[proofplan]
The transition functions are obtained by reading the fibre coordinate of the change-of-trivialization maps coming from local trivializations of a fibre bundle. The identity and inverse identities follow from composing a trivialization with itself and from the fact that inverse coordinate changes are inverse maps. On a triple overlap, the direct change from the $k$-trivialization to the $i$-trivialization factors through the $j$-trivialization, and comparing the fibre coordinate gives the cocycle identity. Because the transition functions take values in $\operatorname{Diff}(F)$, each fibre-coordinate change is a diffeomorphism with smooth inverse, so the same algebraic identities hold in $\operatorname{Diff}(F)$. The smooth vector bundle statement is the same argument after representing fibrewise linear automorphisms of $\mathbb R^k$ by matrices in $GL(k,\mathbb R)$.
[/proofplan]
[step:Compare a trivialization with itself to obtain the identity transition function]
For each $i\in I$, let
\begin{align*}
\Phi_i:\pi^{-1}(U_i)&\to U_i\times F
\end{align*}
be the smooth local trivialization over $U_i$. By definition of a bundle trivialization, the first coordinate of $\Phi_i(e)$ is $\pi(e)$ for every $e\in \pi^{-1}(U_i)$; hence every change-of-trivialization map preserves the base coordinate. Since each $\Phi_i$ is a diffeomorphism onto $U_i\times F$, every overlap change $\Phi_i\circ\Phi_j^{-1}$ is a smooth diffeomorphism whose inverse is $\Phi_j\circ\Phi_i^{-1}$, and the fibre-coordinate maps are the stated $\operatorname{Diff}(F)$-valued transition functions.
Fix $i\in I$. Define the overlap domain $U_{ii}:=U_i$. The transition map from the $i$-trivialization to itself is the map $\Phi_i\circ \Phi_i^{-1}:U_i\times F\to U_i\times F$. Since $\Phi_i^{-1}$ is the inverse map of $\Phi_i$, this composition is the identity map on $U_i\times F$. Therefore, for every $x\in U_i$ and every $y\in F$,
\begin{align*}
(x,g_{ii}(x)(y))=(\Phi_i\circ \Phi_i^{-1})(x,y)=(x,y).
\end{align*}
Equality in the product $U_i\times F$ gives $g_{ii}(x)(y)=y$ for every $y\in F$. Hence $g_{ii}(x)=\operatorname{id}_F$ for every $x\in U_i$.
[/step]
[step:Invert the change of trivialization to obtain the inverse identity]
Fix $i,j\in I$ with $U_i\cap U_j\neq\varnothing$, and define $U_{ij}:=U_i\cap U_j$. The transition map from the $j$-trivialization to the $i$-trivialization is the map $\Phi_i\circ \Phi_j^{-1}:U_{ij}\times F\to U_{ij}\times F$. The transition map from the $i$-trivialization to the $j$-trivialization is the map $\Phi_j\circ \Phi_i^{-1}:U_{ij}\times F\to U_{ij}\times F$. These maps are inverse maps. Indeed, composing in this order gives
\begin{align*}
(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_i^{-1})=\operatorname{id}_{U_{ij}\times F}.
\end{align*}
The middle composition is $\Phi_j^{-1}\circ\Phi_j=\operatorname{id}$ on $\pi^{-1}(U_{ij})$, and the remaining composition is $\Phi_i\circ \Phi_i^{-1}=\operatorname{id}_{U_{ij}\times F}$.
Using the defining form of the transition functions, fix $x\in U_{ij}$ and $y\in F$. Since the displayed composition is the identity on $U_{ij}\times F$,
\begin{align*}
(x,y)=(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_i^{-1})(x,y).
\end{align*}
The first transition sends $(x,y)$ to $(x,g_{ji}(x)(y))$, so
\begin{align*}
(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_i^{-1})(x,y)=(\Phi_i\circ \Phi_j^{-1})(x,g_{ji}(x)(y)).
\end{align*}
Applying the second transition gives
\begin{align*}
(\Phi_i\circ \Phi_j^{-1})(x,g_{ji}(x)(y))=(x,g_{ij}(x)(g_{ji}(x)(y))).
\end{align*}
Therefore
\begin{align*}
(x,y)=(x,g_{ij}(x)(g_{ji}(x)(y))).
\end{align*}
Thus $g_{ij}(x)\circ g_{ji}(x)=\operatorname{id}_F$. Since the transition functions take values in $\operatorname{Diff}(F)$, both $g_{ij}(x)$ and $g_{ji}(x)$ are diffeomorphisms of $F$; the displayed identity therefore identifies $g_{ji}(x)$ as the inverse diffeomorphism of $g_{ij}(x)$. Hence
\begin{align*}
g_{ij}(x)=g_{ji}(x)^{-1}.
\end{align*}
[/step]
[step:Factor a triple change of trivialization through the middle chart]
Fix $i,j,k\in I$ with $U_i\cap U_j\cap U_k\neq\varnothing$, and define $U_{ijk}:=U_i\cap U_j\cap U_k$. On $U_{ijk}\times F$, the change from the $k$-trivialization to the $i$-trivialization factors as
\begin{align*}
\Phi_i\circ \Phi_k^{-1}
=
(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_k^{-1}).
\end{align*}
Indeed, by associativity of composition and the identity $\Phi_j^{-1}\circ\Phi_j=\operatorname{id}$ on $\pi^{-1}(U_{ijk})$, the right-hand side satisfies
\begin{align*}
(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_k^{-1})=\Phi_i\circ(\Phi_j^{-1}\circ\Phi_j)\circ\Phi_k^{-1}=\Phi_i\circ\Phi_k^{-1}.
