[proofplan] We construct $E$ by gluing the products $U_i \times F$ using the prescribed transition maps. The cocycle identities ensure that the gluing relation is an [equivalence relation](/page/Equivalence%20Relation) and that each point of $\pi^{-1}(U_i)$ has a unique representative in the $i$th product chart. These representatives define local trivializations whose changes of coordinates are exactly the maps $\gamma_{ij}$, so the assumed Hausdorff and second-countable topology supports the desired smooth manifold structure. Finally, uniqueness follows by sending a point written in one local trivialization to the point with the same local coordinates in any other bundle with the same transition functions. [/proofplan]

[step:Verify that the transition data define an equivalence relation] Let $X:=\bigsqcup_{i \in I} U_i \times F$ denote the disjoint union, and write an element of its $i$th summand as $(i,b,y)$ with $b \in U_i$ and $y \in F$. Define a relation on $X$ by \begin{align*} (i,b,y)\sim (j,b',y') \end{align*} if and only if $b=b'$, this common point lies in $U_i \cap U_j$, and \begin{align*} (b,y)=\gamma_{ij}(b,y'). \end{align*} The relation is reflexive because $\gamma_{ii}=\operatorname{id}_{U_i \times F}$. It is symmetric because $\gamma_{ij}\circ \gamma_{ji}=\operatorname{id}$ and $\gamma_{ji}\circ \gamma_{ij}=\operatorname{id}$ on $(U_i\cap U_j)\times F$, so $\gamma_{ji}=\gamma_{ij}^{-1}$. It is transitive because if $(i,b,y)\sim (j,b,y')$ and $(j,b,y')\sim (k,b,y'')$, then $b \in U_i \cap U_j \cap U_k$. The first relation gives \begin{align*} (b,y)=\gamma_{ij}(b,y'). \end{align*} The second relation gives \begin{align*} (b,y')=\gamma_{jk}(b,y''). \end{align*} Substituting the second identity into the first and using the cocycle identity on $(U_i\cap U_j\cap U_k)\times F$ gives \begin{align*} (b,y)=\gamma_{ik}(b,y''). \end{align*} Hence $(i,b,y)\sim(k,b,y'')$. Let $E:=X/{\sim}$, and let $q:X\to E$ be the quotient map. [/step]

[step:Define the projection and prove that it is well-defined] Define $\pi:E\to B$ by $\pi([(i,b,y)]):=b$. This definition is independent of the representative: if $(i,b,y)\sim(j,b',y')$, then the definition of $\sim$ gives $b=b'$. Therefore $\pi$ is well-defined. For each $i\in I$, define $q_i:U_i\times F\to E$ by $q_i(b,y):=q(i,b,y)$. Then \begin{align*} q_i(U_i\times F)=\pi^{-1}(U_i). \end{align*} Indeed, if $(b,y)\in U_i\times F$, then $\pi(q_i(b,y))=b\in U_i$, so $q_i(U_i\times F)\subseteq \pi^{-1}(U_i)$. Conversely, if $e\in \pi^{-1}(U_i)$ and $e=[(j,b,y')]$, then $b=\pi(e)\in U_i\cap U_j$. Since $\gamma_{ij}$ is a map from $(U_i\cap U_j)\times F$ to itself, there is a unique point $(b,y)\in (U_i\cap U_j)\times F$ such that \begin{align*} (b,y)=\gamma_{ij}(b,y'). \end{align*} Thus $(i,b,y)\sim(j,b,y')$, so $e=q_i(b,y)$. The map $q_i$ is injective. If $q_i(b,y)=q_i(b',y')$, then $(i,b,y)\sim(i,b',y')$. Hence $b=b'$ and \begin{align*} (b,y)=\gamma_{ii}(b,y')=(b,y'), \end{align*} so $y=y'$. Therefore $q_i$ is a bijection from $U_i\times F$ onto $\pi^{-1}(U_i)$. [/step]

