Spin-c extensions over a rational ball form a metabolizer coset

Since $B$ has finite CW type by [F1] and has zero positive rational homology, each $H_i(B;\mathbb{Z})$ for $i>0$ is finite. Put $a=\mathrm{\mid }{H}_1(B;\mathbb{Z})\mathrm{\mid }$ and $b=\mathrm{\mid }{H}_2(B;\mathbb{Z})\mathrm{\mid }$. The UCT [F3] gives $\mathrm{\mid }{H}^2(B;\mathbb{Z})\mathrm{\mid }=a$ and $\mathrm{\mid }{H}^3(B;\mathbb{Z})\mathrm{\mid }=b$, because the Hom terms from finite groups vanish and ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}m,\mathbb{Z})\cong \mathbb{Z}\mathrm{/}m$. Poincare-Lefschetz duality [F2] gives $\mathrm{\mid }{H}^2(B,Y;\mathbb{Z})\mathrm{\mid }=b$ and $\mathrm{\mid }{H}^3(B,Y;\mathbb{Z})\mathrm{\mid }=a$.

In the pair sequence, $H^1(Y;\mathbb{Z})=0$ because $Y$ is a rational homology sphere and its first integral cohomology is $\mathrm{Hom}(H_1(Y),\mathbb{Z})$. Thus $H^2(B,Y)\to H^2(B)$ is injective. Also $H^3(Y;\mathbb{Z})\cong \mathbb{Z}$, so the map from the finite group $H^3(B;\mathbb{Z})$ to $H^3(Y;\mathbb{Z})$ is zero; exactness makes $H^3(B,Y)\to H^3(B)$ surjective. It follows that the restriction image $M$ has order $a\mathrm{/}b$, and that $H^2(Y)\mathrm{/}M$ has order $a\mathrm{/}b$. Hence $\mathrm{\mid }{H}^2(Y)\mathrm{\mid }=(a\mathrm{/}b{)}^2$ and $\mathrm{\mid }M{\mathrm{\mid }}^2=\mathrm{\mid }{H}^2(Y)\mathrm{\mid }=\mathrm{\mid }{H}_1(Y)\mathrm{\mid }$, the last equality following from UCT and Poincare duality for the rational homology sphere.

For any $m\in M$, choose $u\in H^2(B;\mathbb{Z})$ restricting to $m$, and let $c$ be an integral cocycle representing $u$. Since $H^2(B;\mathbb{Q})=0$, the rational cocycle $c$ is $\delta q$ for a rational $1$-cochain $q$; reducing $q$ modulo $\mathbb{Z}$ gives a cocycle $v\in H^1(B;\mathbb{Q}\mathrm{/}\mathbb{Z})$ with $\beta (v)=u$. It is unique because $H^1(B;\mathbb{Q})=0$. Restriction of this cochain lift gives $\beta (v{\mathrm{\mid }}_Y)=m$. For $Y$, both $H^1(Y;\mathbb{Q})$ and $H^2(Y;\mathbb{Q})$ vanish, so the same coefficient sequence proves that $\beta :H^1(Y;\mathbb{Q}\mathrm{/}\mathbb{Z})\to H^2(Y;\mathbb{Z})$ is an isomorphism. By UCT, $H^1(Y;\mathbb{Q}\mathrm{/}\mathbb{Z})\cong \mathrm{Hom}(H_1(Y),\mathbb{Q}\mathrm{/}\mathbb{Z})$; by Poincare duality, $H^2(Y;\mathbb{Z})\cong H_1(Y;\mathbb{Z})$. Cup evaluation is therefore the perfect character-evaluation pairing under these identifications. If $X,D$ are torsion $1$-cycles with $mX=\mathrm{\partial }F$, the cochain represented by the rational bounding chain $F\mathrm{/}m$, reduced modulo integral cochains, is ${\beta }^{-1}({\mathrm{P}\mathrm{D}}^{-1}[X])$; its evaluation against ${\mathrm{P}\mathrm{D}}^{-1}[D]$ is $F\cdot D\mathrm{/}m$ modulo $\mathbb{Z}$. Thus this cohomological formula agrees with the chain linking form up to the global orientation sign, which does not change annihilators.

If $x\in M$ is the restriction of $w\in H^2(B;\mathbb{Z})$, naturality of cup product [F5] gives ${\lambda }_Y(m,x)=⟨v{\mathrm{\mid }}_Y⌣w{\mathrm{\mid }}_Y,[Y]⟩=⟨v⌣w,i_{\ast }[Y]⟩=0$, since $i_{\ast }[Y]=0$ in $H_3(B;\mathbb{Z})$ by the homology pair sequence: $[Y]$ is the boundary of the relative fundamental class $[B,Y]$. Therefore $M\subseteq M^{\perp }$. For a nonsingular finite pairing on a group $G$, restriction of characters $G\to \mathrm{Hom}(G,\mathbb{Q}\mathrm{/}\mathbb{Z})$ to $\mathrm{Hom}(M,\mathbb{Q}\mathrm{/}\mathbb{Z})$ is onto. To see this, extend a character from a subgroup $H$ one generator $g$ at a time: if the least $n>0$ with $ng\in H$ is fixed, choose $z\in \mathbb{Q}\mathrm{/}\mathbb{Z}$ with $nz=f(ng)$ (possible because multiplication by $n$ is onto), and set $f^{\mathrm{\prime }}(h+kg)=f(h)+kz$; the relation $ng\in H$ makes $f^{\mathrm{\prime }}$ well defined. Since $G$ is finite this process terminates. Hence $\mathrm{\mid }{M}^{\perp }\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }M\mathrm{\mid }$. Apply this with $G=H^2(Y;\mathbb{Z})$ and use step 2.1 to obtain $\mathrm{\mid }{M}^{\perp }\mathrm{\mid }=\mathrm{\mid }M\mathrm{\mid }$. Thus $M=M^{\perp }$.

By [F7], choose a Spin${}^c$ structure on $B$. Then [F6] makes the Spin${}^c$ structures on $B$ an affine $H^2(B;\mathbb{Z})$-space. Restriction is equivariant, so the image of this nonempty affine space is one affine coset of the image subgroup $M$. This proves the extension-set assertion. If $M=0$, the same sequence gives $\mathrm{\mid }{H}^2(Y)\mathrm{\mid }=1$, so the trivial subgroup is its own annihilator; the empty-boundary case is excluded by the connected-boundary hypothesis.