[proofplan]
We compute every transition function directly from the induced local trivializations. On an overlap $U_i \cap U_j$, the $E$-coordinate changes by $g_{ij}^E$ and the $F$-coordinate changes by $g_{ij}^F$, so the functorial operations on vector spaces give the direct sum, [tensor product](/page/Tensor%20Product), and exterior power formulas. The dual formula is forced by invariance of the evaluation pairing, and the Hom formula is forced by conjugating the coordinate representation of a fiberwise [linear map](/page/Linear%20Map). Finally, the cocycle identities follow from the original cocycle identities and functoriality of the corresponding linear-algebra operations, so the reconstructed bundles have exactly the same transition data as the operation bundles.
[/proofplan]
[step:Fix the transition convention and induced operation trivializations]
For each $i \in I$, let $v_i: \pi_E^{-1}(U_i) \to \mathbb{R}^r$ and $w_i: \pi_F^{-1}(U_i) \to \mathbb{R}^s$ be the coordinate maps determined by the chosen trivializations of the vector bundles $E$ and $F$. Thus $\Phi_i^E(e)=(\pi_E(e),v_i(e))$ for $e \in \pi_E^{-1}(U_i)$, and $\Phi_i^F(f)=(\pi_F(f),w_i(f))$ for $f \in \pi_F^{-1}(U_i)$. The stated convention means that for $p \in U_i \cap U_j$ and $e \in E_p$,
\begin{align*}
v_i(e) = g_{ij}^E(p)v_j(e),
\end{align*}
and for $f \in F_p$,
\begin{align*}
w_i(f) = g_{ij}^F(p)w_j(f).
\end{align*}
The induced trivialization of the direct sum bundle $E \oplus F$ over $U_i$ is the smooth map $\Phi_i^{E \oplus F}: (E \oplus F)|_{U_i} \to U_i \times (\mathbb{R}^r \oplus \mathbb{R}^s)$ given by $(e,f) \mapsto (\pi_E(e), v_i(e) \oplus w_i(f))$.
The induced trivialization of the tensor product bundle $E \otimes F$ over $U_i$ is the smooth map $\Phi_i^{E \otimes F}: (E \otimes F)|_{U_i} \to U_i \times (\mathbb{R}^r \otimes \mathbb{R}^s)$ defined fiberwise by sending $e \otimes f \in E_p \otimes F_p$ to $(p, v_i(e) \otimes w_i(f))$ and then extending linearly on each fiber. For $0 \leq k \leq r$, the induced trivialization of the exterior power bundle $\Lambda^k E$ over $U_i$ is the smooth map $\Phi_i^{\Lambda^k E}: (\Lambda^k E)|_{U_i} \to U_i \times \Lambda^k \mathbb{R}^r$ defined fiberwise by sending $e_1 \wedge \cdots \wedge e_k$ to $(p, v_i(e_1) \wedge \cdots \wedge v_i(e_k))$ and extending linearly. The induced trivialization of the dual bundle $E^*$ over $U_i$ is the smooth map $\Phi_i^{E^*}: (E^*)|_{U_i} \to U_i \times (\mathbb{R}^r)^*$ that sends $\lambda \in E_p^*$ to the covector $\lambda_i \in (\mathbb{R}^r)^*$ defined by
\begin{align*}
\lambda_i(v_i(e)) = \lambda(e)
\end{align*}
for every $e \in E_p$. The induced trivialization of the Hom bundle $\operatorname{Hom}(E,F)$ over $U_i$ is the smooth map $\Phi_i^{\operatorname{Hom}(E,F)}: \operatorname{Hom}(E,F)|_{U_i} \to U_i \times \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s)$ that sends $T \in \operatorname{Hom}(E_p,F_p)$ to the linear map $A_i \in \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s)$ characterized by
\begin{align*}
A_i(v_i(e)) = w_i(T(e))
\end{align*}
for every $e \in E_p$.
