[proofplan]
Choose a local section $s:U\to P$ and use it to write the principal connection as the $\mathfrak g$-valued local connection form $A_s=s^*A$. The induced connection on the associated bundle has local connection matrix $\rho_*(A_s)$, so its curvature is computed by the usual vector bundle formula. Since $\rho_*:\mathfrak g\to\mathfrak{gl}(V)$ is a [Lie algebra](/page/Lie%20Algebra) homomorphism and commutes with exterior differentiation as a constant [linear map](/page/Linear%20Map), the curvature formula becomes $F_E=\rho_*(F_{A,s})$. Evaluating a $GL(V)$-invariant polynomial then gives the pulled-back invariant polynomial on $\mathfrak g$, and the local identities glue because both sides are the local representatives of globally defined forms.
[/proofplan]
[step:Write the induced connection in the local frame determined by the section]
Fix an [open set](/page/Open%20Set) $U\subset M$ and a smooth local section $s:U\to P$. Define the local principal connection form
\begin{align*}
A_s:=s^*A\in\Omega^1(U;\mathfrak g).
\end{align*}
The section $s$ determines the vector bundle trivialization
\begin{align*}
\Phi_s:E|_U\to U\times V
\end{align*}
by sending the equivalence class $[s(x),v]$ to $(x,v)$. This uses the associated-bundle convention $(pg,v)\sim(p,\rho(g)v)$ from the theorem statement.
Let
\begin{align*}
\sigma:U\to E|_U
\end{align*}
be a smooth local section. In the trivialization $\Phi_s$, write $\sigma(x)=[s(x),f(x)]$ for a smooth map
\begin{align*}
f:U\to V.
\end{align*}
By the construction of the associated connection from the horizontal distribution of $A$, equivalently by the local connection formula in [citetheorem:6271] with $Q=P$ and $\omega=A$, the induced covariant derivative has local expression
\begin{align*}
\nabla^E\sigma=df+\rho_*(A_s)f.
\end{align*}
Thus the local connection matrix of $\nabla^E$ in the frame determined by $s$ is
\begin{align*}
B_s:=\rho_*(A_s)\in\Omega^1(U;\mathfrak{gl}(V)).
\end{align*}
[guided]
The purpose of choosing $s$ is to compare two objects that naturally live in different places. The principal curvature $F_A$ is a $\mathfrak g$-valued form on $P$, while the vector bundle curvature is locally a $\mathfrak{gl}(V)$-valued form on $U$. Pulling back by
\begin{align*}
s:U\to P
\end{align*}
puts the principal connection on the same base open set:
\begin{align*}
A_s:=s^*A\in\Omega^1(U;\mathfrak g).
\end{align*}
The same local section gives a trivialization
\begin{align*}
\Phi_s:E|_U\to U\times V
\end{align*}
by the rule $\Phi_s([s(x),v])=(x,v)$, using the quotient convention $(pg,v)\sim(p,\rho(g)v)$. Therefore every smooth section
\begin{align*}
\sigma:U\to E|_U
\end{align*}
can be written uniquely as $\sigma(x)=[s(x),f(x)]$ for a smooth map
\begin{align*}
f:U\to V.
\end{align*}
In this trivialization, the local connection formula for an associated vector bundle, applied as in [citetheorem:6271] with $Q=P$ and $\omega=A$, says that the connection induced from the principal connection differentiates the $V$-valued representative $f$ and adds the infinitesimal action of the local principal connection form. Hence
\begin{align*}
\nabla^E\sigma=df+\rho_*(A_s)f.
\end{align*}
This identifies the local connection matrix of $\nabla^E$ as
\begin{align*}
B_s:=\rho_*(A_s)\in\Omega^1(U;\mathfrak{gl}(V)).
\end{align*}
This is the only place where the construction of the associated connection is used: after this identification, the proof is a curvature computation.
[/guided]
[/step]
[step:Compute the associated curvature using that $\rho_*$ preserves brackets]
The curvature formula for a vector bundle connection with local connection matrix $B_s$, again the local curvature formula from [citetheorem:6271] in the associated-bundle setting, is
\begin{align*}
F_E=dB_s+\frac{1}{2}[B_s\wedge B_s].
\end{align*}
Substituting $B_s=\rho_*(A_s)$ gives
\begin{align*}
F_E=d(\rho_*(A_s))+\frac{1}{2}[\rho_*(A_s)\wedge\rho_*(A_s)].
