[proofplan] We first prove that the displayed relation is an [equivalence relation](/page/Equivalence%20Relation), using exactly the identity, inverse, and triple-overlap parts of the cocycle identities. We then construct local trivializations over each $U_i$ by transporting every representative to the $i$-th chart, and we verify that these maps are well-defined, bijective, and homeomorphisms. The transition maps between the resulting local models are precisely $(x,y) \mapsto (x,g_{ij}(x)(y))$, so the smoothness of the cocycle gives a smooth atlas and a smooth fibre bundle projection. Finally, in the linear case, the vector-space operations are defined in local trivializations and agree on overlaps because the transition functions lie in $GL(k,\mathbb{R})$. [/proofplan]

[step:Use the cocycle identities to prove that the gluing relation is an equivalence relation]Let $X := \bigsqcup_{i \in I} U_i \times F$. For $i,j \in I$, $x \in U_i \cap U_j$, and $y \in F$, the relation is \begin{align*} (i,x,y) \sim (j,x,g_{ji}(x)(y)). \end{align*} Reflexivity follows from $g_{ii}(x)=\operatorname{id}_F$: for every $(i,x,y) \in X$, \begin{align*} (i,x,y) \sim (i,x,g_{ii}(x)(y)) = (i,x,y). \end{align*} For symmetry, suppose \begin{align*} (i,x,y) \sim (j,x,z). \end{align*} Then $z=g_{ji}(x)(y)$. The cocycle identity with indices $i,j,i$ gives \begin{align*} g_{ij}(x)\circ g_{ji}(x)=g_{ii}(x)=\operatorname{id}_F, \end{align*} so $g_{ij}(x)(z)=y$. Hence \begin{align*} (j,x,z) \sim (i,x,g_{ij}(x)(z))=(i,x,y). \end{align*} For transitivity, suppose \begin{align*} (i,x,y) \sim (j,x,z), \qquad (j,x,z) \sim (l,x,w). \end{align*} Then $z=g_{ji}(x)(y)$ and $w=g_{lj}(x)(z)$. Therefore \begin{align*} w = g_{lj}(x)(g_{ji}(x)(y)) = (g_{lj}(x)\circ g_{ji}(x))(y) = g_{li}(x)(y), \end{align*} where the last equality is the cocycle identity on $U_i \cap U_j \cap U_l$. Hence \begin{align*} (i,x,y) \sim (l,x,g_{li}(x)(y))=(l,x,w). \end{align*} Thus $\sim$ is an equivalence relation.[/step]

[guided]The relation identifies two local descriptions of the same point in the future bundle. The rule says that if a point has fibre coordinate $y$ in the $i$-th local product piece, then its fibre coordinate in the $j$-th local product piece is $g_{ji}(x)(y)$. We verify the three equivalence-relation properties directly. First, reflexivity uses the identity part of the cocycle. Since $g_{ii}(x)=\operatorname{id}_F$ for every $x \in U_i$, we have \begin{align*} (i,x,y) \sim (i,x,g_{ii}(x)(y))=(i,x,y). \end{align*} Next, suppose $(i,x,y) \sim (j,x,z)$. By the definition of the relation, \begin{align*} z=g_{ji}(x)(y). \end{align*} To reverse the relation, we must recover $y$ from $z$ using the transition in the opposite direction. The cocycle identity gives \begin{align*} g_{ij}(x)\circ g_{ji}(x)=g_{ii}(x)=\operatorname{id}_F. \end{align*} Applying both sides to $y$ yields \begin{align*} g_{ij}(x)(z)=g_{ij}(x)(g_{ji}(x)(y))=y. \end{align*} Therefore \begin{align*} (j,x,z) \sim (i,x,g_{ij}(x)(z))=(i,x,y). \end{align*} Finally, suppose $(i,x,y) \sim (j,x,z)$ and $(j,x,z) \sim (l,x,w)$. Then the fibre coordinates satisfy \begin{align*} z = g_{ji}(x)(y). \end{align*} Also, \begin{align*} w = g_{lj}(x)(z). \end{align*} Substituting the first equality into the second gives \begin{align*} w = g_{lj}(x)(g_{ji}(x)(y)) = (g_{lj}(x)\circ g_{ji}(x))(y). \end{align*} The triple-overlap cocycle identity gives \begin{align*} g_{lj}(x)\circ g_{ji}(x)=g_{li}(x), \end{align*} so \begin{align*} w=g_{li}(x)(y). \end{align*} This is exactly the condition saying \begin{align*} (i,x,y)\sim(l,x,w). \end{align*} Thus the relation is reflexive, symmetric, and transitive.[/guided]

