[proofplan] We construct the total space by gluing the trivial bundles $U_i \times F$ using the prescribed cocycle. The cocycle identities make the gluing relation an [equivalence relation](/page/Equivalence%20Relation), so the quotient is a well-defined [topological space](/page/Topological%20Space). We then define local maps from the quotient back to $U_i \times F$ and prove directly from the quotient topology that they are homeomorphisms. Finally, comparing two such local trivializations recovers exactly the transition functions $t_{ij}$. [/proofplan]

[step:Glue the local product spaces by the cocycle relation]Let $X$ denote the disjoint union \begin{align*} X := \bigsqcup_{i \in I} (U_i \times F). \end{align*} We write an element of the $i$-th summand as $(i,b,v)$, where $b \in U_i$ and $v \in F$. Define a relation $\sim$ on $X$ as follows: for $(i,b,v),(j,c,w) \in X$, set \begin{align*} (i,b,v) \sim (j,c,w) \end{align*} if and only if $b=c$, $b \in U_i \cap U_j$, and \begin{align*} v = t_{ij}(b)(w). \end{align*} This relation is reflexive because $t_{ii}(b)=\operatorname{id}_F$ for every $b \in U_i$. The hypothesis $t_{ij}(b)\in\operatorname{Homeo}(F)$ is the standard transition-function hypothesis; in the equivalence-relation verification below, the inverse relation is also forced by the cocycle identities. It is symmetric because if $v=t_{ij}(b)(w)$, then the cocycle identity with indices $j,i,j$ gives \begin{align*} t_{ji}(b)\circ t_{ij}(b)=t_{jj}(b)=\operatorname{id}_F, \end{align*} hence \begin{align*} w=t_{ji}(b)(v). \end{align*} It is transitive because if \begin{align*} (i,b,v) \sim (j,b,w) \end{align*} and \begin{align*} (j,b,w) \sim (k,b,z), \end{align*} then \begin{align*} v=t_{ij}(b)(w)=t_{ij}(b)(t_{jk}(b)(z))=(t_{ij}(b)\circ t_{jk}(b))(z)=t_{ik}(b)(z), \end{align*} so \begin{align*} (i,b,v)\sim(k,b,z). \end{align*} Thus $\sim$ is an equivalence relation on $X$. Let \begin{align*} E := X/{\sim} \end{align*} be the [quotient topological space](/page/Quotient%20Topology), and let \begin{align*} q:X \to E \end{align*} be the quotient map.[/step]

[guided]The construction begins with the most literal candidate for a total space: take one copy of the product model $U_i \times F$ over each [open set](/page/Open%20Set) $U_i$, and then identify points that are meant to represent the same point of the desired bundle. Thus we set \begin{align*} X := \bigsqcup_{i \in I} (U_i \times F), \end{align*} and write points in the $i$-th summand as triples $(i,b,v)$. The intended meaning of the transition function $t_{ij}$ is that, over a point $b \in U_i \cap U_j$, a fibre coordinate $w$ in the $j$-trivialization corresponds to the fibre coordinate $t_{ij}(b)(w)$ in the $i$-trivialization. Therefore we define \begin{align*} (i,b,v) \sim (j,c,w) \end{align*} exactly when $b=c$, this common point lies in $U_i \cap U_j$, and \begin{align*} v=t_{ij}(b)(w). \end{align*} We verify that this is an equivalence relation. Reflexivity follows because the identity cocycle condition says $t_{ii}(b)=\operatorname{id}_F$, so \begin{align*} v=t_{ii}(b)(v) \end{align*} for every $(i,b,v) \in X$. For symmetry, suppose $(i,b,v)\sim(j,b,w)$. Then $v=t_{ij}(b)(w)$. The cocycle identity with indices $j,i,j$ gives \begin{align*} t_{ji}(b)\circ t_{ij}(b)=t_{jj}(b)=\operatorname{id}_F. \end{align*} Applying this identity to $w$ gives \begin{align*} t_{ji}(b)(v)=t_{ji}(b)(t_{ij}(b)(w))=w. \end{align*} Hence $w=t_{ji}(b)(v)$, which is precisely $(j,b,w)\sim(i,b,v)$. For transitivity, suppose $(i,b,v)\sim(j,b,w)$ and $(j,b,w)\sim(k,b,z)$. Then \begin{align*} v=t_{ij}(b)(w) \end{align*} and \begin{align*} w=t_{jk}(b)(z). \end{align*} Substituting the second identity into the first and using the cocycle identity with indices $i,j,k$, we obtain \begin{align*} v=t_{ij}(b)(t_{jk}(b)(z))=(t_{ij}(b)\circ t_{jk}(b))(z)=t_{ik}(b)(z). \end{align*} Thus $(i,b,v)\sim(k,b,z)$. This proves transitivity. Therefore the quotient \begin{align*} E:=X/{\sim} \end{align*} is a well-defined topological space equipped with the [quotient topology](/page/Quotient%20Topology). We denote the quotient map by \begin{align*} q:X \to E. \end{align*}[/guided]

