Connector compatibility for a two-arc cover of the circle

Choose $a\in W_0$ and $b\in W_1$. Let $u$ be the increasing arc in $U$ from $a$ to $b$, and $v$ the increasing arc in $V$ from $b$ through the quotient seam to $a$; $z=u+v$ is the positively oriented circle cycle. For the de Rham lift set $\alpha =-{\rho }_{V}c$ on $U$ and $\beta ={\rho }_{U}c$ on $V$. Their difference on the overlap is $\beta -\alpha =({\rho }_U+{\rho }_V)c=c$, and their derivatives glue to the connecting one-form $\zeta$.

Since $\zeta =d\alpha$ on $U$ and $\zeta =d\beta$ on $V$, endpoint evaluation gives ${\int }_z\zeta =\alpha (b)-\alpha (a)+\beta (a)-\beta (b)\mathrm{.}$ At $b\in W_1$, $c(b)=0$, hence $\alpha (b)=\beta (b)=0$; at $a\in W_0$, $\beta (a)-\alpha (a)=c(a)=1$. Therefore ${\int }_z\zeta =1$.

On the singular side use the lift $e_0=(-E_{U}c,0)$ from [F1]. Its differential glues to the connector cocycle $z_0$. On $u:a\to b$, $z_0(u)=\delta (-E_{U}c)(u)=(-c(b))-(-c(a))=1,$ while $z_0(v)=0$ on the $V$ lift. Thus $z_0(z)=1$, exactly the value in step 2.1, and [F1] identifies the two connector classes.

Replacing $c$ by $-c$ or reversing $z$ reverses both answers. The zero cocycle gives zero on both sides; if an overlap component were absent, this particular nonzero witness would not exist. Both endpoints of each arc occur in the displayed coboundary differences, and degenerate simplices contribute zero. Apart from [F2]'s partition existence, every lift, path, sign and evaluation is finite and explicit.