= Solution
For a smooth complex <vector bundle> $E$, a connection is a complex-linear map $D:\Gamma(E)\to A^1(M;E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth complex functions $f$. Local trivializations supply flat local connections. A locally finite smooth <partition of unity> $(\rho_i)$ subordinate to them gives $D=\sum_i\rho_iD_i$; local finiteness makes the sum smooth and $\sum_i\rho_i=1$ gives the <Leibniz rule>. Thus <connections on a vector bundle> always exist on the usual paracompact smooth manifold.
Extend $D$ to bundle-valued forms by $D(\alpha\otimes s)=d\alpha\otimes s+(-1)^{\deg\alpha}\alpha\wedge Ds$. Squaring shows that $D^2(fs)=fD^2s$, so the <curvature form> is the endomorphism-valued two-form $\Theta=D^2$. With a row frame $e=(e_1,\ldots,e_r)$, write $De=eA$; for coefficient columns $s=ev$, $Ds=e(dv+Av)$. Applying $D$ again and using the graded <Leibniz rule> gives
$$
D^2(ev)=e(dA+A\wedge A)v.
$$
The cross terms $-A\wedge dv$ and $A\wedge dv$ cancel. Hence the <Cartan curvature matrix equation> in this convention is
$$
\boxed{\Theta=dA+A\wedge A.}
$$
For a frame change $e'=eg$, the <connection matrix> and curvature transform as $A'=g^{-1}Ag+g^{-1}dg$ and $\Theta'=g^{-1}\Theta g$. Trace is invariant under conjugation, so the local forms $\operatorname{Tr}\Theta$ patch into the global <trace of vector-bundle curvature>.
The induced <determinant connection> on $\det E=\Lambda^rE$ is
$$
D_r(s_1\wedge\cdots\wedge s_r)=\sum_{j=1}^r s_1\wedge\cdots\wedge Ds_j\wedge\cdots\wedge s_r,
$$
where the one-form factor of each $Ds_j$ is placed in front of the section factors. In the local determinant frame $e_1\wedge\cdots\wedge e_r$, only the diagonal terms survive, so its connection form is $\operatorname{Tr}A$. Its curvature is $d\operatorname{Tr}A$. Since
$$
\operatorname{Tr}(A\wedge A)=\sum_{i,j}A_{ij}\wedge A_{ji}=0
$$
by pairing off-diagonal terms and using $A_{ii}\wedge A_{ii}=0$, we obtain
$$
\boxed{\Theta_{D_r}=d\operatorname{Tr}A=\operatorname{Tr}\Theta_D.}
$$
It is closed locally by $d^2=0$, and thus closed globally.
For two connections, $B=D_1-D_0$ is a global endomorphism-valued one-form: the derivative terms cancel in their Leibniz rules. In a local frame $A_1=A_0+B$, so
$$
\Theta_1-\Theta_0=dB+A_0\wedge B+B\wedge A_0+B\wedge B.
$$
Taking traces cancels the mixed terms, because both factors are one-forms, and cancels $\operatorname{Tr}(B\wedge B)$. Therefore the <trace curvature transgression> formula is
$$
\boxed{\operatorname{Tr}\Theta_1-\operatorname{Tr}\Theta_0=d\operatorname{Tr}B.}
$$
The right-hand side is globally exact, proving independence of the <de Rham cohomology> class without needing a sheaf-cohomology identification.
Finally let $e$ be a holomorphic frame for the <holomorphic line bundle>, with $h=\|e\|^2>0$. Define $De=e\,\partial\log h$, extending by the Leibniz rule. If $e'=eg$ for a nowhere-zero holomorphic function $g$, then $h'=|g|^2h$ and
$$
\partial\log h'=\partial\log h+g^{-1}dg.
$$
This is exactly the line-bundle frame-change law for a connection, so the local definitions glue globally. Its $(0,1)$ part is the <Dolbeault operator>. Metric compatibility follows from $d\log h=\partial\log h+\bar\partial\log h$, or $dh=h(A+\overline A)$ for $A=\partial\log h$. These conditions uniquely force this $A$, identifying the <Chern connection>. Since a scalar one-form wedges with itself to zero,
$$
\boxed{\Theta=d(\partial\log h)=\bar\partial\partial\log h=-\partial\bar\partial\log h.}
$$
This agrees with the <local formula for the Chern connection on a line bundle>; keeping the order of the two Dolbeault differentials fixes the sign.
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