Bordant cycles have equal intersection numbers

First assume both transversality conditions. By [F1], $S=F^{-1}(B)$ is a neat one-manifold with $\mathrm{\partial }S=(F{\mathrm{\mid }}_{\mathrm{\partial }W}{)}^{-1}(B)$: its dimension is $(a+1)-(n-b)=1$. It is compact because $B$ is closed and $W$ is compact. Its boundary consists of the finite transverse intersections with $A_1$ and $A_0$.

For arbitrary $F$, choose a collar $c(x,t)$ by [F4]. Choose a smooth function $h:[0,1)\to [0,1)$ with $h=0$ near zero and $h(t)=t$ outside a smaller collar. Replace $F(c(x,t))$ there by $F(c(x,h(t)))$. Interpolation between $t$ and $h(t)$ gives a homotopy fixed on the boundary, and the new map $F^{\mathrm{\prime }}$ is constant in the normal coordinate near the boundary. Consequently using $F^{\mathrm{\prime }}$ on both halves extends smoothly to a map $H:DW\to M$. If $\mathrm{\partial }W$ is empty, use the disjoint double without this modification.

Orient $Q=TM{\mathrm{\mid }}_B\mathrm{/}TB$ by normal-first order, so a positive quotient determinant followed by a positive determinant of $TB$ is positive in $TM$. Orient $S$ by $\det TW=\det TS\otimes \det F^{\ast }Q$. At an endpoint choose an outward vector $r\in TS$; neatness makes it outward also in $W$, and it is transverse to $\mathrm{\partial }W$, not tangent to it. A positive boundary determinant $u$ then makes $(r,u)$ positive in $TW$. The quotient image of $u$ has sign equal to the local intersection sign of the boundary map with $B$, by the normal-first definition of $Q$. Thus the boundary point sign of $S$ is that local sign. On $A_1$ this is ${\epsilon }_{A_1}$, and on the oppositely oriented $A_0$ it is $-{\epsilon }_{A_0}$. This determinant-ray argument includes zero-dimensional boundary factors.

By [F2] the signed boundary sum is zero; step 2.1 identifies it with $I(A_1,B)-I(A_0,B)$. Reducing the same finite sum modulo two gives the parity equality.

A submersive parameter family for $H$ from [F4] remains submersive in its parameter directions after restriction to the seam $\mathrm{\partial }W$. Parametric transversality on $DW$ and on $\mathrm{\partial }W$ therefore excludes only two null sets of parameters. Their union is null; choose a good parameter arbitrarily near zero (in parameter dimension zero the bad sets are empty). Its restriction to $W$ and to $\mathrm{\partial }W$ is transverse, and its boundary maps are homotopic to the original inclusions along the parameter segment. Apply step 3.1 to the perturbed maps and then [F3] to recover the original numbers. If the original boundary map was transverse, $H$ is transverse on a seam neighbourhood because it is constant in the collar direction; relative transversality in [F4] instead fixes that neighbourhood. Countable Choice is inherited from [F2] and [F4]; the finite sign calculations add none.