Morse Inequalities and the Handle Chain Complex

If $N<0$, the vanishing hypotheses force every $A_k,C_k$ to be zero, and exactness forces every $B_k$ to be zero; all formulas then hold with $Q=0$. Henceforth assume $N\ge 0$. Exactness gives $\mathrm{im}{\alpha }_k=\ker {\beta }_k$, $\mathrm{im}{\beta }_k=\ker {\gamma }_k$, and $\mathrm{im}{\gamma }_k=\ker {\alpha }_{k-1}$. Put $r_k:={\dim }_F\mathrm{im}{\alpha }_k$ and $s_k:={\dim }_F\ker {\gamma }_k$; these are finite by [L1] and [L2], since $A_k,C_k$ are finite-dimensional. The maps retain the names ${\alpha }_k,{\beta }_k,{\gamma }_k$.

Rank-nullity for ${\alpha }_k:A_k\to B_k$ and ${\gamma }_k:C_k\to A_{k-1}$ gives ${\dim }_F{A}_k=a_k+r_k$ and ${\dim }_F{C}_k=s_k+a_{k-1}$.

Choose a finite basis $y_1,\dots ,y_m$ of $\mathrm{im}{\beta }_k=\ker {\gamma }_k$ and lifts $x_i\in B_k$ with ${\beta }_k(x_i)=y_i$. Then $B_k=\ker {\beta }_k+\mathrm{span}(x_1,\dots ,x_m)$: subtract the corresponding linear combination of the lifts from any element. The kernel has dimension $r_k$ by exactness and step 2.1, so this sum is finite-dimensional by [L3]. Rank-nullity now applies to ${\beta }_k$ and gives ${\dim }_F{B}_k=r_k+s_k$. Only finitely many lifts are chosen.

The three dimension identities give ${\dim }_F{A}_k+{\dim }_F{C}_k-{\dim }_F{B}_k=a_k+a_{k-1}\ge 0$. In particular the weak inequality holds. Outside $0\le k\le N$, exactness and $A_k=C_k=0$ force $B_k=0$. Also $a_N=0$, since exactness identifies $\ker {\alpha }_N$ with the image of the zero space $C_{N+1}$.

Set $Q(t)={\sum }_{k=0}^N{a}_k{t}^k$. With $a_k=0$ outside this range, the coefficient of $(1+t)Q$ in degree $k$ is $a_k+a_{k-1}$; at degree $N+1$ it vanishes because $a_N=0$. Thus step 4.1 and coefficient comparison give $P_A+P_C=P_B+(1+t)Q$. Its coefficients are nonnegative integers.

Telescoping gives ${\sum }_{i=0}^k(-1{)}^{k-i}(a_i+a_{i-1})=a_k$, since $a_{-1}=0$. This proves the partial-sum formula and identifies $q_k=a_k$. For negative $k$ read the sum as empty and set $q_k=0$.

Finally, uniqueness of $Q$: if $(1+t)Q=0$ with $Q={\sum }_k{q}_k{t}^k\in \mathbb{Z}[t]$, then by [F2] the coefficients satisfy $q_0=0$ and $q_k=-q_{k-1}$ for every $k\ge 1$, so all $q_k=0$ and $Q=0$; hence two polynomials with $(1+t)Q=(1+t)Q^{\mathrm{\prime }}$ agree.

In each degree the sequence $0\to C_n(A,B;G)\to C_n(X,B;G)\to C_n(X,A;G)\to 0$ is the sequence of quotients $C_n(A)\mathrm{/}{C}_n(B)\to C_n(X)\mathrm{/}{C}_n(B)\to C_n(X)\mathrm{/}{C}_n(A)$ induced by $C_n(B)\subseteq C_n(A)\subseteq C_n(X)$ by [F1]; it is exact because the first map is injective (the inclusion $C_n(A)\to C_n(X)$ descends injectively after dividing by the common subgroup $C_n(B)$), the second is surjective, and its kernel is exactly $C_n(A)\mathrm{/}{C}_n(B)$.

By [F2] all three boundary maps descend to the quotients, so the degreewise maps of step 1.1 commute with the boundaries and form chain maps; hence they constitute a short exact sequence of complexes in the sense of [F3].

Applying [F4] to the short exact sequence of step 2.1 gives the long exact sequence of the statement; the homology groups are $H_n(X,A;G)$, $H_n(X,B;G)$, $H_n(A,B;G)$ by [L1], and the first two maps are induced by inclusions because the chain maps of step 1.1 are.

For the factorization, let $c$ be a relative cycle for $(X,A)$ with $\mathrm{\partial }c\in C_{n-1}(A;G)$, representing a class in $H_n(X,A;G)$. The connecting map of the triple sequence sends its class to the class of $\mathrm{\partial }c$ in $H_{n-1}(A,B;G)$: this is the same cycle formula as in [L2] read in the middle complex $C_{\bullet }(X,B;G)$, where the role of the subspace is played by $A$ modulo $B$. The pair connector of $(X,A)$ sends the same class to $[\mathrm{\partial }c]\in H_{n-1}(A;G)$ by [L2], and the quotient map $H_{n-1}(A;G)\to H_{n-1}(A,B;G)$ is induced by the quotient chain map; composing gives the triple connector. Hence $\delta$ factors as the pair connector followed by the quotient map.

