[proofplan] We use the smooth cocycle reconstruction result recorded in the Cocycle reference: the quotient models $E_g$ and $E_{g'}$ come with quotient maps, projections, and smooth local trivializations. On the $i$-th local piece, the map sends the fibre coordinate $y$ to $a_i(x)(y)$; the cocycle relation between $g$ and $g'$ is exactly the compatibility condition needed for this rule to be independent of the chosen representative. Smoothness and preservation of the base are checked in those local quotient charts, and the inverse is obtained by replacing each $a_i(x)$ with $a_i(x)^{-1}$, giving a bundle isomorphism. [/proofplan] [step:Define the candidate map on representatives] Let $q_g: \bigsqcup_{i \in I} U_i \times F \to E_g$ and $q_{g'}: \bigsqcup_{i \in I} U_i \times F \to E_{g'}$ be the quotient maps supplied by the reconstructed bundles associated to the smooth cocycles $(g_{ij})$ and $(g'_{ij})$. For $i \in I$, define the smooth local map $A_i: U_i \times F \to U_i \times F$ by $A_i(x,y)=(x,a_i(x)(y))$. The smoothness of $A_i$ follows from the smoothness of $a_i:U_i \to G$ and the smooth action map $G \times F \to F$. We define a set map candidate $\Phi: E_g \to E_{g'}$ by the rule $\Phi(q_g(i,x,y)):=q_{g'}(i,x,a_i(x)(y))$. It remains to prove that this rule is independent of the representative $(i,x,y)$. [/step] [step:Use the cocycle change formula to prove well-definedness] Suppose $(i,x,y)\sim_g(j,x,z)$ with $x \in U_i \cap U_j$ and $z=g_{ji}(x)(y)$. The assumed relation for the ordered pair $(j,i)$ gives \begin{align*} g'_{ji}(x)=a_j(x)\circ g_{ji}(x)\circ a_i(x)^{-1}. \end{align*} Composing on the right by $a_i(x)$ gives the equivalent identity \begin{align*} a_j(x)\circ g_{ji}(x)=g'_{ji}(x)\circ a_i(x). \end{align*} Since $z=g_{ji}(x)(y)$, \begin{align*} a_j(x)(z)=a_j(x)(g_{ji}(x)(y)). \end{align*} Using $a_j(x)\circ g_{ji}(x)=g'_{ji}(x)\circ a_i(x)$ and evaluating at $y \in F$ gives \begin{align*} a_j(x)(g_{ji}(x)(y))=g'_{ji}(x)(a_i(x)(y)). \end{align*} Thus \begin{align*} a_j(x)(z)=g'_{ji}(x)(a_i(x)(y)). \end{align*} By the defining relation $\sim_{g'}$, this means \begin{align*} (i,x,a_i(x)(y))\sim_{g'}(j,x,a_j(x)(z)). \end{align*} Hence \begin{align*} q_{g'}(i,x,a_i(x)(y))=q_{g'}(j,x,a_j(x)(z)), \end{align*} so $\Phi$ is well-defined. [guided] The only possible obstruction is that a point of $E_g$ has many representatives. If the same point is written as $q_g(i,x,y)$ and also as $q_g(j,x,z)$, then by the definition of the reconstructed bundle there is a transition relation \begin{align*} z=g_{ji}(x)(y). \end{align*} For $\Phi$ to be well-defined, the two proposed images must represent the same point of $E_{g'}$: \begin{align*} q_{g'}(i,x,a_i(x)(y)) \quad\text{and}\quad q_{g'}(j,x,a_j(x)(z)). \end{align*} The transition rule in $E_{g'}$ says that these two classes agree exactly when \begin{align*} a_j(x)(z)=g'_{ji}(x)(a_i(x)(y)). \end{align*} Using $z=g_{ji}(x)(y)$, the left-hand side becomes \begin{align*} a_j(x)(z)=a_j(x)(g_{ji}(x)(y)). \end{align*} Now apply the assumed cohomology relation with indices $(j,i)$: \begin{align*} g'_{ji}(x)=a_j(x)\circ g_{ji}(x)\circ a_i(x)^{-1}. \end{align*} After composing this identity on the right by $a_i(x)$, we obtain \begin{align*} a_j(x)\circ g_{ji}(x)=g'_{ji}(x)\circ a_i(x). \end{align*} Evaluating both sides at $y \in F$ gives \begin{align*} a_j(x)(g_{ji}(x)(y))=g'_{ji}(x)(a_i(x)(y)). \end{align*} Thus the two proposed images are related by the transition relation for $g'$, and so they define the same point of $E_{g'}$. This proves that $\Phi$ is independent of the chosen representative. [/guided] [/step] [step:Check that the map is smooth