Let $G$ and $H$ be torsion abelian groups.
Are the following are equivalent:


1. $\mathrm{Hom}(G, H) = 0$
2. $\mathrm{map}\_\*( K(G, m), K(H,n)) \sim \*$ for all $n, m\geq 1$


?


Clearly (2) implies (1).



Contractible space of maps between Eilenberg-Mac Lane spaces, 2
No, we have $Hom(\mathbb{Z}/n,\mathbb{Z})=0$ but there is a nontrivial Bockstein $K(\mathbb{Z}/n,1)\to K(\mathbb{Z},2)$.


For finite abelian groups we have (1) iff (2) iff ($|G|$ and $|H|$ are coprime).


