The de Rham map on the angular form

Differentiation using [F2] gives $x^{\mathrm{\prime }}(t)=-2\pi \sin (2\pi t)$ and $y^{\mathrm{\prime }}(t)=2\pi \cos (2\pi t)$. Hence ${\gamma }^{\ast }(-ydx+xdy)=(-y(t)x^{\mathrm{\prime }}(t)+x(t)y^{\mathrm{\prime }}(t))dt=2\pi ({\sin }^2(2\pi t)+{\cos }^2(2\pi t))dt=2\pi dt,$ and therefore ${\gamma }^{\ast }\alpha =dt$.

Therefore [F1] gives $I_{S^1}^1(\alpha )(\gamma )={\int }_0^{1}dt=1.$ More generally, composing with $P_m(z)=z^m$ gives the lift $t\mapsto mt$ and $P_m^{\ast }\alpha =m\alpha$, so the same calculation yields $I^1(\alpha )(P_m\gamma )=m$, in agreement with $\mathrm{deg}(P_m)=m$ from [F3], including negative $m$ and $m=0$.

Reversing the loop changes the value to $-1$; the constant loop has value zero. Both endpoints of $[0,1]$ map to the same circle point, so the path is a cycle and there is no seam contribution. The calculation uses explicit maps and no choice principle.