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Suppose $A$ is an abelian torsion group, with no elements of order $p$, and let $P$ be an abelian $p$-group (i.e., the order of each element is a power of $p$). It sure seems to me that $$ \mathrm{map}_*( K(A,n), K(P,m) ) \sim * $$ for all $n, m\geq 1$.

Unfortunately, I want to prove this without homology or (explicitly) cohomology.

If $A$ is finite, then we can boil the question down to the case of a cyclic group $\mathbb{Z}/a$ with $(a,p) = 1$; then there is a cofiber sequence $M\to B\mathbb{Z}/a \to X$ where $M = S^1 \cup_a D^2$ is the "$1$-dimensional Moore space" for $\mathbb{Z}/a$ and $X$ has a cone decomposition using Moore spaces $M(\mathbb{Z}/a,k)$ for $k\geq 2$.

Since (as is easily shown) $\mathrm{map}_{*}( M(\mathbb{Z}/a,n), K(P,m) ) \sim \star$, we get $ \mathrm{map}_{*}( B\mathbb{Z}/a, K(P,m)) \sim *$, and then homotopy colimit stuff proves the assertion for finite $A$.

Question: Is there such an "elementary" argument for the general case?

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    $\begingroup$ Just come across this. "Unfortunately, I want to prove this without homology or (explicitly) cohomology." Why? I ask this as my 1962 PhD had to do with mapping spaces and Eilenberg-Mac Lane spaces. But I did use cohomology, chain complexes, Dold-Kan, ... $\endgroup$
    – Ronnie Brown
    Feb 1, 2012 at 16:53

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Couldn't one filter $A$ by finite subgroups $A_j$ and use $$ \text{map}_*(K(A,n),K(P,m)) = \lim_j \quad \text{map}_*(K(A_j,n),K(P,m)) , $$ the fact that the limit and homotopy limit coincide in this case, and finally that you are taking the homotopy limit of contractible spaces over a diagram which has contractible shape?

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  • $\begingroup$ Doesn't this just get the result for locally finite groups $A$? $\endgroup$
    – Jeff Strom
    Jan 24, 2011 at 20:25
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    $\begingroup$ Or is this not a problem because I'm working only with abelian groups? $\endgroup$
    – Jeff Strom
    Jan 24, 2011 at 20:29
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    $\begingroup$ This gives it for countably generated ones. But a similar argument works I think by filtering $A$ by its poset $\cal P_A$ of finite subgroups and inclusions, and then noting that $\cal P_A$ is contractible (since if $B_\alpha \subset A$ is a finite diagram of finite subgroups, then the subgroup generated by all of these is still finite; this shows that $\cal P_A$ is contractible. $\endgroup$
    – John Klein
    Jan 24, 2011 at 21:15

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