[proofplan] We first verify that the transition prescription defines an [equivalence relation](/page/Equivalence%20Relation) and hence a quotient set $E$. Then we use the representatives with a fixed index $i$ to build local bijections between $\pi^{-1}(U_i)$ and $U_i \times F$, and we declare the topology and smooth structure on $E$ by requiring these maps to be coordinate trivializations. The stated Hausdorff and second-countability hypothesis supplies the topological manifold conditions. The cocycle identity makes the local descriptions compatible, because the changes of trivialization are exactly the given smooth maps $G_{ji}$. In the vector bundle case, the linear structure on each fibre is transported from $\mathbb{R}^k$ through the local trivializations, and the linearity of the transition maps makes this independent of the chosen trivialization. [/proofplan]

[step:Verify that the transition rule defines an equivalence relation]Let \begin{align*} X := \bigsqcup_{i \in I} (U_i \times F) \end{align*} and write an element of the summand indexed by $i$ as $(x,v,i)$, where $x \in U_i$ and $v \in F$. Define a binary relation $\sim$ on $X$ by \begin{align*} (x,v,i) \sim (y,w,j) \end{align*} if and only if $x=y \in U_i \cap U_j$ and \begin{align*} w = g_{ji}(x)(v). \end{align*} Reflexivity follows from $g_{ii}(x)=\operatorname{id}_F$: for $x \in U_i$ and $v \in F$, \begin{align*} g_{ii}(x)(v)=v. \end{align*} To prove symmetry, suppose $(x,v,i) \sim (x,w,j)$. Then $w=g_{ji}(x)(v)$. 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*} Thus \begin{align*} g_{ij}(x)(w)=g_{ij}(x)(g_{ji}(x)(v))=v, \end{align*} so $(x,w,j) \sim (x,v,i)$. To prove transitivity, suppose \begin{align*} (x,v,i) \sim (x,w,j), \qquad (x,w,j) \sim (x,z,k). \end{align*} Then $w = g_{ji}(x)(v)$ and $z = g_{kj}(x)(w)$. Using the cocycle identity with indices $(k,j,i)$, \begin{align*} g_{kj}(x) \circ g_{ji}(x)=g_{ki}(x). \end{align*} Therefore \begin{align*} z = g_{kj}(x)(g_{ji}(x)(v)) = g_{ki}(x)(v), \end{align*} so $(x,v,i)\sim (x,z,k)$. Hence $\sim$ is an equivalence relation.[/step]

[guided]The relation is designed to say that the point with local fibre coordinate $v$ in the $i$-th trivialization is the same point as the one with local fibre coordinate $g_{ji}(x)(v)$ in the $j$-th trivialization. To be a quotient construction, this prescription must be an equivalence relation. Let \begin{align*} X := \bigsqcup_{i \in I} (U_i \times F) \end{align*} and write elements as triples $(x,v,i)$ with $x \in U_i$ and $v \in F$. Define \begin{align*} (x,v,i) \sim (y,w,j) \end{align*} precisely when $x=y \in U_i \cap U_j$ and \begin{align*} w = g_{ji}(x)(v). \end{align*} Reflexivity uses the identity transition function. If $(x,v,i) \in X$, then $x \in U_i$, and the hypothesis gives $g_{ii}(x)=\operatorname{id}_F$. Hence \begin{align*} g_{ii}(x)(v)=v, \end{align*} so $(x,v,i)\sim (x,v,i)$. For symmetry, assume $(x,v,i)\sim (x,w,j)$. This means \begin{align*} w=g_{ji}(x)(v). \end{align*} We need to recover $v$ from $w$ using the reverse transition map. The cocycle identity applied on $U_i \cap U_j$ with indices $(i,j,i)$ gives \begin{align*} g_{ij}(x)\circ g_{ji}(x)=g_{ii}(x)=\operatorname{id}_F. \end{align*} Applying both sides to $v$ yields \begin{align*} g_{ij}(x)(w) = g_{ij}(x)(g_{ji}(x)(v)) = v. \end{align*} This is exactly the condition that $(x,w,j)\sim (x,v,i)$. For transitivity, assume \begin{align*} (x,v,i)\sim (x,w,j), \qquad (x,w,j)\sim (x,z,k). \end{align*} By definition of the relation, $w = g_{ji}(x)(v)$ and $z = g_{kj}(x)(w)$. Substituting the first equality into the second gives \begin{align*} z=g_{kj}(x)(g_{ji}(x)(v)). \end{align*} The cocycle identity on $U_i\cap U_j\cap U_k$, with indices $(k,j,i)$, says \begin{align*} g_{kj}(x)\circ g_{ji}(x)=g_{ki}(x). \end{align*} Therefore \begin{align*} z=g_{ki}(x)(v), \end{align*} which is exactly the condition $(x,v,i)\sim (x,z,k)$. Thus the relation is reflexive, symmetric, and transitive.[/guided]

