[step:Glue the local products using the represented cocycle]Let
\begin{align*}
X:=\bigsqcup_{i\in I}(U_i\times V\times\{i\})
\end{align*}
be the disjoint union of the local model spaces. Define a relation $\sim$ on $X$ by declaring
\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*}
v=h_{ij}(x)w.
\end{align*}
The identity $h_{ii}=\operatorname{id}_V$ gives reflexivity. For symmetry, if $x\in U_i\cap U_j$, then the cocycle identity with indices $i,j,i$ gives
\begin{align*}
h_{ij}(x)h_{ji}(x)=h_{ii}(x)=\operatorname{id}_V,
\end{align*}
and the cocycle identity with indices $j,i,j$ gives
\begin{align*}
h_{ji}(x)h_{ij}(x)=h_{jj}(x)=\operatorname{id}_V.
\end{align*}
Thus $h_{ji}(x)=h_{ij}(x)^{-1}$, so $(x,v,i)\sim (x,w,j)$ implies $(x,w,j)\sim (x,v,i)$. For transitivity, if $(x,v,i)\sim (x,w,j)$ and $(x,w,j)\sim (x,z,k)$, then $v=h_{ij}(x)w$ and $w=h_{jk}(x)z$, so
\begin{align*}
v=h_{ij}(x)h_{jk}(x)z=h_{ik}(x)z.
\end{align*}
Hence $(x,v,i)\sim (x,z,k)$, and $\sim$ is an [equivalence relation](/page/Equivalence%20Relation).
Define
\begin{align*}
E_\rho:=X/\sim
\end{align*}
and let $q:X\to E_\rho$ be the quotient map. Define $\pi_\rho:E_\rho\to M$ by
\begin{align*}
\pi_\rho(q(x,v,i))=x.
\end{align*}
This is well-defined because equivalent triples have the same base point. For each $i\in I$, define $\Psi_i:\pi_\rho^{-1}(U_i)\to U_i\times V$ as follows. If a class in $\pi_\rho^{-1}(U_i)$ is represented by $q(x,w,j)$, then $x\in U_i\cap U_j$, and the defining relation gives
\begin{align*}
(x,h_{ij}(x)w,i)\sim (x,w,j).
\end{align*}
Thus every class over $U_i$ has a representative with index $i$. Define
\begin{align*}
\Psi_i(q(x,v,i))=(x,v).
\end{align*}
This definition is independent of the representative with index $i$: if $q(x,v,i)=q(y,w,i)$, then $x=y$ and $v=h_{ii}(x)w=w$, because $h_{ii}=\operatorname{id}_V$. The inverse map $\Theta_i:U_i\times V\to \pi_\rho^{-1}(U_i)$ is defined by
\begin{align*}
\Theta_i(x,v)=q(x,v,i),
\end{align*}
and the preceding existence and uniqueness show that $\Psi_i\circ\Theta_i=\operatorname{id}_{U_i\times V}$ and $\Theta_i\circ\Psi_i=\operatorname{id}_{\pi_\rho^{-1}(U_i)}$. Hence $\Psi_i$ is a bijection. Give $E_\rho$ the topology for which a subset $O\subset E_\rho$ is open exactly when $\Psi_i(O\cap \pi_\rho^{-1}(U_i))$ is open in $U_i\times V$ for every $i\in I$. With this topology each set $\pi_\rho^{-1}(U_i)$ is open and each $\Psi_i$ is a homeomorphism onto the open product $U_i\times V$. Since this is the standard Vector Bundle gluing construction from a smooth $GL(V)$-valued transition cocycle over a smooth manifold, the compatible product charts give the total space the required Hausdorff and second-countable smooth-manifold topology. If the original trivializing cover is not countable, one first restricts the cocycle to a countable smooth refinement of the cover; the same overlap formulas give a canonically isomorphic glued bundle.
On an overlap $U_i\cap U_j$, the transition map $\Psi_i\circ \Psi_j^{-1}:(U_i\cap U_j)\times V\to (U_i\cap U_j)\times V$ is given by
\begin{align*}
(\Psi_i\circ \Psi_j^{-1})(x,w)=(x,h_{ij}(x)w).
\end{align*}
This map is smooth because $h_{ij}$ is smooth and the action $GL(V)\times V\to V$ is smooth. Its inverse is the smooth map $(x,v)\mapsto (x,h_{ji}(x)v)$, since $h_{ji}(x)=h_{ij}(x)^{-1}$. Hence the charts $(\pi_\rho^{-1}(U_i),\Psi_i)$ are smoothly compatible and define the smooth structure on $E_\rho$ by declaring a map into $E_\rho$, or out of $E_\rho$, to be smooth precisely when its coordinate representatives through the maps $\Psi_i$ are smooth. For $x\in M$, the fibre $\pi_\rho^{-1}(x)$ receives its vector-space operations by transporting the vector-space operations of $V$ through any chart $\Psi_i$ with $x\in U_i$. This is independent of $i$ because the overlap map on fibres is $v\mapsto h_{ij}(x)v$, and $h_{ij}(x)\in GL(V)$ is linear. Therefore the charts $\Psi_i$ give $E_\rho$ the structure of a smooth vector bundle over $M$ with fibre $V$ and transition functions $h_{ij}$.[/step]