[proofplan] We first verify that the determinant transition functions are holomorphic, nowhere zero, and satisfy the line-bundle cocycle identity. Then we compare the line bundle obtained from this cocycle with the top exterior power $\Lambda^r E$ using the wedge of a local frame. The key computation is that changing a frame by the matrix $g_{ij}$ changes the top wedge by multiplication with $\det(g_{ij})$, so the transition functions of $\Lambda^r E$ are exactly the determinant cocycle. [/proofplan]

[step:Show that the determinant transition functions are holomorphic and nowhere zero] For each pair $i,j \in I$ with $U_i \cap U_j \neq \varnothing$, define \begin{align*} h_{ij}: U_i \cap U_j \to \mathbb{C}^* \end{align*} by $h_{ij}(x) = \det(g_{ij}(x))$. Since $g_{ij}: U_i \cap U_j \to GL(r,\mathbb{C})$ is holomorphic, each matrix entry of $g_{ij}$ is a [holomorphic function](/page/Holomorphic%20Function) on $U_i \cap U_j$. The determinant is a polynomial in the matrix entries, so $h_{ij}$ is holomorphic. Since $g_{ij}(x) \in GL(r,\mathbb{C})$ for every $x \in U_i \cap U_j$, the matrix $g_{ij}(x)$ is invertible, hence $\det(g_{ij}(x)) \neq 0$. Therefore $h_{ij}$ takes values in $\mathbb{C}^*$. [/step]

[step:Transfer the vector-bundle cocycle identity through the determinant] Let $i,j,k \in I$, and write $U_{ijk} := U_i \cap U_j \cap U_k$. On $U_{ijk}$, the transition functions of $E$ satisfy \begin{align*} g_{ij}(x)g_{jk}(x) = g_{ik}(x) \end{align*} for every $x \in U_{ijk}$. Applying multiplicativity of the determinant to this matrix identity gives \begin{align*} \det(g_{ij}(x))\det(g_{jk}(x)) = \det(g_{ik}(x)). \end{align*} Thus \begin{align*} h_{ij}(x)h_{jk}(x) = h_{ik}(x) \end{align*} for every $x \in U_{ijk}$. Also $g_{ii}(x) = I_r$ on $U_i$, so $h_{ii}(x) = \det(I_r) = 1$. Hence $(h_{ij})_{i,j \in I}$ is a holomorphic line-bundle cocycle. [/step]

[step:Compute the transition functions of the top exterior power]For each $i \in I$, let $e_i = (e_{i,1},\dots,e_{i,r})$ be the chosen local holomorphic frame of $E$ over $U_i$. Define a local holomorphic frame \begin{align*} s_i: U_i \to \Lambda^r E \end{align*} by \begin{align*} s_i(x) = e_{i,1}(x) \wedge \cdots \wedge e_{i,r}(x). \end{align*} On $U_i \cap U_j$, the frame-change convention $e_j = e_i g_{ij}$ means that, if $(g_{ij})_{ab}: U_i \cap U_j \to \mathbb{C}$ denotes the $(a,b)$ entry of $g_{ij}$, then \begin{align*} e_{j,b}(x) = \sum_{a=1}^r e_{i,a}(x)(g_{ij})_{ab}(x) \end{align*} for every $b \in \{1,\dots,r\}$ and every $x \in U_i \cap U_j$. By multilinearity and alternation of the exterior product, \begin{align*} e_{j,1}(x) \wedge \cdots \wedge e_{j,r}(x) = \det(g_{ij}(x)) e_{i,1}(x) \wedge \cdots \wedge e_{i,r}(x). \end{align*} Therefore \begin{align*} s_j(x) = h_{ij}(x)s_i(x) \end{align*} on $U_i \cap U_j$.[/step]

[guided]The purpose of this step is to identify the cocycle of $\Lambda^r E$ in the frames induced from the original frames of $E$. For each $i \in I$, the local frame $e_i = (e_{i,1},\dots,e_{i,r})$ of $E$ over $U_i$ determines a local holomorphic frame \begin{align*} s_i: U_i \to \Lambda^r E \end{align*} by \begin{align*} s_i(x) = e_{i,1}(x) \wedge \cdots \wedge e_{i,r}(x). \end{align*} This is a frame because $e_{i,1}(x),\dots,e_{i,r}(x)$ form a basis of the fiber $E_x$, and the top exterior power $\Lambda^r E_x$ is one-dimensional with basis $e_{i,1}(x) \wedge \cdots \wedge e_{i,r}(x)$. Now fix $i,j \in I$ and a point $x \in U_i \cap U_j$. The convention in the theorem is $e_j = e_i g_{ij}$. Writing $(g_{ij})_{ab}: U_i \cap U_j \to \mathbb{C}$ for the $(a,b)$ entry of the transition matrix, this means \begin{align*} e_{j,b}(x) = \sum_{a=1}^r e_{i,a}(x)(g_{ij})_{ab}(x) \end{align*} for each column index $b \in \{1,\dots,r\}$. We now wedge these $r$ equations together. Multilinearity of the exterior product expands the wedge as a sum over all choices of one row index for each column. Alternation kills every term in which two row indices are equal, because then two equal basis vectors $e_{i,a}(x)$ appear in the wedge. The surviving terms are exactly the terms indexed by permutations of $\{1,\dots,r\}$, and their signed sum is the determinant of $g_{ij}(x)$. Hence \begin{align*} e_{j,1}(x) \wedge \cdots \wedge e_{j,r}(x) = \det(g_{ij}(x)) e_{i,1}(x) \wedge \cdots \wedge e_{i,r}(x). \end{align*} Since $h_{ij}(x)$ was defined to be $\det(g_{ij}(x))$, this becomes \begin{align*} s_j(x) = h_{ij}(x)s_i(x). \end{align*} Thus the transition function of $\Lambda^r E$ from the frame $s_j$ to the frame $s_i$ is exactly $h_{ij}$.[/guided]

[step:Identify the determinant cocycle line bundle with $\Lambda^r E$] Let $L \to X$ be the holomorphic line bundle obtained from the cocycle $(h_{ij})_{i,j \in I}$. By construction, $L$ has local holomorphic frames $\ell_i: U_i \to L$ satisfying \begin{align*} \ell_j(x) = h_{ij}(x)\ell_i(x) \end{align*} on $U_i \cap U_j$. The preceding step shows that the induced frames $s_i$ of $\Lambda^r E$ satisfy the same transition rule: \begin{align*} s_j(x) = h_{ij}(x)s_i(x). \end{align*} Therefore the local frame-preserving maps $\ell_i(x) \mapsto s_i(x)$ agree on overlaps, because both sides transform by the same scalar $h_{ij}(x)$. They glue to a holomorphic line-bundle isomorphism \begin{align*} L \to \Lambda^r E. \end{align*} Hence the line bundle defined by the determinant cocycle is canonically isomorphic to $\Lambda^r E$. This line bundle is denoted $\det E$. [/step]