[step:Define the quotient projection and local trivialization maps] Let \begin{align*} E := X/{\sim} \end{align*} and denote the equivalence class of $(x,v,i)$ by $[(x,v,i)]$. Define the map $\pi:E \to M$ by \begin{align*} \pi([(x,v,i)]) = x. \end{align*} This is well-defined because equivalent triples have the same base point by definition of $\sim$. For each $i \in I$, define \begin{align*} E_i := \pi^{-1}(U_i). \end{align*} Every class in $E_i$ has an $i$-representative: if $[(x,w,j)] \in E_i$, then $x \in U_i \cap U_j$, and \begin{align*} [(x,w,j)] = [(x,g_{ij}(x)(w),i)]. \end{align*} Define the map $\Phi_i:E_i \to U_i \times F$ by \begin{align*} \Phi_i([(x,v,i)]) = (x,v). \end{align*} This map is well-defined: if $[(x,v,i)]=[(x,w,i)]$, then $w=g_{ii}(x)(v)=v$. It is bijective, with inverse $\Psi_i:U_i \times F \to E_i$ given by \begin{align*} \Psi_i(x,v)=[(x,v,i)]. \end{align*} Thus each $E_i$ is identified set-theoretically with $U_i \times F$. [/step]

[step:Put the unique topology and smooth structure making the local trivializations smooth]Declare a subset $A \subset E$ to be open if and only if $\Phi_i(A\cap E_i)$ is open in the product manifold $U_i \times F$ for every $i \in I$. With this topology, each $E_i$ is open in $E$, since for every $j \in I$, \begin{align*} \Phi_j(E_i\cap E_j)=(U_i\cap U_j)\times F, \end{align*} which is open in $U_j\times F$. For each $i \in I$, the map \begin{align*} \Phi_i:E_i \to U_i \times F \end{align*} is continuous by the definition of the topology. Its inverse is the map $\Psi_i:U_i\times F\to E_i$ defined above. To prove that $\Psi_i$ is continuous, let $B\subset U_i\times F$ be open. For every $j\in I$, define the transition diffeomorphism \begin{align*} G_{ji}:(U_i\cap U_j)\times F &\to (U_i\cap U_j)\times F \end{align*} by \begin{align*} G_{ji}(x,v)=(x,g_{ji}(x)(v)). \end{align*} Then \begin{align*} \Phi_j(\Psi_i(B)\cap E_j)=G_{ji}\bigl(B\cap((U_i\cap U_j)\times F)\bigr), \end{align*} which is open in $U_j\times F$ because $B\cap((U_i\cap U_j)\times F)$ is open in $(U_i\cap U_j)\times F$ and $G_{ji}$ is a diffeomorphism. Hence $\Psi_i(B)$ is open in $E$, so $\Psi_i$ is continuous. Therefore each $\Phi_i$ is a homeomorphism. Moreover, $\pi$ is continuous: for an [open set](/page/Open%20Set) $V \subset M$, \begin{align*} \Phi_i(\pi^{-1}(V)\cap E_i)=(V\cap U_i)\times F, \end{align*} which is open in $U_i\times F$ for every $i \in I$. The domains $(E_i)_{i \in I}$ cover $E$, because every class $[(x,v,j)]$ has base point $x \in M$ and the cover $(U_i)_{i \in I}$ contains some $U_i$ with $x \in U_i$. On overlaps, for $i,j \in I$, the common chart domain is $E_i \cap E_j = \pi^{-1}(U_i \cap U_j)$, and the change of local coordinates is the map $\Phi_j \circ \Phi_i^{-1}: (U_i \cap U_j)\times F \to (U_i \cap U_j)\times F$ given by \begin{align*} (\Phi_j \circ \Phi_i^{-1})(x,v)=(x,g_{ji}(x)(v)). \end{align*} This is precisely the map $G_{ji}$ restricted to $(U_i\cap U_j)\times F$, and it is a smooth diffeomorphism by hypothesis. Thus the chart domains cover $E$ and all coordinate changes on overlaps are smooth diffeomorphisms. Since each model space $U_i\times F$ is a smooth manifold, it is locally Euclidean. By the standard atlas construction theorem for smooth manifolds, the product smooth structures on the charts $U_i \times F$ are smoothly compatible and generate an atlas on $E$; taking the maximal smooth atlas containing these charts gives a unique smooth structure on $E$ for which every $\Phi_i$ is a diffeomorphism onto $U_i\times F$. By the topological hypothesis in the statement, this atlas topology is Hausdorff and second-countable. Therefore the atlas just constructed makes $E$ a smooth manifold in the usual sense.[/step]

