[proofplan]
We pass back and forth between a section and its coordinate representatives under the two trivializations. Starting from a section, the representatives are obtained by projecting the trivialized section to the $\mathbb{R}^r$ factor, and the compatibility condition is exactly the definition of the transition function. Conversely, compatible local coordinate maps define local sections by applying the inverse trivializations; the compatibility equation makes these local sections agree on the overlap. Continuity of the glued section follows because continuity is local on the [open cover](/page/Open%20Cover) $\{U,V\}$ of $U\cup V$.
[/proofplan]
[step:Extract the local coordinate maps from a section]
Let
\begin{align*}
s:U\cup V\to E
\end{align*}
be a continuous section of $\pi$ over $U\cup V$, so $\pi\circ s=\operatorname{id}_{U\cup V}$. For $W\in\{U,V\}$, let
\begin{align*}
\operatorname{pr}_{\mathbb{R}^r}:W\times \mathbb{R}^r\to \mathbb{R}^r
\end{align*}
denote the projection onto the second factor. Define
\begin{align*}
s_U:U\to \mathbb{R}^r,\qquad s_U(b)=\operatorname{pr}_{\mathbb{R}^r}(\Phi_U(s(b))).
\end{align*}
Define
\begin{align*}
s_V:V\to \mathbb{R}^r,\qquad s_V(b)=\operatorname{pr}_{\mathbb{R}^r}(\Phi_V(s(b))).
\end{align*}
Because $s|_U$, $s|_V$, $\Phi_U$, $\Phi_V$, and the projection maps are continuous, both $s_U$ and $s_V$ are continuous. Since $\Phi_U$ and $\Phi_V$ preserve base points, these definitions are equivalently characterized by
\begin{align*}
\Phi_U(s(b))=(b,s_U(b))
\end{align*}
for $b\in U$, and
\begin{align*}
\Phi_V(s(b))=(b,s_V(b))
\end{align*}
for $b\in V$.
[/step]
[step:Derive the overlap compatibility from the transition function]
Fix $b\in U\cap V$. From the definition of $s_U$, we have
\begin{align*}
s(b)=\Phi_U^{-1}(b,s_U(b)).
\end{align*}
Applying $\Phi_V$ and using the defining convention for $g_{VU}$ gives
\begin{align*}
\Phi_V(s(b))=(\Phi_V\circ \Phi_U^{-1})(b,s_U(b))=(b,g_{VU}(b)s_U(b)).
\end{align*}
On the other hand, the definition of $s_V$ gives
\begin{align*}
\Phi_V(s(b))=(b,s_V(b)).
\end{align*}
Equality in $V\times \mathbb{R}^r$ therefore implies
\begin{align*}
s_V(b)=g_{VU}(b)s_U(b).
\end{align*}
Since $b\in U\cap V$ was arbitrary, the compatibility equation holds on the whole overlap.
[/step]
[step:Build local sections from compatible coordinate maps]
Conversely, suppose continuous maps
\begin{align*}
s_U:U\to \mathbb{R}^r
\end{align*}
and
\begin{align*}
s_V:V\to \mathbb{R}^r
\end{align*}
satisfy
\begin{align*}
s_V(b)=g_{VU}(b)s_U(b)
\end{align*}
for every $b\in U\cap V$. Define
\begin{align*}
\sigma_U:U\to E,\qquad \sigma_U(b)=\Phi_U^{-1}(b,s_U(b)).
\end{align*}
Define
\begin{align*}
\sigma_V:V\to E,\qquad \sigma_V(b)=\Phi_V^{-1}(b,s_V(b)).
\end{align*}
The maps $\sigma_U$ and $\sigma_V$ are continuous because they are compositions of the continuous maps $b\mapsto (b,s_U(b))$, $b\mapsto (b,s_V(b))$, and the homeomorphisms $\Phi_U^{-1}$ and $\Phi_V^{-1}$.
[guided]
We now reverse the construction. The input is not a section of $E$ yet; it is only a pair of coordinate functions into the model fiber $\mathbb{R}^r$. To turn those coordinates into actual points of $E$, we must use the inverse trivializations.
For the map over $U$, define
\begin{align*}
\sigma_U:U\to E,\qquad \sigma_U(b)=\Phi_U^{-1}(b,s_U(b)).