[step:Define the projection to the base] Define a map $\pi:E \to B$ by the rule \begin{align*} \pi(q(i,b,v)) := b \end{align*} for every $(i,b,v) \in X$. This map is well-defined: if $(i,b,v)\sim(j,c,w)$, then by definition $b=c$. Let $p:X \to B$ be the map defined by \begin{align*} p(i,b,v) := b \end{align*} for every $(i,b,v) \in X$. This map restricts on each summand $U_i \times F$ to the first projection. Since each first projection $U_i \times F \to U_i \subset B$ is continuous, the defining property of the disjoint union topology implies that $p$ is continuous. Also $p=\pi \circ q$. Since $q$ is a quotient map, the equality $p=\pi \circ q$ implies that $\pi$ is continuous. [/step]

[step:Construct local trivializations from quotient representatives] Fix $i \in I$. Define a map $\Phi_i:\pi^{-1}(U_i) \to U_i \times F$ by \begin{align*} \Phi_i(q(j,b,w)) := (b,t_{ij}(b)(w)) \end{align*} for every representative $(j,b,w) \in X$ with $b \in U_i \cap U_j$. This is well-defined. Indeed, if $q(j,b,w)=q(k,b,z)$, then $(j,b,w)\sim(k,b,z)$, so \begin{align*} w=t_{jk}(b)(z). \end{align*} Using the cocycle identity on $U_i \cap U_j \cap U_k$, \begin{align*} t_{ij}(b)(w)=t_{ij}(b)(t_{jk}(b)(z))=t_{ik}(b)(z). \end{align*} Thus the value of $\Phi_i$ is independent of the chosen representative. Define also a map $\Psi_i:U_i \times F \to \pi^{-1}(U_i)$ by \begin{align*} \Psi_i(b,v) := q(i,b,v). \end{align*} Then \begin{align*} \Phi_i(\Psi_i(b,v))=\Phi_i(q(i,b,v))=(b,t_{ii}(b)(v))=(b,v), \end{align*} and, for $q(j,b,w)\in\pi^{-1}(U_i)$, \begin{align*} \Psi_i(\Phi_i(q(j,b,w)))=\Psi_i(b,t_{ij}(b)(w))=q(i,b,t_{ij}(b)(w))=q(j,b,w), \end{align*} where the last equality is exactly the defining equivalence relation. Hence $\Phi_i$ and $\Psi_i$ are inverse bijections. [/step]

[step:Prove the local trivializations are homeomorphisms] The map $\Psi_i:U_i \times F \to \pi^{-1}(U_i)$ is continuous because it is the restriction of the quotient map $q$ to the $i$-th summand. It remains to prove that $\Phi_i$ is continuous. Let \begin{align*} Y_i:=q^{-1}(\pi^{-1}(U_i))\subset X. \end{align*} The set $Y_i$ is saturated with respect to $q$, because $Y_i=q^{-1}(\pi^{-1}(U_i))$. It is open in $X$ because $Y_i$ is the disjoint union of the open subspaces $(U_i\cap U_j)\times F\subset U_j\times F$ over $j\in I$, using the disjoint union topology. We now use the restriction property for quotient maps: if $q:X\to E$ is a quotient map and $Y\subset X$ is open and saturated, then the restricted map $q|_Y:Y\to q(Y)$ is a quotient map for the [subspace topology](/page/Subspace%20Topology) on $q(Y)$. Its hypotheses are satisfied here with $Y=Y_i$, and $q(Y_i)=\pi^{-1}(U_i)$. Therefore $q|_{Y_i}:Y_i\to \pi^{-1}(U_i)$ is a quotient map. Hence it is enough to prove that the map $\Phi_i\circ q|_{Y_i}:Y_i \to U_i\times F$ is continuous. For each $j\in I$, the restriction of this composite to the summand \begin{align*} (U_i\cap U_j)\times F \subset U_j\times F \end{align*} is the map $(U_i\cap U_j)\times F \to U_i\times F$ defined by \begin{align*} (b,w) \mapsto (b,t_{ij}(b)(w)). \end{align*} Its first component is the continuous inclusion $U_i\cap U_j\hookrightarrow U_i$, and its second component is continuous by the stated continuity assumption on the associated map $(b,w)\mapsto t_{ij}(b)(w)$. Therefore the restriction to each summand is continuous. By the definition of the disjoint union topology, $\Phi_i\circ q|_{Y_i}$ is continuous. Hence $\Phi_i$ is continuous. Thus each map $\Phi_i:\pi^{-1}(U_i)\to U_i\times F$ is a homeomorphism with inverse $\Psi_i$. Let $\operatorname{pr}_1:U_i\times F\to U_i$ denote the first projection map. Moreover, $\operatorname{pr}_1\circ\Phi_i=\pi|_{\pi^{-1}(U_i)}$, so $\Phi_i$ is fibre-preserving over $U_i$. [/step]

[step:Recover the prescribed transition functions] Let $i,j\in I$ and let $(b,v)\in (U_i\cap U_j)\times F$. Using the formula for $\Psi_j=\Phi_j^{-1}$, we have \begin{align*} \Phi_j^{-1}(b,v)=q(j,b,v). \end{align*} Therefore \begin{align*} (\Phi_i\circ\Phi_j^{-1})(b,v) =\Phi_i(q(j,b,v)) =(b,t_{ij}(b)(v)). \end{align*} Thus the transition function from the $j$-trivialization to the $i$-trivialization is precisely $t_{ij}$. Since $(U_i)_{i\in I}$ covers $B$, the maps $\Phi_i$ give local trivializations of the continuous projection $\pi:E\to B$ over an open cover of $B$. Hence $\pi:E\to B$ is a fibre bundle with fibre $F$ and with the prescribed transition functions. [/step]