[proofplan] We construct the bundle by gluing the local products $U_i \times \mathbb{C}^r$ using the matrices $g_{ij}$ as transition maps. The identities $g_{ii}=I_r$ and $g_{ij}g_{jk}=g_{ik}$ make the gluing relation an [equivalence relation](/page/Equivalence%20Relation) and ensure that the quotient has compatible holomorphic product charts. A change of cocycle by holomorphic matrices $h_i$ gives a well-defined map on the quotient, and conversely any bundle isomorphism is represented in the chosen local frames by such matrices. For different covers, both cocycles are pulled back to a common refinement, reducing the comparison to the same-cover case. [/proofplan]

[step:Glue the local products by the cocycle relation] For each $i \in I$, let \begin{align*} S_i := U_i \times \mathbb{C}^r \end{align*} and let \begin{align*} S := \bigsqcup_{i \in I} S_i \end{align*} be the disjoint union. We write an element of the $i$-th summand as $(i,x,v)$, where $x \in U_i$ and $v \in \mathbb{C}^r$. Define a relation $\sim$ on $S$ by declaring \begin{align*} (i,x,v) \sim (j,y,w) \end{align*} if and only if $x=y$, this common point belongs to $U_i \cap U_j$, and \begin{align*} v = g_{ij}(x)w. \end{align*} The identity $g_{ii}(x)=I_r$ gives reflexivity. If $(i,x,v)\sim (j,x,w)$, then $v=g_{ij}(x)w$. Applying the cocycle identity with indices $i,j,i$ gives \begin{align*} g_{ij}(x)g_{ji}(x)=g_{ii}(x)=I_r, \end{align*} so $g_{ji}(x)=g_{ij}(x)^{-1}$ and hence $w=g_{ji}(x)v$; therefore $(j,x,w)\sim(i,x,v)$. If $(i,x,v)\sim(j,x,w)$ and $(j,x,w)\sim(k,x,u)$, then \begin{align*} v = g_{ij}(x)w \end{align*} and \begin{align*} w = g_{jk}(x)u. \end{align*} Thus \begin{align*} v = g_{ij}(x)g_{jk}(x)u = g_{ik}(x)u, \end{align*} where the last equality is the cocycle identity on $U_i \cap U_j \cap U_k$. Hence $(i,x,v)\sim(k,x,u)$. Therefore $\sim$ is an equivalence relation. Let \begin{align*} E := S/{\sim} \end{align*} and denote the equivalence class of $(i,x,v)$ by $[i,x,v]$. Define \begin{align*} \pi: E \to X \end{align*} by \begin{align*} \pi([i,x,v]) := x. \end{align*} This is well-defined because equivalent triples have the same base point. [/step]

[step:Build holomorphic product charts on the quotient]For each $i \in I$, define \begin{align*} \Phi_i: \pi^{-1}(U_i) \to U_i \times \mathbb{C}^r \end{align*} by \begin{align*} \Phi_i([j,x,w]) := (x,g_{ij}(x)w). \end{align*} This formula is well-defined. Indeed, suppose $[j,x,w]=[k,x,u]$. Then $w=g_{jk}(x)u$, and therefore \begin{align*} g_{ij}(x)w = g_{ij}(x)g_{jk}(x)u = g_{ik}(x)u \end{align*} by the cocycle identity. Thus the value of $\Phi_i$ does not depend on the representative. Define \begin{align*} \Psi_i: U_i \times \mathbb{C}^r \to \pi^{-1}(U_i) \end{align*} by \begin{align*} \Psi_i(x,v) := [i,x,v]. \end{align*} Then $\Phi_i \circ \Psi_i$ is the identity map on $U_i \times \mathbb{C}^r$, since $g_{ii}(x)=I_r$. Also, for $[j,x,w]\in \pi^{-1}(U_i)$, \begin{align*} \Psi_i(\Phi_i([j,x,w])) = [i,x,g_{ij}(x)w] = [j,x,w], \end{align*} by the definition of $\sim$. Hence $\Phi_i$ is a bijection with inverse $\Psi_i$. On overlaps, the transition map \begin{align*} \Phi_i \circ \Phi_j^{-1}: (U_i \cap U_j)\times \mathbb{C}^r \to (U_i \cap U_j)\times \mathbb{C}^r \end{align*} is \begin{align*} (x,v) \mapsto (x,g_{ij}(x)v). \end{align*} This map is holomorphic because $g_{ij}:U_i\cap U_j\to GL(r,\mathbb{C})$ is holomorphic and matrix multiplication defines a holomorphic map $GL(r,\mathbb{C})\times \mathbb{C}^r\to \mathbb{C}^r$.[/step]

