TITLE: Moore space, induced map in homology
QUESTION [9 upvotes]: Let $A$ be a finitely generated abelian group and $n$ a positive integer. I have built a connected space $M(A,n)$ such that all its reduced homology groups are zero but the i-th reduced homology group that is equal to $A$ (Moore space).
Let $A_0$ and $A_1$ be finitely generated abelian groups and
let $F: A_0 \to A_1$ a homomorphism.  How could I construct a map
$$f: M(A_0,n) \to M(A_1,n)$$
such that $f_*=F$.
Context
For me, if $A \simeq Z$, $M(A,n)=S^n$. If $A \simeq Z_m$, then $M(A,n)$ is given by the mapping cone $C(S^n) \cup_f S^n$, where $C(S^n)$ is $S^n\times [0,1]/((x,0)\simeq(x',0))$ and $f: S^n \times 1 \to S^n$ comes from a map from $S^n$ to $S^n$ of degree $m$.

REPLY [3 votes]: In general, such map will not exist. Take for instance the 3-sphere and an aspherical 3-manifold which is an integer homology 3-sphere. Then each map from former to later is null-homotopic and hence cannot induce an isomorphism of top homology groups. Take a look at the n-Lab article about lack of functoriality of Moore spaces.