A map with two preimages but degree zero

Under [F1] the displayed map is $G([t])=\left[\frac{\sin (2\pi t)}{2\pi }\right]\mathrm{.}$ It is well defined because replacing $t$ by an integer translate does not change the sine by [F2]. In local increasing angular coordinates its lifts differ only by integer constants and have derivative $\cos (2\pi t)$ by [F3]; repeated differentiation cycles through sine and cosine, so $G$ is smooth. If $K$ is compact in the target, [F4] makes $K$ closed, hence $G^{-1}(K)$ closed in the compact source and therefore compact. Thus $G$, equivalently $F$, is proper.

The fibre of $[0]$, corresponding to $1\in S^1$, satisfies $\sin (2\pi t)\in 2\pi \mathbb{Z}$. Since $\mathrm{\mid }\sin (2\pi t)\mathrm{\mid }\le 1<2\pi$ by [F2], this is equivalent to $\sin (2\pi t)=0$. The zero-set formula in [F2] gives exactly $[t]=[0]$ or $[t]=[1\mathrm{/}2]$. By [F3] their derivatives are respectively $+1$ and $-1$, so $[0]$ is regular and [F5] gives $\mathrm{deg}(F)=(+1)+(-1)=0,$ although this fibre has two points.

This witnesses the failed unsigned-count conclusion. For comparison, the target value $[1\mathrm{/}2]$ has empty fibre because every lifted value of $G$ has absolute value at most $1\mathrm{/}(2\pi )<1\mathrm{/}2$, and the empty regular-fibre sum again gives zero. The extreme target $[1\mathrm{/}(2\pi )]$ has the singleton preimage $[1\mathrm{/}4]$, but its derivative is $\cos (\pi \mathrm{/}2)=0$, so it is critical rather than a counterexample to the regular-value formula. Quotient seams are handled by local lifts, and every fibre point used above is explicitly listed; no choice principle is used.