[guided]The quotient $E$ has been defined as a set, but a holomorphic vector bundle requires local holomorphic product charts. The point of the maps $\Phi_i$ is to use the $i$-th trivialization as a coordinate system over $U_i$. For each $i\in I$, define \begin{align*} \Phi_i: \pi^{-1}(U_i) \to U_i \times \mathbb{C}^r \end{align*} by \begin{align*} \Phi_i([j,x,w]) := (x,g_{ij}(x)w). \end{align*} Why is this the correct formula? The vector $w$ is written in the $j$-th local frame, and $g_{ij}(x)$ converts it into the $i$-th local frame. We must first check that the formula does not depend on the representative of the equivalence class. Suppose $[j,x,w]=[k,x,u]$. By definition of the relation, this means \begin{align*} w = g_{jk}(x)u. \end{align*} Then \begin{align*} g_{ij}(x)w = g_{ij}(x)g_{jk}(x)u. \end{align*} Since $x\in U_i\cap U_j\cap U_k$, the cocycle identity applies and gives \begin{align*} g_{ij}(x)g_{jk}(x)=g_{ik}(x). \end{align*} Therefore \begin{align*} g_{ij}(x)w = g_{ik}(x)u. \end{align*} This is exactly the statement that $\Phi_i([j,x,w])=\Phi_i([k,x,u])$. Now define \begin{align*} \Psi_i: U_i \times \mathbb{C}^r \to \pi^{-1}(U_i) \end{align*} by \begin{align*} \Psi_i(x,v) := [i,x,v]. \end{align*} Then \begin{align*} \Phi_i(\Psi_i(x,v)) = \Phi_i([i,x,v]) = (x,g_{ii}(x)v) = (x,v), \end{align*} because $g_{ii}(x)=I_r$. Conversely, if $[j,x,w]\in\pi^{-1}(U_i)$, then \begin{align*} \Psi_i(\Phi_i([j,x,w])) = \Psi_i(x,g_{ij}(x)w) = [i,x,g_{ij}(x)w]. \end{align*} The defining relation says precisely that $[i,x,g_{ij}(x)w]=[j,x,w]$. Hence $\Phi_i$ and $\Psi_i$ are inverse bijections. The transition function between the $j$-th and $i$-th charts is therefore \begin{align*} \Phi_i\circ\Phi_j^{-1}: (U_i\cap U_j)\times\mathbb{C}^r \to (U_i\cap U_j)\times\mathbb{C}^r, \end{align*} and evaluating it on $(x,v)$ gives \begin{align*} (\Phi_i\circ\Phi_j^{-1})(x,v) = \Phi_i([j,x,v]) = (x,g_{ij}(x)v). \end{align*} This map is holomorphic because $g_{ij}$ is holomorphic and multiplication of a complex matrix by a vector is holomorphic in all entries. Thus the quotient carries compatible holomorphic product charts.[/guided]

[step:Identify the quotient as a holomorphic vector bundle of rank $r$] Equip $E$ with the unique complex-manifold structure for which all maps $\Phi_i$ are holomorphic charts. In these charts the projection \begin{align*} \pi: E \to X \end{align*} is represented by \begin{align*} U_i \times \mathbb{C}^r \to U_i,\qquad (x,v)\mapsto x, \end{align*} so $\pi$ is holomorphic. For each $x\in X$, the fiber \begin{align*} E_x := \pi^{-1}(\{x\}) \end{align*} is identified by any chart $\Phi_i$ with $\{x\}\times\mathbb{C}^r$ whenever $x\in U_i$. The transition maps on overlaps are linear in the fiber variable, because \begin{align*} (x,v)\mapsto (x,g_{ij}(x)v) \end{align*} is linear in $v$. Hence the local product charts $\Phi_i$ define a holomorphic vector bundle structure of rank $r$ on $\pi:E\to X$, and its transition functions are exactly the given maps $g_{ij}$. [/step]

[step:Descend a change of cocycle to a bundle isomorphism] Assume that $(g'_{ij})$ is another cocycle on the same cover and that there are holomorphic maps \begin{align*} h_i: U_i \to GL(r,\mathbb{C}) \end{align*} satisfying \begin{align*} g'_{ij}(x)=h_i(x)g_{ij}(x)h_j(x)^{-1} \end{align*} for all $x\in U_i\cap U_j$. Let $E$ and $E'$ be the quotient bundles constructed from $(g_{ij})$ and $(g'_{ij})$, respectively. Define \begin{align*} F:E\to E' \end{align*} by \begin{align*} F([i,x,v]) := [i,x,h_i(x)v]'. \end{align*} Here $[\cdot]'$ denotes the equivalence class in $E'$. To check that $F$ is well-defined, suppose $[i,x,v]=[j,x,w]$ in $E$. Then $v=g_{ij}(x)w$, so \begin{align*} h_i(x)v = h_i(x)g_{ij}(x)w = g'_{ij}(x)h_j(x)w. \end{align*} Thus $[i,x,h_i(x)v]'=[j,x,h_j(x)w]'$ in $E'$, proving well-definedness. In local trivializations, $F$ is represented by \begin{align*} U_i\times\mathbb{C}^r\to U_i\times\mathbb{C}^r,\qquad (x,v)\mapsto (x,h_i(x)v). \end{align*} This map is holomorphic and fiberwise complex-linear. Its inverse is represented locally by \begin{align*} (x,v)\mapsto (x,h_i(x)^{-1}v), \end{align*} which is holomorphic because inversion in $GL(r,\mathbb{C})$ is holomorphic. Therefore $F:E\to E'$ is a holomorphic vector bundle isomorphism over $X$. [/step]

