The second Betti number of a group

Question

First I can't find the definition of second Betti number of a group. (Can you tell me any reference about this definition?)

Also I don't know why $b_2(M)\ge b_2(G)$, where $M$ is a manifold with fundamental group $G$.

Answer 1

$b_2(G)$ here means the rank of $H_2(G;\Bbb Z)$, the second group homology; this is the same as the rank of the second group cohomology. A perhaps comfortable definition is that if $BG$ is a space with specified isomorphism $\pi_1 BG \cong G$ and whose universal cover is contractible, we set $$H_n(G;\Bbb Z) := H_n(BG;\Bbb Z).$$ This space $BG$ is unique up to unique homotopy equivalence; in fact if $X$ is a space equipped with a homomorphism $\pi_1 X \to G$, there is a unique map $X \to BG$ up to homotopy inducing this homomorphism. Then because the first homology group is the abelianization of the fundamental group, we have $$H_1(G;\Bbb Z) = G^{\text{ab}}.$$

Some discussion of Eilenberg-MacLane spaces (written $K(G,1)$, these are the same as the "$BG$"s above) appears in the appendix to chapter 1 of Hatcher's book on algebraic topology.

Let $X$ be a space with an isomorphism $\pi_1 X \cong G$. Then one may form a space $BG$ by attaching cells to $X$ of degree $3$ or greater: you already have the desired fundamental group, and you attach cells to kill off higher homotopy groups. In particular, by an inductive Mayer-Vietoris argument (you cannot be adding to second homology!) you find that the natural map $X \to BG$ induced by the isomorphism $\pi_1(X) \cong G$, $$H_2(X;\Bbb Z) \to H_2(BG;\Bbb Z),$$ is surjective. In particular, $b_2(X) \geq b_2(G)$.

What's more, this argument actually identifies $H_2(BG) = H_2(X)/\pi_2(X)$, quotienting by the image under the Hurewicz homomorphism.

This argument is that suggested by exercise 4.23 in Hatcher's book.