Handle boundary coefficients are attaching-belt intersection numbers

In the middle level $N_k={\mathrm{\partial }}_{+}{W}_k$ the attaching sphere $A_i$ of the $(k+1)$-handle has dimension $k$ and the belt sphere $B_j$ of the $k$-handle has dimension $n-k-1$ (K handle core cocore attaching region and belt sphere); the two dimensions sum to $\dim N_k=n-1$, and by hypothesis the spheres are transverse, so $A_i\cap B_j$ is finite (compactness of $N_k$). The ambient orientation of $N_k$ is the boundary orientation induced by that of $W_k$ [F3, F7].

By [F1] the handle boundary is ${\mathrm{\partial }}_{k+1}=q_k\circ {\delta }_{k+1}$, where ${\delta }_{k+1}:C_{k+1}\to H_k(W_k;F)$ is the connecting map of the pair $(W_{k+1},W_k)$ and $q_k:H_k(W_k;F)\to C_k$ is the relative quotient map; by [F2] this composite is the triple boundary.

For $k\ge 1$, define $p_j:W_k\to D^k\mathrm{/}{S}^{k-1}\cong S^k$ by $p_j(x,y)=[x]$ on the $j$th $k$-handle, and send $W_{k-1}$ and every other handle to the basepoint. On the attaching seam $\mathrm{\mid }x\mathrm{\mid }=1$ the formula is the basepoint, so it glues continuously. Choose the sphere orientation so that the core quotient has degree $+1$. Then $(p_j{)}_{\ast }:H_k(W_k,W_{k-1};F)\to H_k(S^k,\ast ;F)$ sends the $j$th core generator to $1$ and the other core generators to zero, by [F1] and the quotient identification [F4]. It therefore extracts the $j$th coefficient.

By [F2] the relative image of the attaching sphere $A_i$ is ${\mathrm{\partial }}_{k+1}$ of the upper core class. Consequently its $j$th coefficient is the degree of $p_j{\mathrm{\mid }}_{A_i}:S^k\to S^k$, interpreted in $F$. Indeed on positive-degree homology the natural map $H_k(S^k;F)\to H_k(S^k,\ast ;F)$ is an isomorphism, also for $k=1$ by the pair sequence and the isomorphism on $H_0$.

The fibre of the interior value $[0]$ of this map is exactly $A_i\cap B_j$. On the outgoing region of the $j$th handle, $p_j(x,y)=[x]$ and $B_j=\{0\}\times S^{n-k-1}$; all other regions map to the basepoint. Near a fibre point the map is the core-coordinate projection. Transversality makes its restriction to $A_i$ a local diffeomorphism. The specified belt orientation makes its local degree equal to the sign comparing $T{A}_i\oplus T{B}_j$ with $T{N}_k$, which is the local intersection sign of [F7]. Thus the projection is onto the core coordinates, rather than the cocore.

The finite-fibre formula [F5] now gives $\mathrm{deg}(p_j{\mathrm{\mid }}_{A_i})={\sum }_{p\in A_i\cap B_j}{\mathrm{sign}}_p(A_i,B_j)=I(A_i,B_j)$. This also includes an empty fibre. Therefore the integer coefficient, and its image in any field, is the claimed intersection number.

Without orientations the same local-excision computation uses coefficient-one generators over $\mathbb{Z}\mathrm{/}2$: every local diffeomorphism contributes $1$, and the global class restricts to the diagonal of these generators as in [F5]. Thus the coefficient is the parity of $A_i\cap B_j$. For oriented handles reducing the integer calculation modulo two agrees with [F7]. The positive-index proof includes $k=n-1$; the belt then has dimension zero, and the same local projection and orientation comparison apply.

For $k=0$ use the chain connector directly: the boundary of the oriented upper interval is its terminal point minus its initial point. The $j$th coefficient in $H_0(W_0;F)$ is therefore $+1$ if its terminal point is on the $j$th disk boundary and $-1$ if its initial point is there, adding both if necessary. Orient that boundary circle or sphere as the belt of the positive zero-dimensional core; these are precisely the local intersection signs. Modulo two count the endpoints. This proves the endpoint formula independently of a sphere-degree assertion in dimension zero.