Welcome to MathOverflow

Question

1

2

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?

flag	asked Jan 24 2011 at 19:16 Jeff Strom 3,514●1●6●19

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, ... – Ronnie Brown Feb 1 2012 at 16:53

John Klein · Answer 1 · 2011-01-24 19:50:27Z UTC

3

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?

answered Jan 24 2011 at 19:50

John Klein
7,256●19●28

Doesn't this just get the result for locally finite groups $A$? – Jeff Strom Jan 24 2011 at 20:25

Or is this not a problem because I'm working only with abelian groups? – Jeff Strom Jan 24 2011 at 20:29

1

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. – John Klein Jan 24 2011 at 21:15

Contractible space of maps between Eilenberg-Mac Lane spaces

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.