[guided]We now turn the quotient set into a smooth manifold by using the proposed local trivializations as charts. Declare a subset $A \subset E$ to be open precisely when $\Phi_i(A\cap E_i)$ is open in the product manifold $U_i \times F$ for every $i \in I$. With this topology, each $E_i$ is open: indeed, for every $j \in I$, \begin{align*} \Phi_j(E_i \cap E_j)=(U_i\cap U_j)\times F, \end{align*} which is open in $U_j\times F$ because $U_i\cap U_j$ is open in $U_j$. By the definition of the topology, each map \begin{align*} \Phi_i:E_i \to U_i\times F \end{align*} is continuous. We also need the inverse map to be continuous. Recall that $\Psi_i:U_i\times F\to E_i$ is defined by $\Psi_i(x,v)=[(x,v,i)]$. Let $B\subset U_i\times F$ be open. To test whether $\Psi_i(B)$ is open in $E$, the topology asks us to inspect its image under every chart $\Phi_j$. For each $j\in I$, define \begin{align*} G_{ji}:(U_i\cap U_j)\times F &\to (U_i\cap U_j)\times F \end{align*} by \begin{align*} G_{ji}(x,v)=(x,g_{ji}(x)(v)). \end{align*} Then the equivalence relation gives \begin{align*} \Phi_j(\Psi_i(B)\cap E_j)=G_{ji}\bigl(B\cap((U_i\cap U_j)\times F)\bigr). \end{align*} The set $B\cap((U_i\cap U_j)\times F)$ is open in $(U_i\cap U_j)\times F$, and $G_{ji}$ is a diffeomorphism by hypothesis, hence its image is open in $(U_i\cap U_j)\times F$ and therefore open in $U_j\times F$. Thus $\Psi_i(B)$ is open in $E$. This proves that $\Psi_i$ is continuous, and consequently $\Phi_i$ is a homeomorphism. We also verify that the projection is continuous before using it in the separation argument. If $V \subset M$ is open, then for every $i \in I$, \begin{align*} \Phi_i(\pi^{-1}(V)\cap E_i)=(V\cap U_i)\times F, \end{align*} which is open in $U_i\times F$. Hence $\pi^{-1}(V)$ is open in $E$, so $\pi$ is continuous. The atlas compatibility is exactly where the transition-data hypothesis is used. The chart domains $(E_i)_{i \in I}$ cover $E$: if $[(x,v,j)]\in E$, then $x\in M$, so some $U_i$ contains $x$, and the class lies in $E_i=\pi^{-1}(U_i)$. On the overlap $E_i\cap E_j=\pi^{-1}(U_i\cap U_j)$, the coordinate change is the map \begin{align*} \Phi_j\circ \Phi_i^{-1}:(U_i\cap U_j)\times F &\to (U_i\cap U_j)\times F \end{align*} given by \begin{align*} (\Phi_j\circ \Phi_i^{-1})(x,v)=(x,g_{ji}(x)(v)). \end{align*} This is the induced transition map from the hypotheses, restricted to the overlap, and therefore it is a smooth diffeomorphism. Thus the chart domains cover $E$ and all coordinate changes are smooth diffeomorphisms. Each target $U_i\times F$ is a product of smooth manifolds, hence is itself locally Euclidean. By the standard atlas construction theorem for smooth manifolds, the compatible charts therefore form an atlas on $E$; taking the maximal smooth atlas generated by these charts gives the unique smooth structure on $E$ for which each $\Phi_i$ is a diffeomorphism onto $U_i\times F$. The remaining manifold separation and countability conditions are exactly the topological hypothesis in the statement. Since this atlas topology is assumed Hausdorff and second-countable, the compatible product charts make $E$ a smooth manifold in the usual sense.[/guided]