[step:Define the projection and local trivializations on the quotient] Let $E := X/{\sim}$ and let $q: X \to E$ be the quotient map given by $q(i,x,y)=[(i,x,y)]$. Define $\pi:E\to M$ by $\pi([(i,x,y)])=x$. This is well-defined because the relation only identifies points with the same base point $x$. The map $\pi$ is continuous. Indeed, the composite $\pi\circ q:X\to M$ restricts on the $i$-th summand $U_i\times F$ to the projection $(x,y)\mapsto x$, which is continuous. By the [universal property of the quotient topology](/theorems/1031), $\pi$ is continuous. Hence, for each $i\in I$, the set \begin{align*} E_i := \pi^{-1}(U_i) \end{align*} is open in $E$ because $U_i$ is open in $M$. For each $i\in I$, define $\Phi_i:E_i\to U_i\times F$ by \begin{align*} \Phi_i([(j,x,y)])=(x,g_{ij}(x)(y)). \end{align*} This formula is meaningful because $[(j,x,y)]\in E_i$ implies $x\in U_i\cap U_j$. We check that $\Phi_i$ is well-defined. Suppose $[(j,x,y)]=[(l,x,z)]$. Then $z=g_{lj}(x)(y)$, and the cocycle identity on $U_i\cap U_j\cap U_l$ gives \begin{align*} g_{il}(x)(z)=g_{il}(x)(g_{lj}(x)(y))=(g_{il}(x)\circ g_{lj}(x))(y)=g_{ij}(x)(y). \end{align*} Therefore $\Phi_i([(l,x,z)])=\Phi_i([(j,x,y)])$. [/step]

[step:Show that the local trivializations are homeomorphisms] For each $i\in I$, define $\Psi_i:U_i\times F\to E_i$ by $\Psi_i(x,y)=[(i,x,y)]$. Then \begin{align*} \Phi_i(\Psi_i(x,y))=\Phi_i([(i,x,y)])=(x,g_{ii}(x)(y))=(x,y), \end{align*} so $\Phi_i\circ\Psi_i=\operatorname{id}_{U_i\times F}$. Conversely, for $[(j,x,y)]\in E_i$, \begin{align*} \Psi_i(\Phi_i([(j,x,y)]))=\Psi_i(x,g_{ij}(x)(y))=[(i,x,g_{ij}(x)(y))]. \end{align*} Since $(j,x,y)\sim(i,x,g_{ij}(x)(y))$, we have $\Psi_i\circ\Phi_i=\operatorname{id}_{E_i}$. Thus $\Phi_i$ is bijective with inverse $\Psi_i$. The map $\Psi_i$ is continuous because it is the restriction of the quotient map $q$ to the open summand $U_i\times F\subset X$. To prove continuity of $\Phi_i$, let $X_i:=q^{-1}(E_i)$ with the [subspace topology](/page/Subspace%20Topology) from $X$, and let $q_i:X_i\to E_i$ be the restriction of $q$. Since $E_i$ is open in $E$, the standard restriction property for quotient maps says that $q_i$ is a quotient map. For each $j\in I$, the restriction of $\Phi_i\circ q_i$ to $(U_i\cap U_j)\times F$ is \begin{align*} (x,y)\mapsto (x,g_{ij}(x)(y)). \end{align*} This map is continuous because the cocycle is smooth as a $\operatorname{Diff}(F)$-valued cocycle, meaning its evaluation map $(x,y)\mapsto g_{ij}(x)(y)$ is smooth and hence continuous. Since the summands $(U_i\cap U_j)\times F$ cover $X_i$ with the disjoint-union topology, $\Phi_i\circ q_i$ is continuous. Because $q_i$ is quotient, $\Phi_i$ is continuous. Therefore $\Phi_i$ is a homeomorphism. [/step]