and preserves the base] Let $\pi_g: E_g \to M$ be the bundle projection induced by $(i,x,y)\mapsto x$ in the reconstructed bundle $E_g$. Let $\pi_{g'}: E_{g'} \to M$ be the corresponding bundle projection induced by $(i,x,y)\mapsto x$ in the reconstructed bundle $E_{g'}$. For every representative $(i,x,y)$, the definition of the bundle projections gives \begin{align*} \pi_{g'}(\Phi(q_g(i,x,y)))=\pi_{g'}(q_{g'}(i,x,a_i(x)(y)))=x=\pi_g(q_g(i,x,y)). \end{align*} Hence $\pi_{g'}\circ \Phi=\pi_g$. For each $i \in I$, let $\theta_i^g: \pi_g^{-1}(U_i) \to U_i \times F$ and $\theta_i^{g'}: \pi_{g'}^{-1}(U_i) \to U_i \times F$ be the standard smooth local trivializations induced by the quotient constructions. These charts are available because the smooth cocycle reconstruction result recorded in the Cocycle reference gives smooth bundle structures on $E_g$ and $E_{g'}$ whose transition maps are $g_{ij}$ and $g'_{ij}$, respectively. In these local coordinates, \begin{align*} \theta_i^{g'}\circ \Phi \circ (\theta_i^g)^{-1}=A_i, \end{align*} where \begin{align*} A_i(x,y)=(x,a_i(x)(y)). \end{align*} Since $A_i$ is smooth, $\Phi$ is smooth on each $\pi_g^{-1}(U_i)$. The sets $\pi_g^{-1}(U_i)$ cover $E_g$, so $\Phi$ is smooth. [/step] [step:Construct the inverse by using the inverse local changes] For each $i \in I$, define $b_i: U_i \to G$ by $b_i(x)=a_i(x)^{-1}$. Since $G$ is a Lie structure group, inversion $G \to G$ is smooth, and therefore each $b_i$ is smooth. Define the map $\Psi: E_{g'} \to E_g$ by $\Psi(q_{g'}(i,x,y)):=q_g(i,x,a_i(x)^{-1}(y))$. We prove that $\Psi$ is well-defined. Suppose $(i,x,y)\sim_{g'}(j,x,z)$ with $x \in U_i \cap U_j$ and $z=g'_{ji}(x)(y)$. From \begin{align*} g'_{ji}(x)=a_j(x)\circ g_{ji}(x)\circ a_i(x)^{-1} \end{align*} we obtain, by composing on the left with $a_j(x)^{-1}$, the identity \begin{align*} g_{ji}(x)\circ a_i(x)^{-1}=a_j(x)^{-1}\circ g'_{ji}(x). \end{align*} Evaluating this identity at $y \in F$ gives \begin{align*} g_{ji}(x)(a_i(x)^{-1}(y))=a_j(x)^{-1}(g'_{ji}(x)(y))=a_j(x)^{-1}(z). \end{align*} By the defining relation $\sim_g$, this means \begin{align*} (i,x,a_i(x)^{-1}(y))\sim_g(j,x,a_j(x)^{-1}(z)). \end{align*} Hence \begin{align*} q_g(i,x,a_i(x)^{-1}(y))=q_g(j,x,a_j(x)^{-1}(z)), \end{align*} so $\Psi$ is independent of the representative. In the local trivializations, $\Psi$ is represented over $U_i$ by $(x,y)\mapsto (x,a_i(x)^{-1}(y))$, which is smooth by smoothness of $b_i$ and of the $G$-action on $F$. The same projection computation as for $\Phi$ gives $\pi_g\circ\Psi=\pi_{g'}$, so $\Psi$ is a smooth bundle map over $M$. For every representative $(i,x,y)$ of a point in $E_g$, \begin{align*} \Psi(\Phi(q_g(i,x,y)))=\Psi(q_{g'}(i,x,a_i(x)(y)))=q_g(i,x,a_i(x)^{-1}(a_i(x)(y)))=q_g(i,x,y). \end{align*} Similarly, for every representative $(i,x,y)$ of a point in $E_{g'}$, \begin{align*} \Phi(\Psi(q_{g'}(i,x,y)))=\Phi(q_g(i,x,a_i(x)^{-1}(y)))=q_{g'}(i,x,a_i(x)(a_i(x)^{-1}(y)))=q_{g'}(i,x,y). \end{align*} Therefore $\Psi=\Phi^{-1}$, and $\Phi$ is a bundle isomorphism over $M$. [/step] [step:Specialize the construction to vector bundles] Assume now that $F$ is a [vector space](/page/Vector%20Space) and $G \subseteq GL(F)$. Then each value $a_i(x)$ is a linear automorphism of $F$, and the local representative of $\Phi$ over $U_i$ is the map $A_i: U_i \times F \to U_i \times F$ defined by \begin{align*} A_i(x,y)=(x,a_i(x)y). \end{align*} For each fixed $x \in U_i$, the fibre map $F \to F$ given by $y \mapsto a_i(x)y$ is linear. Hence the bundle isomorphism $\Phi:E_g \to E_{g'}$ is fibrewise linear. This proves the vector-bundle assertion and completes the proof. [/step]