Talk:Betti numbers of 3-manifolds (Ex)

Now let $M$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}M$ be a closed, orientable $3$$3$-manifold with $\pi_1(M)=G$$\pi_1(M)=G$. View $M$$M$ as a finite CW complex, with cells in dimensions-$0,1,2,3$$0,1,2,3$, as usual. Attach cells of dimension $\ge 3$$\ge 3$ to $M$$M$ to obtain $N$$N$, a $K(G,1)$$K(G,1)$ space. Then $b_1(G)$$b_1(G)$ is the rank of $H_1(N;\mathbb{Z})$$H_1(N;\mathbb{Z})$ and $b_2(G)$$b_2(G)$ is the rank of $H_2(N;\mathbb{Z})$$H_2(N;\mathbb{Z})$. By construction of $N$$N$, $H_1(N;\mathbb{Z})\cong H_1(M;\mathbb{Z})$$H_1(N;\mathbb{Z})\cong H_1(M;\mathbb{Z})$ and the rank of $H_2(N;\mathbb{Z})$$H_2(N;\mathbb{Z})$ is bounded above by the rank of $H_2(M;\mathbb{Z})$$H_2(M;\mathbb{Z})$. Moreover, $M$$M$ is a $3$$3$-manifold, so Poincaré duality says the ranks of $H_1(M;\mathbb{Z})$$H_1(M;\mathbb{Z})$ and $H_2(M;\mathbb{Z})$$H_2(M;\mathbb{Z})$ agree.

Thus, $b_1(G)=\text{rk} H_1(N;\mathbb{Z})=\text{rk}H_1(M;\mathbb{Z})=\text{rk}H_2(M;\mathbb{Z})\ge \text{rk} H_2(N;\mathbb{Z})=b_2(G)$$b_1(G)=\text{rk} H_1(N;\mathbb{Z})=\text{rk}H_1(M;\mathbb{Z})=\text{rk}H_2(M;\mathbb{Z})\ge \text{rk} H_2(N;\mathbb{Z})=b_2(G)$, so $b_1(G)\ge b_2(G)$$b_1(G)\ge b_2(G)$.