Oriented two-plane bundles over the two-sphere by winding number

The map $R:\mathbb{R}\mathrm{/}\mathbb{Z}\to \mathrm{SO}(2)$ defined by $R([t])=\left(\begin{array}{cc}\cos (2\pi t)& -\sin (2\pi t)\\ \sin (2\pi t)& \cos (2\pi t)\end{array}\right)$ is a continuous group isomorphism with continuous inverse obtained from the oriented angle of the first column. Hence [F2] gives ${\pi }_1(\mathrm{SO}(2),I)\cong \mathbb{Z}$, with the class of $t\mapsto R([mt])$ corresponding to $m$.

Apply [F1] with $n=k=2$. Since $\mathrm{SO}(2)$ is path connected, the unbased set $[S^1,\mathrm{SO}(2)]$ is identified with its fundamental group; it is abelian, so changing the path used to the basepoint causes no conjugacy ambiguity. Step 1.1 therefore assigns exactly one integer $m$ to each oriented bundle, and every $m$ is realized by clutching with $t\mapsto R([mt])$. In particular $m=0$ gives the trivial oriented bundle.

Take the reflection $r=\mathrm{diag}(1,-1)$. Direct multiplication gives $rR([t])r^{-1}=R([-t])$. Thus the orientation-reversal action from [F1] sends the loop of winding $m$ to the loop of winding $-m$, as asserted. The calculation uses no choice principle.