[step:Recover the change of cocycle from a bundle isomorphism] Conversely, suppose \begin{align*} F:E\to E' \end{align*} is a holomorphic vector bundle isomorphism over $X$. Let \begin{align*} \Phi_i: \pi^{-1}(U_i)\to U_i\times\mathbb{C}^r \end{align*} and \begin{align*} \Phi'_i: (\pi')^{-1}(U_i)\to U_i\times\mathbb{C}^r \end{align*} be the local trivializations determined by $(g_{ij})$ and $(g'_{ij})$. For each $i\in I$, define \begin{align*} A_i: U_i\times\mathbb{C}^r\to U_i\times\mathbb{C}^r \end{align*} by \begin{align*} A_i := \Phi'_i\circ F\circ \Phi_i^{-1}. \end{align*} Because $F$ is a vector bundle isomorphism over $X$, there is a unique holomorphic map \begin{align*} h_i: U_i\to GL(r,\mathbb{C}) \end{align*} such that \begin{align*} A_i(x,v)=(x,h_i(x)v) \end{align*} for every $x\in U_i$ and $v\in\mathbb{C}^r$. The map $h_i$ is obtained by applying $A_i$ to the standard basis vectors of $\mathbb{C}^r$, so its matrix entries are holomorphic. For $x\in U_i\cap U_j$ and $v\in\mathbb{C}^r$, compute the same point of $E$ in the two local frames. In the bundle $E$, the transition rule is \begin{align*} \Phi_i\circ\Phi_j^{-1}(x,v)=(x,g_{ij}(x)v). \end{align*} In the bundle $E'$, the transition rule is \begin{align*} \Phi'_i\circ(\Phi'_j)^{-1}(x,v)=(x,g'_{ij}(x)v). \end{align*} The identity $A_i\circ(\Phi_i\circ\Phi_j^{-1})=(\Phi'_i\circ(\Phi'_j)^{-1})\circ A_j$ gives \begin{align*} (x,h_i(x)g_{ij}(x)v)=(x,g'_{ij}(x)h_j(x)v) \end{align*} for every $v\in\mathbb{C}^r$. Hence \begin{align*} h_i(x)g_{ij}(x)=g'_{ij}(x)h_j(x). \end{align*} Multiplying on the right by $h_j(x)^{-1}$ gives \begin{align*} g'_{ij}(x)=h_i(x)g_{ij}(x)h_j(x)^{-1}. \end{align*} Thus every holomorphic vector bundle isomorphism is represented in local frames by a family of holomorphic maps $h_i$ satisfying the stated transformation law. [/step]

[step:Reduce different covers to a common refinement] Let $(U_i)_{i\in I}$ and $(V_a)_{a\in A}$ be two open covers of $X$ carrying cocycles $(g_{ij})$ and $(\gamma_{ab})$, respectively. A common refinement is an open cover $(W_\lambda)_{\lambda\in\Lambda}$ together with maps $\alpha:\Lambda\to I$ and $\beta:\Lambda\to A$ such that \begin{align*} W_\lambda\subset U_{\alpha(\lambda)}\cap V_{\beta(\lambda)} \end{align*} for every $\lambda\in\Lambda$. The refined cocycles are defined by restriction: \begin{align*} g^\alpha_{\lambda\mu}:=g_{\alpha(\lambda)\alpha(\mu)}\big|_{W_\lambda\cap W_\mu} \end{align*} and \begin{align*} \gamma^\beta_{\lambda\mu}:=\gamma_{\beta(\lambda)\beta(\mu)}\big|_{W_\lambda\cap W_\mu}. \end{align*} The preceding same-cover criterion applies to $(g^\alpha_{\lambda\mu})$ and $(\gamma^\beta_{\lambda\mu})$. Therefore cocycles on different covers define isomorphic holomorphic vector bundles exactly when their pullbacks to a common refinement are related by holomorphic gauge transformations on the refined cover. This completes the classification. [/step]