[step:Use the local representatives to define the trivializations]For each $i\in I$, define \begin{align*} \Phi_i:\pi^{-1}(U_i)&\to U_i\times F \end{align*} to be the inverse of the bijection $q_i:U_i\times F\to \pi^{-1}(U_i)$. Equivalently, \begin{align*} \Phi_i([(i,b,y)])=(b,y). \end{align*} For $b\in U_i\cap U_j$ and $y'\in F$, the definition of $\sim$ gives \begin{align*} (\Phi_i\circ \Phi_j^{-1})(b,y') =\gamma_{ij}(b,y'). \end{align*} Thus the transition map from the $j$th trivialization to the $i$th trivialization is precisely $\gamma_{ij}$.[/step]

[guided]We first reprove the representative statement needed to define $\Phi_i$ without ambiguity. A point $e\in \pi^{-1}(U_i)$ may initially be represented as $[(j,b,y')]$, not as an $i$th-chart representative. Since $b=\pi(e)\in U_i$ and also $b\in U_j$, we may transfer the representative to the $i$th chart by applying the transition map: \begin{align*} (b,y):=\gamma_{ij}(b,y'). \end{align*} Then $(i,b,y)\sim(j,b,y')$, so $e=[(i,b,y)]$. If also $e=[(i,b,z)]$ for some $z\in F$, then $(i,b,y)\sim(i,b,z)$, and the identity condition $\gamma_{ii}=\operatorname{id}_{U_i\times F}$ gives $(b,y)=(b,z)$, hence $y=z$. Therefore every $e\in\pi^{-1}(U_i)$ has a unique $i$th representative, and $\Phi_i(e)$ is defined as that unique pair $(b,y)\in U_i\times F$. Now compute the coordinate change. Let $b\in U_i\cap U_j$ and $y'\in F$. Then \begin{align*} \Phi_j^{-1}(b,y')=[(j,b,y')]. \end{align*} To express this same point in the $i$th trivialization, we use the defining equivalence relation: \begin{align*} (i,b,y)\sim(j,b,y') \quad\Longleftrightarrow\quad (b,y)=\gamma_{ij}(b,y'). \end{align*} Hence \begin{align*} (\Phi_i\circ \Phi_j^{-1})(b,y') =\Phi_i([(j,b,y')]) =\gamma_{ij}(b,y'). \end{align*} Thus the changes of trivialization are exactly the prescribed transition maps.[/guided]

[step:Equip the quotient with the smooth manifold structure determined by the trivializations] By hypothesis, $E$ carries the [quotient topology](/page/Quotient%20Topology) induced by $q:X\to E$, and this topology is Hausdorff and second-countable. We first prove that the proposed chart domains are open and that the proposed coordinate maps are homeomorphisms for this topology. Fix $i\in I$. The set $\pi^{-1}(U_i)$ is open in $E$ because its full preimage under the quotient map is \begin{align*} q^{-1}(\pi^{-1}(U_i)) = \bigsqcup_{j\in I} (U_i\cap U_j)\times F, \end{align*} which is open in $X$ since each $U_i\cap U_j$ is open in the smooth manifold $B$. The map $q_i:U_i\times F\to\pi^{-1}(U_i)$ is continuous as the restriction of the quotient map $q$ to the $i$th summand. To prove that $q_i$ is open, let $W\subset U_i\times F$ be open. For each $j\in I$, the preimage of $q_i(W)$ under $q_j$ is \begin{align*} q_j^{-1}(q_i(W)) = \gamma_{ij}^{-1}\bigl(W\cap ((U_i\cap U_j)\times F)\bigr), \end{align*} where $\gamma_{ij}$ is restricted to $(U_i\cap U_j)\times F$. The set $W\cap ((U_i\cap U_j)\times F)$ is open in $(U_i\cap U_j)\times F$, and $\gamma_{ij}$ is a diffeomorphism of this overlap product, so the displayed preimage is open in $(U_i\cap U_j)\times F$ and hence open in $U_j\times F$. Therefore $q^{-1}(q_i(W))$ is open in the disjoint union $X$, and the quotient topology implies that $q_i(W)$ is open in $E$. Thus $q_i$ is a homeomorphism from $U_i\times F$ onto the open subspace $\pi^{-1}(U_i)$, and $\Phi_i=q_i^{-1}$ is a homeomorphism from $\pi^{-1}(U_i)$ to $U_i\times F$. The transition maps between these coordinate domains are \begin{align*} \Phi_i\circ \Phi_j^{-1}=\gamma_{ij}, \end{align*} and each $\gamma_{ij}$ is smooth by hypothesis. Since $\gamma_{ji}=\gamma_{ij}^{-1}$ is also smooth, the charts are smoothly compatible. Each product $U_i\times F$ is a smooth manifold of dimension $\dim B+\dim F$, because $U_i$ is an open submanifold of $B$ and products of smooth manifolds carry the product smooth structure. The chart domains are open, the coordinate maps are homeomorphisms onto these product manifolds, and $E$ is Hausdorff and second-countable by hypothesis. Therefore the smooth manifold atlas construction applies: the compatible local homeomorphisms $(\Phi_i)_{i\in I}$ make $E$ locally Euclidean of dimension $\dim B+\dim F$ and define a smooth manifold structure on the given [topological space](/page/Topological%20Space) $E$. With this structure, each $\Phi_i:\pi^{-1}(U_i)\to U_i\times F$ is a diffeomorphism by construction. Moreover, \begin{align*} \operatorname{pr}_1\circ \Phi_i([(i,b,y)])=b=\pi([(i,b,y)]), \end{align*} where $\operatorname{pr}_1:U_i\times F\to U_i$ is the projection onto the first factor. Hence \begin{align*} \pi|_{\pi^{-1}(U_i)}=\operatorname{pr}_1\circ \Phi_i. \end{align*} Since $\operatorname{pr}_1$ and $\Phi_i$ are smooth, $\pi$ is smooth locally on the cover $(\pi^{-1}(U_i))_{i\in I}$. Thus $\pi:E\to B$ is a smooth fibre bundle with fibre $F$ and local trivializations $\Phi_i$. [/step]