[/step]
[step:Compute the direct sum, tensor product, and exterior power transitions]
Let $p \in U_i \cap U_j$. For $(e,f) \in E_p \oplus F_p$, the coordinate change in $E \oplus F$ is
\begin{align*}
v_i(e) \oplus w_i(f) = \bigl(g_{ij}^E(p) \oplus g_{ij}^F(p)\bigr)\bigl(v_j(e) \oplus w_j(f)\bigr).
\end{align*}
Thus
\begin{align*}
g_{ij}^{E \oplus F}(p) = g_{ij}^E(p) \oplus g_{ij}^F(p).
\end{align*}
For a decomposable tensor $e \otimes f \in E_p \otimes F_p$, the coordinate change in $E \otimes F$ is
\begin{align*}
v_i(e) \otimes w_i(f) = \bigl(g_{ij}^E(p) \otimes g_{ij}^F(p)\bigr)\bigl(v_j(e) \otimes w_j(f)\bigr).
\end{align*}
Since decomposable tensors span $E_p \otimes F_p$, the transition map is
\begin{align*}
g_{ij}^{E \otimes F}(p) = g_{ij}^E(p) \otimes g_{ij}^F(p).
\end{align*}
For $e_1,\dots,e_k \in E_p$, the coordinate change in $\Lambda^k E$ is
\begin{align*}
v_i(e_1) \wedge \cdots \wedge v_i(e_k) = \bigl(\Lambda^k g_{ij}^E(p)\bigr)\bigl(v_j(e_1) \wedge \cdots \wedge v_j(e_k)\bigr).
\end{align*}
Since decomposable alternating tensors span $\Lambda^k E_p$, this gives
\begin{align*}
g_{ij}^{\Lambda^k E}(p)=\Lambda^k g_{ij}^E(p).
\end{align*}
[/step]
[step:Force the dual transition by invariance of evaluation]
Let $p \in U_i \cap U_j$ and let $\lambda \in E_p^*$. Define $\lambda_i,\lambda_j \in (\mathbb{R}^r)^*$ by requiring $\lambda_i(v_i(e))=\lambda(e)$ and $\lambda_j(v_j(e))=\lambda(e)$ for every $e \in E_p$. Since $v_i(e)=g_{ij}^E(p)v_j(e)$, for every $a \in \mathbb{R}^r$ we may choose the unique $e \in E_p$ with $a=v_j(e)$ and obtain
\begin{align*}
\lambda_j(a)=\lambda(e)=\lambda_i(v_i(e))=\lambda_i(g_{ij}^E(p)a).
\end{align*}
Thus, as covectors,
\begin{align*}
\lambda_j = \lambda_i \circ g_{ij}^E(p).
\end{align*}
Equivalently,
\begin{align*}
\lambda_i = \lambda_j \circ \bigl(g_{ij}^E(p)\bigr)^{-1}.
\end{align*}
Representing covectors as column coordinate vectors relative to the dual standard basis, the pullback action $\lambda_j \mapsto \lambda_j \circ (g_{ij}^E(p))^{-1}$ is represented by the matrix $\bigl(g_{ij}^E(p)\bigr)^{-\top}$. Hence
\begin{align*}
g_{ij}^{E^*}(p)=\bigl(g_{ij}^E(p)\bigr)^{-\top}.
\end{align*}
[guided]
The dual bundle is the one operation where the direction of the transition is easiest to get wrong. The defining invariant object is the evaluation pairing $E_p^* \times E_p \to \mathbb{R}$, $(\lambda,e) \mapsto \lambda(e)$. This number must not depend on which local frame is used.
Let $\lambda \in E_p^*$ and let $\lambda_i,\lambda_j \in (\mathbb{R}^r)^*$ be its coordinate covectors in the $i$-th and $j$-th trivializations. By definition of these coordinate covectors, $\lambda_i(v_i(e))=\lambda(e)$ and $\lambda_j(v_j(e))=\lambda(e)$ for every $e \in E_p$. The vector coordinates satisfy
\begin{align*}
v_i(e)=g_{ij}^E(p)v_j(e).