[step:Show that the projection is a smooth fibre bundle with fibre $F$] For each $i \in I$, define the local trivialization over $U_i$ by \begin{align*} \Phi_i:E_i \to U_i\times F. \end{align*} The projection satisfies \begin{align*} \operatorname{pr}_1\circ \Phi_i([(x,v,i)]) = \operatorname{pr}_1(x,v) = x = \pi([(x,v,i)]), \end{align*} where $\operatorname{pr}_1:U_i\times F\to U_i$ is the first-coordinate projection. Thus \begin{align*} \pi|_{E_i}=\operatorname{pr}_1\circ \Phi_i. \end{align*} Because $\Phi_i$ is a diffeomorphism and $\operatorname{pr}_1$ is smooth, $\pi$ is smooth on each $E_i$, hence smooth on $E$. The transition map between trivializations over $U_i\cap U_j$ is \begin{align*} \Phi_j\circ \Phi_i^{-1}(x,v)=(x,g_{ji}(x)(v)), \end{align*} which preserves the base point and is a smooth diffeomorphism. Therefore $\pi:E\to M$ is a smooth fibre bundle with fibre $F$ and transition functions $g_{ji}$ relative to the trivializations $\Phi_i$. [/step]

[step:Transport the vector space structure when the fibre is $\mathbb{R}^k$] Assume now that $F=\mathbb{R}^k$ and each $g_{ij}(x)$ lies in the general linear group $GL(k,\mathbb{R})$, the group of invertible linear maps $\mathbb{R}^k \to \mathbb{R}^k$. For $x\in M$, define the fibre \begin{align*} E_x:=\pi^{-1}(\{x\}). \end{align*} Choose $i\in I$ with $x\in U_i$. Transport the [vector space](/page/Vector%20Space) structure of $\mathbb{R}^k$ to $E_x$ through the bijection \begin{align*} \Phi_i|_{E_x}:E_x\to \{x\}\times \mathbb{R}^k. \end{align*} Thus, for $e,e'\in E_x$ and $a\in \mathbb{R}$, suppose $\Phi_i(e)=(x,v)$ and $\Phi_i(e')=(x,w)$. Define \begin{align*} e+e' := \Phi_i^{-1}(x,v+w) \end{align*} and \begin{align*} a e := \Phi_i^{-1}(x,av). \end{align*} This definition is independent of the chosen index. Indeed, if $j\in I$ also satisfies $x\in U_j$, then \begin{align*} \Phi_j\circ \Phi_i^{-1}(x,v)=(x,g_{ji}(x)(v)). \end{align*} Since $g_{ji}(x):\mathbb{R}^k\to\mathbb{R}^k$ is linear, it satisfies \begin{align*} g_{ji}(x)(v+w)=g_{ji}(x)(v)+g_{ji}(x)(w) \end{align*} and \begin{align*} g_{ji}(x)(av)=a\,g_{ji}(x)(v). \end{align*} Hence addition and scalar multiplication transported through $\Phi_i$ agree with those transported through $\Phi_j$. In each trivialization $\Phi_i$, the fibrewise addition becomes the standard map $U_i\times \mathbb{R}^k\times \mathbb{R}^k \to U_i\times \mathbb{R}^k$ given by \begin{align*} (x,v,w)\mapsto (x,v+w). \end{align*} The fibrewise scalar multiplication becomes the standard map $\mathbb{R}\times U_i\times \mathbb{R}^k \to U_i\times \mathbb{R}^k$ given by \begin{align*} (a,x,v)\mapsto (x,av). \end{align*} The zero section becomes the standard map $U_i\to U_i\times \mathbb{R}^k$ given by \begin{align*} x\mapsto (x,0). \end{align*} These maps are smooth. Therefore $E$ is a smooth vector bundle of rank $k$ over $M$. [/step]