Intuition for killing homology group

Question

I am working through Hatcher's Algebraic Topology section 4.2, and landed on exercise 24, which says that an $M(G,1)$ with fundamental group $G$ exists if and only if $H_2(K(G,1); \mathbb{Z})=0$. The left to right direction is easy enough, however the other direction is giving me problems.

In particular I am having trouble with intuition on what it takes to be able to "kill" homology groups, whereas killing homotopy groups is relatively easy. More specifically, I mean killing higher homology groups, without affecting "lower dimensional" information such as homotopy groups and homology groups in lower degree than the degree of the group we are trying to kill.

In fact for homology I think it isn't always possible to kill homology groups by adding finitely many cells. For example I don't think we can kill the second homology of the torus, without modifiying $\pi_1$

However I need some intuition on when this can be done, as the hint for the exercise leads me to believe that is what is used. Hence my questions:

1. Does anybody want to share heuristics on when we can kill of the homology groups of a space
2. Does anybody know how to show in my particular case that I can kill of the homology in degree $\geq 2$ the two skeleton of a $K(G,1)$ under the assumption $H_2(K(G,1), \mathbb{Z})=0$ (following Hatcher's hint for 4.2.24).

Thank you in advance

Answer 1

In general if $Y$ is obtained from $X$ by attaching $(n+1)$-cells, then the characteristic maps induce an isomorphism $$\bigoplus_{\alpha\in A}H_{k}(D^{n+1}_\alpha, S^{n}_\alpha)\cong H_{k}(Y,X)$$ for all $k$. It follows that $H_k(X)\cong H_k(Y)$ if $k\neq n, n+1$ and that there is an exact sequence $$0\to H_{n+1}(X)\to H_{n+1}(Y)\to H_{n+1}(Y,X)\to H_{n}(X)\to H_n(Y)\to 0.$$ Furthermore, by comparing long exact sequences we see that $H_n(Y)$ is the cokernel of the map $\bigoplus_{\alpha\in A} H_n(S^n_\alpha)\to H_n(X)$ induced by the attaching maps. Thus, in order to kill $H_n(X)$ we need to find a map $\coprod_{\alpha\in A} S^n_\alpha\to X$ that induces a surjection on $H_n$, which is equivalent to the Hurewicz map $h\colon \pi_n(X)\to H_n(X)$ being surjective. We also see that given the surjectivity of the Hurewicz map we can additionally arrange $H_{n+1}(Y)=0$ iff $H_{n+1}(X) = 0$ and $H_n(X)$ is free abelian, which is for example the case if $X$ is an $n$-dimensional CW-complex.

If we apply this to $K^2$, the $2$-skeleton of $K(G,1)$, then we can attach 3-cells to $K^2$ to obtain $M(G,1)$ iff the Hurewicz map $h\colon \pi_2(K^2)\to H_2(K^2)$ is surjective. Hatcher 4.2.23 says (in a special case) that if $X$ is path connected, there is an isomorphism $H_2(X)/h(\pi_2(X))\cong H_2(K(\pi_1(X),1)$. Now $\pi_1(K^2)=\pi_1(K(G,1)) = G$, hence the Hurewicz map $h\colon \pi_2(K^2)\to H_2(K^2)$ is surjective iff $H_2(K(G,1))=0$.