\end{align*}
Now fix an arbitrary vector $a \in \mathbb{R}^r$. Because $v_j: E_p \to \mathbb{R}^r$ is a linear isomorphism, there is a unique $e \in E_p$ such that $a=v_j(e)$. Substituting this $e$ gives
\begin{align*}
\lambda_j(a)=\lambda_j(v_j(e))=\lambda(e)=\lambda_i(v_i(e))=\lambda_i(g_{ij}^E(p)v_j(e))=\lambda_i(g_{ij}^E(p)a).
\end{align*}
Since this holds for every $a \in \mathbb{R}^r$, we have
\begin{align*}
\lambda_j = \lambda_i \circ g_{ij}^E(p).
\end{align*}
Solving for $\lambda_i$ gives
\begin{align*}
\lambda_i = \lambda_j \circ \bigl(g_{ij}^E(p)\bigr)^{-1}.
\end{align*}
Therefore the dual transition sends the $j$-coordinate covector to the $i$-coordinate covector by pullback through $\bigl(g_{ij}^E(p)\bigr)^{-1}$.
In matrix notation, if covectors are represented as column vectors in the dual standard basis, precomposition by $\bigl(g_{ij}^E(p)\bigr)^{-1}$ is represented by the transpose inverse matrix:
\begin{align*}
\lambda_i = \bigl(g_{ij}^E(p)\bigr)^{-\top}\lambda_j.
\end{align*}
Thus the transition function for $E^*$ is
\begin{align*}
g_{ij}^{E^*}(p)=\bigl(g_{ij}^E(p)\bigr)^{-\top}.
\end{align*}
[/guided]
[/step]
[step:Compute the Hom transition by conjugating coordinate representatives]
Let $p \in U_i \cap U_j$ and let $T \in \operatorname{Hom}(E_p,F_p)$. Let $A_i,A_j: \mathbb{R}^r \to \mathbb{R}^s$ be the coordinate representatives of $T$ in the $i$-th and $j$-th trivializations, so that $A_i(v_i(e))=w_i(T(e))$ and $A_j(v_j(e))=w_j(T(e))$ for every $e \in E_p$. For $a \in \mathbb{R}^r$, choose the unique $e \in E_p$ with $a=v_j(e)$. Then
\begin{align*}
A_i(g_{ij}^E(p)a)=A_i(v_i(e))=w_i(T(e))=g_{ij}^F(p)w_j(T(e))=g_{ij}^F(p)A_j(a).
\end{align*}
Replacing $a$ by $\bigl(g_{ij}^E(p)\bigr)^{-1}b$ gives, for every $b \in \mathbb{R}^r$,
\begin{align*}
A_i(b)=g_{ij}^F(p)A_j\bigl((g_{ij}^E(p))^{-1}b\bigr).
\end{align*}
Hence
\begin{align*}
A_i = g_{ij}^F(p)A_j\bigl(g_{ij}^E(p)\bigr)^{-1},
\end{align*}
so
\begin{align*}
g_{ij}^{\operatorname{Hom}(E,F)}(p)(A)
=
g_{ij}^F(p)A\bigl(g_{ij}^E(p)\bigr)^{-1}.
\end{align*}
[/step]
[step:Verify smoothness and the cocycle identities]
Each displayed transition map is smooth because it is obtained from the smooth maps $g_{ij}^E$ and $g_{ij}^F$ by smooth finite-dimensional operations: block sum, tensor product, matrix inversion, transpose, exterior power, and composition in $\operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s)$.
