 Solution

Write the flat connection as $d+A_{x}dx+A_{y}dy$ in a global unitary frame. Solve the matrix ordinary differential equationSmooth dependence on parameters gives a smooth solution on the entire open square. Since $A_x$ is skew-Hermitian and trace free, differentiating $u^{\ast }u$ and $\det u$ shows that $u(x,y)\in SU(2)$. Under the bundle gauge transformation $u$, the transformed connection hasIts vector-bundle curvature is still zero. The remaining vector-bundle curvature equation is ${\partial }_x{A}_y^{\prime }=0$, so $A_y^{\prime }=B(y)$ is independent of $x$. Now solve $v^{\prime }(y)=-B(y)v(y)$ with $v(0)=I$. Again $v\in SU(2)$; this second bundle gauge transformation, independent of $x$, leaves $A_x^{\prime }=0$ and makes $A_y^{\prime \prime }=0$. Thus the original connection is gauge equivalent to the trivial connection. Both transformations exist on the whole square, not just in a small coordinate neighborhood.