Let $G$ and $H$ be torsion abelian groups. Are the following are equivalent:
$\mathrm{Hom}(G, H) = 0$
$\mathrm{map}_*( K(G, m), K(H,n)) \sim *$ for all $n, m\geq 1$
?
Clearly (2) implies (1).
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1 Answer
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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).
answered Jan 24, 2011 at 20:24
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$\begingroup$ Of course. I'll edit the question so that it makes more sense. $\endgroup$ Commented Jan 24, 2011 at 20:35
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$\begingroup$ I see that I haven't gotten around to editing the question, and I don't quite remember what I was driving at; and since you did answer the question, you get the cigar. $\endgroup$ Commented Apr 29, 2014 at 0:34