The original transition functions satisfy $g_{ij}^E(p)g_{j\ell}^E(p)=g_{i\ell}^E(p)$ and $g_{ij}^F(p)g_{j\ell}^F(p)=g_{i\ell}^F(p)$ for every $p \in U_i \cap U_j \cap U_\ell$. Therefore
\begin{align*}
(g_{ij}^E \oplus g_{ij}^F)(p)(g_{j\ell}^E \oplus g_{j\ell}^F)(p)
&= g_{i\ell}^E(p) \oplus g_{i\ell}^F(p),
\end{align*}
and
\begin{align*}
(g_{ij}^E \otimes g_{ij}^F)(p)(g_{j\ell}^E \otimes g_{j\ell}^F)(p)
&= g_{i\ell}^E(p) \otimes g_{i\ell}^F(p).
\end{align*}
For exterior powers,
\begin{align*}
\Lambda^k g_{ij}^E(p)\,\Lambda^k g_{j\ell}^E(p)=\Lambda^k\bigl(g_{ij}^E(p)g_{j\ell}^E(p)\bigr)=\Lambda^k g_{i\ell}^E(p).
\end{align*}
For duals,
\begin{align*}
\bigl(g_{ij}^E(p)\bigr)^{-\top}\bigl(g_{j\ell}^E(p)\bigr)^{-\top}=\bigl(g_{ij}^E(p)g_{j\ell}^E(p)\bigr)^{-\top}=\bigl(g_{i\ell}^E(p)\bigr)^{-\top},
\end{align*}
where the first equality uses $(AB)^{-\top}=A^{-\top}B^{-\top}$. For Hom, if $A \in \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s)$, then
\begin{align*}
g_{ij}^{\operatorname{Hom}(E,F)}(p)\bigl(g_{j\ell}^{\operatorname{Hom}(E,F)}(p)(A)\bigr)=g_{ij}^F(p)\bigl(g_{j\ell}^F(p)A(g_{j\ell}^E(p))^{-1}\bigr)(g_{ij}^E(p))^{-1}=g_{i\ell}^F(p)A\bigl(g_{ij}^E(p)g_{j\ell}^E(p)\bigr)^{-1}=g_{i\ell}^F(p)A\bigl(g_{i\ell}^E(p)\bigr)^{-1}=g_{i\ell}^{\operatorname{Hom}(E,F)}(p)(A).
\end{align*}
Thus every displayed family satisfies the cocycle identity.
[/step]
[step:Identify the reconstructed bundles with the operation bundles]
For each operation, the preceding computations show that the induced local trivializations on the actual operation bundle have exactly the displayed transition maps. A vector bundle reconstructed from a smooth cocycle is obtained by taking the quotient of the disjoint union of the local models $\bigsqcup_{i \in I} U_i \times V$ by the [equivalence relation](/page/Equivalence%20Relation) $(p,a)_j \sim (p,g_{ij}(p)a)_i$ on overlaps, where $V$ is the relevant model [vector space](/page/Vector%20Space) for the chosen operation. The map from this quotient to the actual operation bundle sends the class of $(p,a)_i$ to the unique element of the operation fiber over $p$ whose $i$-th local coordinate is $a$. This map is well-defined because, on $U_i \cap U_j$, replacing $(p,a)_j$ by $(p,g_{ij}(p)a)_i$ gives the same fiber element exactly by the transition identity computed above.
Consequently the glued bundle determined by $g_{ij}^{E \oplus F}$ is naturally isomorphic to $E \oplus F$, the glued bundle determined by $g_{ij}^{E \otimes F}$ is naturally isomorphic to $E \otimes F$, the glued bundle determined by $g_{ij}^{E^*}$ is naturally isomorphic to $E^*$, the glued bundle determined by $g_{ij}^{\Lambda^k E}$ is naturally isomorphic to $\Lambda^k E$, and the glued bundle determined by $g_{ij}^{\operatorname{Hom}(E,F)}$ is naturally isomorphic to $\operatorname{Hom}(E,F)$. This proves both the transition formulas and the reconstruction statement.
[/step]