\end{align*}
Since $\rho_*:\mathfrak g\to\mathfrak{gl}(V)$ is a constant linear map, exterior differentiation commutes with applying $\rho_*$:
\begin{align*}
d(\rho_*(A_s))=\rho_*(dA_s).
\end{align*}
Since $\rho_*$ is a Lie algebra homomorphism, for all $X,Y\in\mathfrak g$ one has
\begin{align*}
\rho_*([X,Y])=[\rho_*X,\rho_*Y].
\end{align*}
Applying this pointwise to the bracket-wedge product gives
\begin{align*}
[\rho_*(A_s)\wedge\rho_*(A_s)]=\rho_*([A_s\wedge A_s]).
\end{align*}
Therefore
\begin{align*}
F_E=\rho_*(dA_s)+\frac{1}{2}\rho_*([A_s\wedge A_s]).
\end{align*}
By linearity of $\rho_*$,
\begin{align*}
F_E=\rho_*\left(dA_s+\frac{1}{2}[A_s\wedge A_s]\right).
\end{align*}
Pullback commutes with exterior differentiation and with the bracket-wedge product of forms, so
\begin{align*}
dA_s+\frac{1}{2}[A_s\wedge A_s]=s^*\left(dA+\frac{1}{2}[A\wedge A]\right)=s^*F_A=F_{A,s}.
\end{align*}
Hence
\begin{align*}
F_E=\rho_*(F_{A,s}).
\end{align*}
[/step]
[step:Evaluate invariant polynomials after pulling them back along $\rho_*$]
Let $q:\mathfrak{gl}(V)^k\to\mathbb K$ be a $GL(V)$-invariant symmetric $k$-linear polynomial. Define
\begin{align*}
\rho^*q:\mathfrak g^k\to\mathbb K
\end{align*}
by
\begin{align*}
(\rho^*q)(X_1,\dots,X_k):=q(\rho_*X_1,\dots,\rho_*X_k).
\end{align*}
For $g\in G$, let $\operatorname{Ad}_g:\mathfrak g\to\mathfrak g$ denote the adjoint action of $g$ on the Lie algebra $\mathfrak g$. For $T\in GL(V)$, let $\operatorname{Ad}_T:\mathfrak{gl}(V)\to\mathfrak{gl}(V)$ denote conjugation, so $\operatorname{Ad}_T(S)=TST^{-1}$ for $S\in\mathfrak{gl}(V)$. For $g\in G$ and $X_1,\dots,X_k\in\mathfrak g$, the identity
\begin{align*}
\rho_*(\operatorname{Ad}_g X_i)=\operatorname{Ad}_{\rho(g)}(\rho_*X_i)
\end{align*}
and the $GL(V)$-invariance of $q$ imply
\begin{align*}
(\rho^*q)(\operatorname{Ad}_g X_1,\dots,\operatorname{Ad}_g X_k)=(\rho^*q)(X_1,\dots,X_k).
\end{align*}
Thus $\rho^*q$ is $\operatorname{Ad}(G)$-invariant.
Using the curvature identity from the previous step and the definition of evaluation of a symmetric multilinear polynomial on curvature forms, we obtain
\begin{align*}
q(F_E)=q(\rho_*(F_{A,s}),\dots,\rho_*(F_{A,s})).
\end{align*}
By the definition of $\rho^*q$, this is exactly
\begin{align*}
q(F_E)=(\rho^*q)(F_{A,s}).
\end{align*}
[/step]
[step:Glue the local identities to obtain the global Chern-Weil identity]
The form $F_E$ is the curvature of the globally defined connection $\nabla^E$, and $F_A$ is the curvature of the globally defined principal connection $A$. On each local section $s:U\to P$, the preceding steps show that the local representative of $q(F_E)$ is equal to the local representative of $(\rho^*q)(F_A)$:
\begin{align*}
q(F_E)=(\rho^*q)(F_{A,s}).
\end{align*}
On overlaps of two such trivializations, both expressions transform by the same change-of-frame rule for Chern-Weil forms, and the invariant polynomial removes the conjugation dependence. Hence the local equalities agree on overlaps. Therefore the globally defined Chern-Weil form obtained from $q$ and the associated vector bundle connection equals the Chern-Weil form obtained from $\rho^*q$ and the principal connection $A$. This proves the claimed compatibility.
[/step]
