Oriented boundary of an intersection trace has opposite end signs

By [F2], $W$ is a compact neat $1$-manifold with boundary the finite disjoint union of the slice preimages. Orient it by [F3]: for a positive quotient determinant $q$ and any lift $\mathrm{\ell }$ to the ambient tangent, a nonzero kernel vector $\tau$ is positive exactly when $\tau \wedge \mathrm{\ell }$ is in the ambient determinant ray. The normal-first quotient ray is the unique ray whose product with the tangent ray of $Z$ is the ambient ray of $M$. Thus this orientation is defined even when the quotient has dimension zero.

At an end point $p=(t,x)$ let $Q=T_{F(p)}M\mathrm{/}{T}_{F(p)}Z$ with the normal-first orientation. The map $d{F}_t:T_{x}X\to Q$ is an isomorphism, so $\tau ={\mathrm{\partial }}_t+v\in T_{p}W$ exists uniquely with $dF(\tau )=0$ in $Q$. Its $t$-component is $1$. The shear replacing ${\mathrm{\partial }}_t$ by $\tau$ preserves the product determinant, and the kernel-first orientation therefore assigns $\tau$ the sign ${\epsilon }_{F_t}(p)$: in determinant elements $\tau \otimes d{F}_t(u)$ has the product ray exactly when the kernel ray has that sign. This argument includes $x=0$, where $u$ and the quotient orientation are signed scalars, rather than empty positive bases. The vector $\tau$ points outward at $t=1$ and inward at $t=0$. Thus the outward-normal-first point sign is $-{\epsilon }_{F_0}(p)$ at the initial end and $+{\epsilon }_{F_1}(p)$ at the terminal end.

Summing these point signs over the finite boundary gives ${\sum }_{p\in \mathrm{\partial }W}{\epsilon }_W(p)=I(F_1,Z)-I(F_0,Z)$ by [F4]. The left side vanishes by [F5]. The determinant comparison itself uses no choice; the boundary-count supplier uses the stated ${\mathrm{A}\mathrm{C}}_{\omega }$.