phi_j(\Phi_i^{-1}(O)\cap \pi^{-1}(U_j))=(\Phi_j\circ \Phi_i^{-1})(O\cap ((U_i\cap U_j)\times F)), \end{align*} and this is open in $U_j\times F$ since $\Phi_j\circ\Phi_i^{-1}$ is a homeomorphism on $(U_i\cap U_j)\times F$. Choose a countable smooth atlas $(V_m,\varphi_m)_{m\in\mathbb N}$ for $B$ and a countable smooth atlas $(W_n,\psi_n)_{n\in\mathbb N}$ for $F$. Since $B$ is second-countable, let $(C_s)_{s\in\mathbb N}$ be a countable basis for its topology. For each $b\in B$, choose $i_b\in I$ with $b\in U_{i_b}$ and then choose a basis element $C_s$ with $b\in C_s\subset U_{i_b}$; the collection of all such $C_s$ is countable, covers $B$, and each of its members is contained in some $U_i$. Enumerate this collection as $(A_r)_{r\in\mathbb N}$ and choose an index $i(r)\in I$ such that $A_r\subset U_{i(r)}$. The charts \begin{align*} \bigl(\Phi_{i(r)}^{-1}((A_r\cap V_m)\times W_n),\, (\varphi_m\times\psi_n)\circ \Phi_{i(r)}\bigr) \end{align*} with $r,m,n\in\mathbb N$ and $A_r\cap V_m\neq\varnothing$ form a countable atlas on $E$. Their chart changes are compositions of product chart changes on $B\times F$ with the transition diffeomorphisms computed above, so all chart changes are smooth. The topology is Hausdorff. If two points of $E$ have the same base point $b$, choose $i\in I$ with $b\in U_i$; both points lie in the chart domain $\pi^{-1}(U_i)$, and the Hausdorff product manifold $U_i\times F$ separates their images under $\Phi_i$. If their base points are distinct, choose disjoint open neighbourhoods in the Hausdorff manifold $B$ and take their inverse images under $\pi$; these inverse images are open because $\pi$ is locally $\operatorname{pr}_1\circ\Phi_i$ in the above charts. Thus $E$ is Hausdorff. The countable atlas above gives second-countability. Therefore these charts define a smooth manifold structure on $E$ for which each $\Phi_i$ is a diffeomorphism.[/step]

[guided]The delicate point is that compatible charts alone do not yet give a smooth manifold in the usual sense; we must also verify the topology, Hausdorffness, and second-countability. We define the topology by declaring $W\subset E$ open exactly when every local representative $\Phi_i(W\cap\pi^{-1}(U_i))$ is open in $U_i\times F$. This prescription is stable under arbitrary unions and finite intersections because those operations are checked inside the ordinary topological spaces $U_i\times F$. First we check that the local domains are genuinely open and that the maps $\Phi_i$ are homeomorphisms. For $i,j\in I$, the part of $\pi^{-1}(U_i)$ seen in the $j$-trivialization is \begin{align*} \Phi_j(\pi^{-1}(U_i)\cap\pi^{-1}(U_j))=(U_i\cap U_j)\times F. \end{align*} This is open in $U_j\times F$, so $\pi^{-1}(U_i)$ is open by the definition of the topology. The inverse $\Psi_i$ of $\Phi_i$ is continuous because an open set $A\subset\pi^{-1}(U_i)$ pulls back to $\Phi_i(A)$, which is open by definition. Conversely, if $O\subset U_i\times F$ is open, then in the $j$-trivialization its image is \begin{align*} \Phi_j(\Phi_i^{-1}(O)\cap\pi^{-1}(U_j))=(\Phi_j\circ\Phi_i^{-1})(O\cap((U_i\cap U_j)\times F)). \end{align*} The set on the right is open because the transition map $\Phi_j\circ\Phi_i^{-1}$ is a homeomorphism. Hence $\Phi_i$ is continuous, and $\Phi_i$ is a homeomorphism. Now we build a countable smooth atlas. Since $B$ and $F$ are smooth manifolds, choose countable smooth atlases $(V_m,\varphi_m)_{m\in\mathbb N}$ for $B$ and $(W_n,\psi_n)_{n\in\mathbb N}$ for $F$. Since $B$ is second-countable, it has a countable basis $(C_s)_{s\in\mathbb N}$. The cover $(U_i)_{i\in I}$ lets us refine this basis as follows: for each $b\in B$, choose an index $i_b\in I$ with $b\in U_{i_b}$, then choose a basis element $C_s$ satisfying $b\in C_s\subset U_{i_b}$. The selected basis elements are still drawn from a countable family, they cover $B$, and each selected element is contained in one member of the original cover. Enumerate them as $(A_r)_{r\in\mathbb N}$ and choose $i(r)\in I$ with $A_r\subset U_{i(r)}$. Transporting product charts through $\Phi_{i(r)}$ gives the countable family \begin{align*} \bigl(\Phi_{i(r)}^{-1}((A_r\cap V_m)\times W_n),\, (\varphi_m\times\psi_n)\circ\Phi_{i(r)}\bigr). \end{align*} The chart changes are obtained by composing ordinary product chart changes with the transition maps $(b,y)\mapsto(b,g_{ji}(b)\cdot y)$, already proved smooth. Hence this countable atlas is smoothly compatible. It remains to verify Hausdorffness. If two points of $E$ project to the same $b\in B$, choose an index $i$ with $b\in U_i$. Both points lie in $\pi^{-1}(U_i)$, and $\Phi_i$ identifies this set with the Hausdorff product manifold $U_i\times F$, so the two points have disjoint neighbourhoods in $E$. If two points project to distinct points of $B$, choose disjoint open neighbourhoods of those base points in the Hausdorff manifold $B$. Their inverse images under $\pi$ are open because locally $\pi$ is the projection $\operatorname{pr}_1\circ\Phi_i$, and these inverse images separate the two points. Therefore $E$ is Hausdorff. Together with the countable atlas, this proves that $E$ is a smooth manifold and that each $\Phi_i$ is a diffeomorphism.[/guided]

[step:Verify the fibre bundle structure and identify the structure group] For each $i \in I$, the map $\Phi_i$ satisfies \begin{align*} \operatorname{pr}_1 \circ \Phi_i = \pi|_{\pi^{-1}(U_i)}, \end{align*} where $\operatorname{pr}_1: U_i \times F \to U_i$ is the first projection, given by $\operatorname{pr}_1(b,y)=b$. Therefore $\Phi_i$ is a local trivialization of $\pi$ over $U_i$. On an overlap $U_i \cap U_j$, the change of trivialization sends a point $(b,y)\in (U_i\cap U_j)\times F$ to $(b,g_{ji}(b)\cdot y)$. Equivalently, the inverse change from the $j$-trivialization to the $i$-trivialization sends $(b,y)$ to $(b,g_{ij}(b)\cdot y)$. Thus every transition function is given by the smooth $G$-valued map $g_{ij}$ through the given smooth Lie [group action](/page/Group%20Action) on the fibre $F$. Consequently $\pi:E\to B$ is a smooth fibre bundle with fibre $F$ and structure group $G$. [/step]