TITLE: Finitely generated singular homology
QUESTION [11 upvotes]: Let G be a finitely generated abelian group and M a compact manifold,  I want to prove that $H_r(M,G)$ is finitely generated for $r\ge 0 $.
First I was thinking if  I could do induction over $r$ because for $r=0$, $H_0(M,G) \cong \bigoplus_{i=1}^nG$ where $n$ denotes the number of connected components of M.But for $r>0$ I can't see a way to relate $H_r(M,G$) and $H_{r+1}(M,G)$.
Then I tried to use the fact that for every compact surface there exists a finite cover $\{A_i\}_{i=1}^m$ such that the finite intersections of $A_i$'s are contractile.Let $\{A_i\}_{i=1}^m$ such cover for $M$.Then,using induction on $m$, we have :
-For $m=0$, $M$ is contractile and $G$ finitely generated then $H_r(M,G) $is finitely generated.
-For $m=r$ it's true by induction hypothesis.
-For $m=r+1$ now putting $A_{r+1}=B$ and $\cup_{i=1}^r A_i=A$, and using Mayer Vietoris we get:
$$ ...\rightarrow H_{r+1}(A\cap B,G)\rightarrow  H_{r+1}(A,G) \oplus H_{r+1}(B,G) \rightarrow H_{r+1}(M,G)\rightarrow H_{r}(A\cap B,G) \rightarrow...$$ 
Here I don't know what to do, maybe I'm missing some algebraic argument It's easy to see that $H_{r+1}(A\cap B,G)$ and $H_{r+1}(A,G) \oplus H_{r+1}(B,G)$ are finitely generated.But how can I argue that $H_{r+1}(M,G)$ is finitely generated?
Any idea would be aprecciated.

REPLY [3 votes]: We recall problem 2.2.33 from Hatcher:

Suppose the space $X$ is the union of open sets $A_1,\ldots,A_m$ such that each intersection $A_{i_1} \cap \ldots \cap A_{i_k}$ is either empty or has trivial reduced homology groups. Then $\tilde{H}_i(X) = 0$ for $i \geq m-1$.

We will use this problem for your problem above. You have already remarked that the base case has been settled. Now suppose it is true that for $m = r-1$ that $H_{r-1}(M,G)$ is finitely generated. Then when $m = r$, we can consider the subspace $A$ of $X$ that is the union of the first $r-1$ open sets. The LES of the pair $(X,A)$ is now
$$\ldots \to H_r(A,G) \to H_r(X,G) \to H_r(X,A;G) \to H_{r-1}(A,G) \to H_{r-1}(X,G) \to  \ldots $$
Now $H_r(A,G) = 0$ and $H_{r-1}(X,G) = 0$ by the problem in Hatcher. By induction $H_{r-1}(A,G)$ is finitely generated and so in effect we are looking at the ses 
$$0 \to H_r(X,G) \to H_r(X,A;G) \to H_{r-1}(A,G) \to 0.$$
I will now leave you to show that $H_r(X,A;G)$ is finitely generated which will prove the problem because $H_r(X,G)$ will be isomorphic to a subgroup of a finitely generated abelian group which is finitely generated.