[step:Construct the unique isomorphism to any other bundle with the same transition maps] Let $\pi':E'\to B$ be another smooth fibre bundle with fibre $F$, and let \begin{align*} \Psi_i:(\pi')^{-1}(U_i)&\to U_i\times F \end{align*} be local trivializations satisfying \begin{align*} \Psi_i\circ \Psi_j^{-1}=\gamma_{ij} \end{align*} on $(U_i\cap U_j)\times F$. Define \begin{align*} A:E&\to E' \end{align*} as follows. If $e=[(i,b,y)]\in E$, set \begin{align*} A(e):=\Psi_i^{-1}(b,y). \end{align*} This is well-defined. If $[(i,b,y)]=[(j,b,y')]$, then \begin{align*} (b,y)=\gamma_{ij}(b,y'). \end{align*} Using $\Psi_i\circ\Psi_j^{-1}=\gamma_{ij}$ gives \begin{align*} \Psi_i^{-1}(b,y) = \Psi_i^{-1}(\gamma_{ij}(b,y')) = \Psi_j^{-1}(b,y'). \end{align*} Thus the value of $A(e)$ is independent of the chosen representative. The map $A$ lies over $B$ because \begin{align*} \pi'(A([(i,b,y)]))=\pi'(\Psi_i^{-1}(b,y))=b=\pi([(i,b,y)]). \end{align*} In local coordinates, \begin{align*} \Psi_i\circ A\circ \Phi_i^{-1} = \operatorname{id}_{U_i\times F}. \end{align*} Therefore $A$ is smooth and locally a diffeomorphism. Its inverse is constructed by the same formula with $\Phi_i$ and $\Psi_i$ interchanged: \begin{align*} A^{-1}(e'):=\Phi_i^{-1}(\Psi_i(e')) \end{align*} for $e'\in(\pi')^{-1}(U_i)$. The same transition-map computation proves that this inverse is well-defined. Hence $A:E\to E'$ is a diffeomorphism over $B$. Finally, suppose $A_1:E\to E'$ is any diffeomorphism over $B$ preserving the given local trivializations, meaning \begin{align*} \Psi_i\circ A_1\circ \Phi_i^{-1} = \operatorname{id}_{U_i\times F} \end{align*} for every $i\in I$. Then for every representative $[(i,b,y)]$, \begin{align*} A_1([(i,b,y)]) = \Psi_i^{-1}(b,y) = A([(i,b,y)]). \end{align*} Thus $A_1=A$, proving uniqueness. This completes the construction and uniqueness of the smooth fibre bundle determined by the smooth bundle atlas. [/step]