= Solution
Use the convention $\nabla=d+A$ with anti-Hermitian <connection one-forms>. For a unitary <complex line bundle>, $A$ is locally an $i\mathbb R$-valued one-form and $F_\nabla=dA$, since the Lie algebra is abelian. Under a change of unitary frame, the <vector-bundle curvature> transforms by conjugation, which is trivial for $U(1)$. Hence $F_\nabla$ is a globally defined imaginary-valued two-form, and \b[$iF_\nabla$ is an ordinary real two-form]. Equivalently, $\operatorname{End}(L)$ has its canonical scalar trivialization, so no nontrivial coefficient bundle remains.
Locally $dF_\nabla=d^2A=0$, and these equalities patch globally. If another <unitary connection> is $\nabla'=\nabla+i\beta$ for a global real one-form $\beta$, then
$$
F_{\nabla'}=F_\nabla+i\,d\beta,\qquad
\boxed{iF_{\nabla'}=iF_\nabla-d\beta.}
$$
Thus the real <vector-bundle curvature> is closed and its <de Rham cohomology> class is independent of the <unitary connection>.
With the <First Chern class> normalization $c_1(L)_{\mathbb R}=[iF_\nabla/(2\pi)]$, the transverse section also identifies this class as $\operatorname{PD}[\Sigma]$. Indeed, the complex orientation of each normal disk makes the section wind positively once around a transverse zero. The <Euler class> of the underlying oriented plane bundle counts these zeros with their induced orientation, and equals $c_1(L)$. This records the role of the section and orientation hypotheses; connection independence itself did not require choosing a section.
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