Finitely generated singular homology 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.
 A: 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.
A: Before we start, we need a useful lemma from algebra:
Lemma: Suppose $G_1\overset{f}{\to} G_2\overset{g}{\to}G_3$ is an exact sequence of abelian groups, and $G_1,G_3$ are finitely generated. Then $G_2$ is finitely generated.
Proof: Note that we can break the exact sequence up into short exact sequences as
$$\begin{align}
0\to \ker f\to &G_1 \to \mathrm{im}\:f \to 0\\\\
0 \to \mathrm{im}\: f\to &G_2 \to \mathrm{im}\: g\to 0\\\\
0\to \mathrm{im}\: g\to &G_3 \to \mathrm{coker}\: g\to 0\\\\
\end{align}$$
Recall that every subgroup of a finitely generated abelian group is finitely generated, so $\mathrm{im}\: g$ is finitely generated, say by $a_1,\ldots,a_n$. Additionally, every homomorphic image of a finitely generated group is finitely generated, so $\mathrm{im}\:f$ is finitely generated, say by $b_1,\ldots,b_k$. A simple calculation shows that $g^{-1}(a_1),\ldots,g^{-1}(a_n),f(b_1),\ldots,f(b_k)$ generate $G_2$ for any choice of inverses.
The following lemma is elementary, but I prove it for completeness:
Lemma: For any compact manifold $M$, there is a collection of closed contractible sets $\{A_i\}_{i=1}^m$ such that the interiors cover $M$.
Proof: Let $(U_x,\phi_x)$ be a family of charts for $M$ such that for all $x\in M$, $x\in \phi_x(U_x)$. Let $B_x\subseteq U_x$ be an open ball and $B_x'\subset B_x$ the open ball with the same center but half the radius. Then $\phi_x(B_x')$ cover $M$, so by compactness we have $x_1,\ldots,x_m\in M$ such that $M=\bigcup\limits_{i=1}^m\phi_{x_i}(B_{x_i}')$. Since $\phi_x:B_x\to \phi_x(B_x)$ is a homeomorphism and $\overline{B_x'}\subset B_x$, the restriction of $\phi_x$ to $\overline{B_x'}$ is a homeomorphism onto its image, and this image is $\overline{\phi_x(B_x')}$. Thus $\overline{\phi_x(B_x')}^\circ=\phi_x\left(\overline{B_x'}^\circ\right)=\phi_x(B_x')$ and $\overline{\phi_x(B_x')}$ is contractible. Letting $A_i=\overline{\phi_{x_i}(B_{x_i}')}$ finishes the proof.
Proof of Theorem: We shall induct on $r$. You have already done the base case $r=0$. Suppose for any compact manifold $M$, $H_r(M,G)$ is finitely generated. Let $\{A_i\}_{i=1}^m$ be as in the lemma. We induct on $m$. If $m=1$ the result is trivial. Suppose $H_{r+1}(M,G)$ is finitely generated for any $M$ which can be covered by $m$ such sets, and let $M'$ be covered by $m+1$ such sets. Let $A=\bigcup\limits_{i=2}^{m+1} A_i$. Then M-V gives a sequence
$$H_{r+1}(A,G)\cong H_{r+1}(A_1,G)\oplus H_{r+1}(A,G)\to H_{r+1}(M',G)\to H_r(A_1\cap A,G)$$
and by our two inductive hypotheses $H_{r+1}(A,G)$ and $H_r(A_1\cap A,G)$ are finitely generated. Thus by the lemma, $H_{r+1}(M',G)$ is finitely generated. Induction on $m$ shows this holds for all $m$, and then induction on $r$ shows this holds for all $r$.
*I originally used open sets $A_i$, but then $A$ and $A_1\cap A$ in the main proof are not actually compact manifolds, so the induction fails. Note that contractibility of intersections is not necessary in this proof. In fact, all that is necessary is that the homology groups of each $A_i$ be finitely generated.
