Degree of the power map on the circle

The formula is well defined: if $[s]=[t]$, then $s-t\in \mathbb{Z}$ by [F1], so $m(s-t)\in \mathbb{Z}$ and $[ms]=[mt]$. On every quotient arc shorter than one, source and target lift coordinates express $P_m$ as $u\mapsto mu+k$ for an integer constant $k$; hence it is smooth with derivative $m$. For any compact $K\subseteq S^1$, Hausdorffness makes $K$ closed, continuity makes $P_m^{-1}(K)$ closed, and [F2] makes this closed subset of the compact circle compact. Thus $P_m$ is proper.

Suppose $m>0$. The fibre over $[0]$ is exactly $P_m^{-1}([0])=\{[k\mathrm{/}m]:0\le k<m\}\mathrm{.}$ Indeed, after taking the unique representative $t\in [0,1)$, the condition $mt\in \mathbb{Z}$ says $mt=k$ for exactly one of these integers. The derivative in positive lift coordinates is $m>0$, so every one of these $m$ points has local sign $+1$. The value is regular, and [F3] gives $\mathrm{deg}(P_m)={\sum }_{k=0}^{m-1}1=m$.

Suppose $m<0$ and put $r=-m>0$. The same representative calculation gives the $r$ distinct preimages $[k\mathrm{/}r]$, $0\le k<r$, of $[0]$. In positive lift coordinates the derivative is $m<0$, so every local sign is $-1$. Hence [F3] gives $\mathrm{deg}(P_m)={\sum }_{k=0}^{r-1}(-1)=-r=m$.

If $m=0$, $P_0$ is the constant map with value $[0]$. The point $[1\mathrm{/}2]$ has empty fibre and is therefore a regular value; [F3] gives degree equal to the empty sum, namely zero. Thus all integers are covered. In particular $P_1$ is the identity and $P_{-1}$ reverses orientation. All fibres used are explicitly finite, no root is selected from a family, and no choice axiom is used.