[step:Use the local trivializations to define the smooth bundle atlas] The sets $E_i=\pi^{-1}(U_i)$ cover $E$ because $(U_i)_{i \in I}$ covers $M$. For $i,j \in I$, the overlap map \begin{align*} \Phi_i \circ \Phi_j^{-1}: (U_i \cap U_j)\times F &\to (U_i \cap U_j)\times F \end{align*} is computed as follows. For $x \in U_i \cap U_j$ and $y \in F$, \begin{align*} \Phi_j^{-1}(x,y)=[(j,x,y)], \end{align*} and hence \begin{align*} (\Phi_i \circ \Phi_j^{-1})(x,y) = \Phi_i([(j,x,y)]) = (x,g_{ij}(x)(y)). \end{align*} By hypothesis, the map \begin{align*} (x,y) \mapsto g_{ij}(x)(y) \end{align*} from $(U_i \cap U_j)\times F$ to $F$ is smooth. Therefore every overlap map $\Phi_i \circ \Phi_j^{-1}$ is smooth. Its inverse is $\Phi_j \circ \Phi_i^{-1}$, which is smooth by the same argument. Thus the charts $\Phi_i$ define a smooth atlas on $E$. Each chart target $U_i\times F$ is an open subset of the smooth manifold product $M\times F$ locally modelled on Euclidean space, and the overlap maps above are smooth diffeomorphisms. Since $E$ is Hausdorff and second-countable by hypothesis, the standard compatible-atlas construction gives $E$ a smooth manifold structure. In these local coordinates, The map $\pi\circ\Phi_i^{-1}:U_i\times F\to U_i$ is given by $(x,y)\mapsto x$. This is the projection onto the first factor, so $\pi:E\to M$ is a smooth fibre bundle with typical fibre $F$ and transition functions $g_{ij}$. [/step]

[step:Prove uniqueness of the smooth fibre bundle structure] Suppose $\mathcal{A}$ is any smooth structure on the same [topological space](/page/Topological%20Space) $E$ for which the canonical maps $\Phi_i:\pi^{-1}(U_i)\to U_i\times F$ defined above are smooth local trivializations with transition functions $g_{ij}$. Then each $\Phi_i$ is a diffeomorphism from $\pi^{-1}(U_i)$ onto $U_i\times F$. Hence the smooth structure $\mathcal{A}$ contains the atlas generated by the charts $\Phi_i$. Conversely, a map from an open subset of $E$ is smooth with respect to the atlas generated by the $\Phi_i$ precisely when its coordinate representatives through the $\Phi_i$ are smooth. Since the $\Phi_i$ are smooth local trivializations for $\mathcal{A}$ and cover $E$, this is exactly the smoothness criterion in $\mathcal{A}$. Therefore $\mathcal{A}$ equals the smooth structure generated by the constructed atlas. The smooth fibre bundle structure is unique. [/step]

[step:Define the vector bundle operations in the linear case] Assume now that $F=\mathbb{R}^k$ and that each map $g_{ij}:U_i\cap U_j\to GL(k,\mathbb{R})$ is smooth. For $p\in M$, choose $i\in I$ with $p\in U_i$. The local trivialization $\Phi_i:\pi^{-1}(U_i)\to U_i\times\mathbb{R}^k$ identifies the fibre $\pi^{-1}(p)$ with $\{p\}\times\mathbb{R}^k$. Define addition and scalar multiplication on $\pi^{-1}(p)$ by transporting the usual vector-space operations on $\mathbb{R}^k$ through this identification. These operations are independent of the chosen index. Indeed, if $p\in U_i\cap U_j$, then the change of fibre coordinates is the map $\mathbb{R}^k\to\mathbb{R}^k$ given by $y\mapsto g_{ij}(p)y$, which is a linear isomorphism because $g_{ij}(p)\in GL(k,\mathbb{R})$. Therefore it preserves addition and scalar multiplication. The local trivializations $\Phi_i$ are fibrewise linear by construction, and their transition functions are the prescribed smooth $GL(k,\mathbb{R})$-valued maps $g_{ij}$. Hence $E$ is a smooth rank $k$ vector bundle over $M$. [/step]