I want to solve this question, which consists of the following 2 parts:
(a) Prove that if $n \geq 1$, d is an integer, there is a map $f\colon S^n \rightarrow S^n$ of degree d.
(b) Let $G$ be a finitely-generated Abelian group. Prove that there is a CW-complex $M(G,n)$ which has $\tilde{H_{k}}(M(G,n)) = G$ if $k=n$ and $0$ otherwise.
My questions are:
1- Is there any relation between the proof of part (a) and part (b)?
2- I know how to solve the following 2 problems:
- Show that for Abelian groups $G$ and $H, \bigl[K(G, n), K(H, n)\bigr] \cong \operatorname{Hom}(G, H).$
- Suppose $X$ is a finite $(n-1)$-connected CW complex, and let $G$ be an abelian group. Show that the function $$D \colon \bigl[X, K(G,n))\bigr] \rightarrow \operatorname{Hom}(\pi_{n}(X), G) $$
given by $D([f]) = f_{*}$ is bijective.
Is the solution of part (b) above similar to the solution of one of those problems? If not, could anyone show me the solution or mention a good reference that contains this proof please.
I gave some hints in the comments, I thought it would be a good idea to put them in an answer and add details later if needed.
Is there any relation between the proof of part (a) and part (b)?
If $f\colon S^n \to S^n$ is a degree $d$ map and we construct the mapping cone $C_f = D^{n+1}\cup_f S^n$, then if $d\neq 0$ this will be a $M(\mathbb{Z}/d, n)$. (You can see this for example by computing the cellular chain complex of $C_f$.) By part (a) there exists such a degree $d$ map for all $n>0$, so this solves part (b) for the class of finite cyclic groups. Note that $S^n$ is a $M(\mathbb{Z}, n)$, so in fact we can prove (b) for all cyclic groups.
Is the solution of part (b) above similar to the solution of one of those problems?
Not really.
Homotopy and Homology tend to have quite different flavours, for example $\pi_n(X\times Y)\cong \pi_n(X) \oplus \pi_n(Y)$ but for the homology groups of a product you need the Künneth theorem, and the homotopy groups of a wedge can be very difficult to compute but $\tilde{H}_n(X\vee Y) \cong \tilde{H}_n(X) \oplus \tilde{H}_n(Y)$. In particular, it follows that $K(G, n) \times K(H, n)$ is a $K(G\oplus H, n)$ whereas $$M(G, n) \vee M(H, n)\text{ is a model for }M(G\oplus H, n).$$
Now to finish (b), invoke the classification theorem for finitely-generated abelian groups.