One handle changes relative homology in one degree only

Work first with the collared gluing model $Y=N{\cup }_{A}H$, $H=D^k\times D^{n-k}$, $A=S^{k-1}\times D^{n-k}$; smoothing transports this model and its core. If $k=0$, then $A=\mathrm{\varnothing }$ and $Y=N\bigsqcup D^n$. The relative chains are exactly those of the disk by the simplex-by-component splitting [F7], so [F3] gives one copy of $F$ in degree zero, represented by its centre, and zero otherwise. Its connector has zero target in degree $-1$.

For $k>0$, $N$ and $A$ are nonempty. The open set $V=N{\cup }_A\{(x,y)\in H:\mathrm{\mid }x\mathrm{\mid }>1\mathrm{/}2\}$ contains all of $N$ and strongly deformation retracts onto it: fix $N$ and send $x$ to $((1-s)+s\mathrm{/}\mathrm{\mid }x\mathrm{\mid })x$ in the handle collar. This agrees with the identity at $\mathrm{\mid }x\mathrm{\mid }=1$ and remains in $V$. Thus $(Y,N)$ is a good pair. The same radial homotopy makes $(H,A)$ a good pair.

By the pushout quotient topology, collapsing $N$ in $Y=N{\cup }_{A}H$ gives $Y\mathrm{/}N\cong H\mathrm{/}A$: in either quotient all of $N$ and $A$ become one point and the rest of the handle is unchanged. This is a quotient-topology identification, not merely a bijection on complements. The inclusion of pairs $(H,A)\to (Y,N)$ induces this homeomorphism on quotients. By the natural quotient isomorphisms [F2], it therefore induces isomorphisms $H_i(H,A;F)\to H_i(Y,N;F)$.

The standard-pair result [F3] computes these groups and identifies the core pair $(D^k,S^{k-1})$ with $(H,A)$ by projection and inclusion. Orient the core disk and represent its relative orientation class by a finite singular fundamental chain (for example map a triangulated disk into the core). It maps to a generator of $H_k(Y,N;F)$. Reversing that orientation reverses the generator; no ambient orientation is required.

For (b), use [F6] to obtain the disjoint handles. To compare pairs, retain a pushed-in lower sublevel below the support of the handle construction: all changes take place in boundary collars and the disjoint critical charts; collar compression retracts both compared lower spaces to this common copy, as in the lower-collar comparison of One critical point handle attachment. The same compression works simultaneously for the finitely many disjoint charts, and [F8] makes the resulting pair homotopies induce homology isomorphisms. Split off the zero-handles, which are disjoint disks, by [F7]. For the remaining positive-index handles the open collars of step 2.1 make both pairs $(Y,N)$ and $(⨆H_j,⨆A_j)$ good; their quotients are homeomorphic by the pushout description of step 3.1. Their relative homology is therefore the same by [F2], while the latter relative chain complex splits by component as in [F7]. Hence $H_i(M^b,M^a;F)\cong {⨁}_j{H}_i(H_j,A_j;F)$, including the zero-handle summands. If there are no positive handles the direct disk splitting alone suffices.

The boundary of the oriented core fundamental chain is its oriented boundary sphere in $N$; for $k=1$ this means the terminal point minus the initial point. It is a relative cycle, and the connector formula [L1] sends its relative class to this boundary class. For $k=0$ the boundary is empty and the connector is zero, as already checked.

Each handle summand is $F$ in its index and zero elsewhere, so the direct sum of step 4.2 has dimension equal to the number of critical points of that index, with their oriented core classes as a basis. If there are no critical points, [F8] identifies the band with a product; compressing that product onto the lower face and fixing the lower sublevel gives a deformation retraction, hence zero relative homology. Inclusion of the lower sublevel is a homotopy equivalence, rather than a diffeomorphism onto the upper space.