Unitary trivialization over Riemann surfaces with boundary

Question

I am puzzled with the proof of Proposition 2.66. in the book "Introduction to Symplectic Topology" by Salamon, McDuff.

The Proposition states, that every Hermitian vector bundle $E \rightarrow \Sigma$ over a compact smooth Riemann surface $\Sigma$ with $\partial \Sigma \neq \emptyset$ admitts a unitary trivialization.

The first part of the proof consists of a Lemma, which states that we can trivialize the bundle along curves. (This part is clear) Then it is claimed that this construction carries over to produce a trivialization on the disc if one just uses trivializations along the rays stating in the origin. I don`t understand why these trivialization should fit smoothly together? Near the origin this follows from the construction, but globally each ray is obtained by different patching procedures..

Answer 1

Unitary vector bundles of rank n over any base B are classified by homotopy classes of maps from B to the classifying space of $U(n)$. Thus, existence if trivialization depends only in the homotopy type of the base. In your case the base is homotopy equivalent to a graph, this reduces the problem to the case if unitary bundles over the bouquet if circles. Existence of unitary trivialization in this case is clear and is checked in the book you are reading.

Answer 2

Perhaps you may use the riemann-cevita connection to move the indenpendent vectors at the origin to the whole disc along radical pathes, which gives you a independent vector field, hence trivalization?