Define $H((x,t),s):=(x,(1-s)t)$ for $(x,t)\in C$ and $s\in [0,1]$. This is continuous, being the restriction of the smooth map $M\times [0,1{]}^2\to C$, and it satisfies $H((x,t),0)=(x,t)$, $H((x,t),1)=(x,0)=r(x,t)$ and $H((x,0),s)=(x,0)$ for every $(x,t)\in C$ and every $s$. Thus $r$ is a retraction of $C$ onto $M_0$ and $H$ is a homotopy from ${\mathrm{id}}_C$ to $i\circ r$ fixing $M_0$ pointwise, so by [F1] the projection $r$ is a strong deformation retraction onto $M_0$.

Since $r\circ i={\mathrm{id}}_{M_0}$ and $H$ displays $i\circ r\simeq {\mathrm{id}}_C$, the maps $r$ and the inclusion $i:M_0\to C$ are homotopy inverses, so $r$ is a homotopy equivalence by [F2]. By [L1], for every $j$ the induced map $H_j(r;G):H_j(C;G)\to H_j(M_0;G)$ is an isomorphism.

The map $r$ is a map of pairs $r:(C,M_0)\to (M_0,M_0)$, so by [F4] it induces a morphism from the exact sequence of [F3] for $(C,M_0)$ to that for $(M_0,M_0)$: $\cdots \to H_j(M_0;G)\to H_j(C;G)\to H_j(C,M_0;G)\to H_{j-1}(M_0;G)\to \cdots$ compared with $\cdots \to H_j(M_0;G)\to H_j(M_0;G)\to H_j(M_0,M_0;G)\to H_{j-1}(M_0;G)\to \cdots \mathrm{.}$ In this morphism the comparison map on $H_j(C;G)$ is the isomorphism of step 2.1, and every comparison map on a copy of $H_j(M_0;G)$ is the identity, because on the subspace $M_0$ the map $r$ is the identity.

Fix $i$ and apply [F5] to the five-term window of this morphism centred on the comparison map $H_i(C,M_0;G)\to H_i(M_0,M_0;G)$: by step 3.1 the four surrounding comparison maps are isomorphisms (each is either an identity or $H_j(r;G)$), so the relative comparison map is an isomorphism.

The relative complex of the pair $(M_0,M_0)$ is the zero complex by [L2], so $H_i(M_0,M_0;G)=0$; hence $H_i(C,M_0;G)=0$ for every $i$, and the map of pairs $r$ induces these isomorphisms on relative homology.

For $(x,y)\in H$ and $s\in [0,1]$ put $H_s(x,y):=(x,(1-s)y)$. This map is continuous, $H_0={\mathrm{id}}_H$, and $H_1(x,y)=(x,0)\in K$; moreover $H_s(x,0)=(x,0)$ for all $s$, so $K$ is fixed pointwise. Hence $H_s$ is a strong deformation retraction of $H$ onto $K$ in the sense of [F2]. Its restriction to the attaching region satisfies $H_s(S^{k-1}\times D^{n-k})=S^{k-1}\times (1-s)D^{n-k}\subseteq S^{k-1}\times D^{n-k}$, and at $s=1$ the image is $S^{k-1}\times \{0\}$, the attaching sphere.

For $(x,y)\in H$ and $s\in [0,1]$ put $G_s(x,y):=((1-s)x,y)$. This map is continuous, $G_0={\mathrm{id}}_H$, and $G_1(x,y)=(0,y)\in C$; moreover $G_s(0,y)=(0,y)$ for all $s$, so $C$ is fixed pointwise and $G_s$ is a strong deformation retraction of $H$ onto the cocore $C$ by [F2]. Its restriction to the outgoing region satisfies $G_s(D^k\times S^{n-k-1})=(1-s)D^k\times S^{n-k-1}\subseteq D^k\times S^{n-k-1}=R$, and at $s=1$ the image is $\{0\}\times S^{n-k-1}=B$, the belt sphere.

The complement of the belt sphere in the outgoing region is $R\setminus B=\{(x,y)\in D^k\times S^{n-k-1}:x\mathrm{\ne }0\}$. Define $K_s(x,y):=((1-s)x+sx\mathrm{/}\mathrm{\mid }x\mathrm{\mid },y)$ on $(R\setminus B)\times [0,1]$; this is well defined and continuous because $x\mathrm{\ne }0$ on the domain and $y\in S^{n-k-1}$ is unchanged, and by [L1] the norm $\mathrm{\mid }x\mathrm{\mid }$ is the Euclidean norm. We have $K_0=\mathrm{id}$, $K_1(x,y)=(x\mathrm{/}\mathrm{\mid }x\mathrm{\mid },y)\in S^{k-1}\times S^{n-k-1}$, and $K_s$ fixes every point of $S^{k-1}\times S^{n-k-1}$ pointwise, because $x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }=x$ there. Hence $K_s$ is a strong deformation retraction of $R\setminus B$ onto $S^{k-1}\times S^{n-k-1}$ in the sense of [F2].

The three formulas are explicit and continuous for all $0\le k\le n$, including the endpoint cases $k=0$ and $k=n$: at $k=0$ the attaching region $S^{-1}\times D^n$ is empty, the core is a point, and $R\setminus B=\mathrm{\varnothing }$ because $B=R$; at $k=n$ the outgoing region and belt sphere are empty while the cocore is a point. Thus (a), (b) and (c) hold as stated, and the standard model is the one glued by [F3].

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.