\end{align*}
Evaluating both sides at $(x,y)\in U_{ijk}\times F$ gives the direct expression
\begin{align*}
(\Phi_i\circ \Phi_k^{-1})(x,y)=(x,g_{ik}(x)(y)).
\end{align*}
The same equality of maps also gives
\begin{align*}
(\Phi_i\circ \Phi_k^{-1})(x,y)=(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_k^{-1})(x,y).
\end{align*}
Using the definition of $g_{jk}$,
\begin{align*}
(\Phi_j\circ \Phi_k^{-1})(x,y)=(x,g_{jk}(x)(y)).
\end{align*}
Using the definition of $g_{ij}$ on that output,
\begin{align*}
(\Phi_i\circ \Phi_j^{-1})(x,g_{jk}(x)(y))=(x,g_{ij}(x)(g_{jk}(x)(y))).
\end{align*}
Combining these equalities gives
\begin{align*}
(x,g_{ik}(x)(y))=(x,g_{ij}(x)(g_{jk}(x)(y))).
\end{align*}
Equality in $U_{ijk}\times F$ gives
\begin{align*}
g_{ik}(x)(y)=\bigl(g_{ij}(x)\circ g_{jk}(x)\bigr)(y)
\end{align*}
for every $y\in F$. Hence
\begin{align*}
g_{ij}(x)\circ g_{jk}(x)=g_{ik}(x)
\end{align*}
for every $x\in U_{ijk}$.
[guided]
The point of the triple-overlap identity is that there are two ways to express the same change of fibre coordinates. Fix $i,j,k\in I$ and set
\begin{align*}
U_{ijk}:=U_i\cap U_j\cap U_k.
\end{align*}
Assume $U_{ijk}\neq\varnothing$. The direct change from the $k$-trivialization to the $i$-trivialization is the map $\Phi_i\circ \Phi_k^{-1}:U_{ijk}\times F\to U_{ijk}\times F$. The indirect change first goes from the $k$-trivialization to the $j$-trivialization and then from the $j$-trivialization to the $i$-trivialization. This indirect map is $(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_k^{-1}):U_{ijk}\times F\to U_{ijk}\times F$. These two maps are equal because associativity of composition gives the intermediate expression $\Phi_i\circ(\Phi_j^{-1}\circ\Phi_j)\circ\Phi_k^{-1}$, and $\Phi_j^{-1}\circ\Phi_j$ is the identity on $\pi^{-1}(U_{ijk})$. Hence
\begin{align*}
(\Phi_i\circ \Phi_j^{-1})\circ(\Phi_j\circ \Phi_k^{-1})=\Phi_i\circ\Phi_k^{-1}.
\end{align*}
Now evaluate this equality at an arbitrary point $(x,y)\in U_{ijk}\times F$. By definition of the transition functions,
\begin{align*}
(\Phi_j\circ \Phi_k^{-1})(x,y)=(x,g_{jk}(x)(y)).
\end{align*}
Applying the next transition map gives
\begin{align*}
(\Phi_i\circ \Phi_j^{-1})(x,g_{jk}(x)(y))
=(x,g_{ij}(x)(g_{jk}(x)(y))).
\end{align*}
On the other hand, the direct transition gives
\begin{align*}
(\Phi_i\circ \Phi_k^{-1})(x,y)=(x,g_{ik}(x)(y)).
\end{align*}
Since the direct and indirect maps are the same map, we have
\begin{align*}
(x,g_{ik}(x)(y))=(x,g_{ij}(x)(g_{jk}(x)(y))).
\end{align*}
Equality in the product $U_{ijk}\times F$ means equality of both coordinates. The base coordinate is already $x$ on both sides, so the fibre coordinates must satisfy
\begin{align*}
g_{ik}(x)(y)=g_{ij}(x)(g_{jk}(x)(y)).
\end{align*}
Because $y\in F$ was arbitrary, this is precisely the equality of maps
\begin{align*}
g_{ij}(x)\circ g_{jk}(x)=g_{ik}(x).
\end{align*}
[/guided]
[/step]
[step:Represent the vector-bundle transition maps by matrices]
Assume now that $\pi:E\to M$ is a smooth vector bundle of rank $k$ and that each $\Phi_i$ is a smooth vector-bundle trivialization, as in the vector-bundle version of the theorem statement. Then $F=\mathbb R^k$, and each fibre map $g_{ij}(x):\mathbb R^k\to\mathbb R^k$ is a linear isomorphism. Define $A_{ij}:U_i\cap U_j\to GL(k,\mathbb R)$ to be the matrix of the [linear map](/page/Linear%20Map) $g_{ij}(x)$ with respect to the standard basis of $\mathbb R^k$. The identity map $\operatorname{id}_{\mathbb R^k}$ is represented by $I_k$, inverses of linear maps are represented by inverse matrices, and composition of linear maps is represented by matrix multiplication in the same order. Therefore the already proved identity transition law becomes
\begin{align*}
A_{ii}(x)=I_k.
\end{align*}
The already proved inverse transition law becomes
\begin{align*}
A_{ij}(x)=A_{ji}(x)^{-1}.
\end{align*}
The already proved cocycle law becomes
\begin{align*}
A_{ij}(x)A_{jk}(x)=A_{ik}(x).
\end{align*}
These identities hold on the corresponding overlaps. This proves the vector-bundle version and completes the proof.
[/step]
