A displayed two-sheeted orientation-preserving covering has degree two

Let $y=[a]\in S^1$. Its fibre is exactly $F^{-1}(y)=\{\left[\frac{a}{2}\right],\left[\frac{a+1}{2}\right]\}\mathrm{.}$ Both displayed classes map to $[a]$, and they are distinct because their difference is $1\mathrm{/}2\mathrm{\notin }\mathbb{Z}$. Conversely, $F([t])=[a]$ means $2t-a=k\in \mathbb{Z}$; according as $k$ is even or odd, $[t]$ is the first or second displayed class. This also shows that changing $a$ by an integer merely permutes the two classes.

Choose quotient arcs about either preimage and $y$, and lift them to increasing real coordinates. On each source arc $F$ has the form $u\mapsto 2u-k$ for an integer $k$, so its derivative is $2>0$. Hence each restriction is an orientation-preserving diffeomorphism onto a sufficiently short target arc; these two restrictions are the two inverse sheets over that arc. Thus every $y$ is regular and both local signs are $+1$.

Since $F$ is proper by [F3], [F2] applies at the arbitrary value $y$ and gives $\mathrm{deg}(F)=(+1)+(+1)=2.$ The fibre is never empty or a singleton, the derivative never degenerates, and quotient seams introduce no boundary endpoints because the calculation uses local lifts. Both inverse branches were displayed explicitly, so no choice principle is used.