\end{align*}
This is well-defined because $s_U(b)\in \mathbb{R}^r$ for every $b\in U$, so $(b,s_U(b))\in U\times \mathbb{R}^r$, which is the domain of $\Phi_U^{-1}$. Similarly, define
\begin{align*}
\sigma_V:V\to E,\qquad \sigma_V(b)=\Phi_V^{-1}(b,s_V(b)).
\end{align*}
This is well-defined because $(b,s_V(b))\in V\times \mathbb{R}^r$ for every $b\in V$.
We also need continuity. The map
\begin{align*}
U\to U\times \mathbb{R}^r,\qquad b\mapsto (b,s_U(b))
\end{align*}
is continuous because its first component is the identity map on $U$ and its second component is the continuous map $s_U$. Since $\Phi_U^{-1}:U\times \mathbb{R}^r\to \pi^{-1}(U)$ is continuous by the definition of a trivialization, the composition defining $\sigma_U$ is continuous. The same argument applies to
\begin{align*}
V\to V\times \mathbb{R}^r,\qquad b\mapsto (b,s_V(b)),
\end{align*}
so $\sigma_V$ is continuous.
[/guided]
[/step]
[step:Show the local sections agree on the overlap]
Let $b\in U\cap V$. Using the compatibility equation and the definition of the transition function, we compute
\begin{align*}
\Phi_V(\sigma_U(b))=(\Phi_V\circ \Phi_U^{-1})(b,s_U(b))=(b,g_{VU}(b)s_U(b)).
\end{align*}
By compatibility, this becomes
\begin{align*}
\Phi_V(\sigma_U(b))=(b,s_V(b)).
\end{align*}
By the definition of $\sigma_V$, we also have
\begin{align*}
\Phi_V(\sigma_V(b))=(b,s_V(b)).
\end{align*}
Since $\Phi_V$ is injective, $\sigma_U(b)=\sigma_V(b)$. Thus $\sigma_U$ and $\sigma_V$ agree on $U\cap V$.
[/step]
[step:Glue the local sections into a global section over $U\cup V$]
Define
\begin{align*}
s:U\cup V\to E
\end{align*}
by setting $s(b)=\sigma_U(b)$ for $b\in U$ and $s(b)=\sigma_V(b)$ for $b\in V$. The preceding step shows that this definition is independent of the choice when $b\in U\cap V$.
The map $s$ is continuous because $U$ and $V$ are open in $U\cup V$, they cover $U\cup V$, and the restrictions $s|_U=\sigma_U$ and $s|_V=\sigma_V$ are continuous. Finally, for $b\in U$,
\begin{align*}
\pi(s(b))=\pi(\Phi_U^{-1}(b,s_U(b)))=b,
\end{align*}
because $\Phi_U$ is a bundle trivialization over $U$. For $b\in V$, the same argument with $\Phi_V$ gives
\begin{align*}
\pi(s(b))=\pi(\Phi_V^{-1}(b,s_V(b)))=b.
\end{align*}
Therefore $\pi\circ s=\operatorname{id}_{U\cup V}$, so $s$ is a section of $\pi$ over $U\cup V$.
[/step]
[step:Verify that the two constructions are inverse to each other]
Starting with a section $s:U\cup V\to E$, constructing $s_U$ and $s_V$, and then rebuilding a section gives, for $b\in U$,
\begin{align*}
\Phi_U^{-1}(b,s_U(b))=\Phi_U^{-1}(\Phi_U(s(b)))=s(b).
\end{align*}
For $b\in V$ the same computation with $\Phi_V$ gives
\begin{align*}
\Phi_V^{-1}(b,s_V(b))=\Phi_V^{-1}(\Phi_V(s(b)))=s(b).
\end{align*}
Thus the rebuilt section is the original section.
Conversely, starting with compatible maps $s_U$ and $s_V$, the section constructed above satisfies
\begin{align*}
\Phi_U(s(b))=(b,s_U(b))
\end{align*}
for every $b\in U$, and
\begin{align*}
\Phi_V(s(b))=(b,s_V(b))
\end{align*}
for every $b\in V$, by construction. Hence extracting local coordinate representatives recovers the original pair. The correspondence is therefore bijective, completing the proof.
[/step]
