most recent 30 from http://mathoverflow.net 2013-05-21T09:13:32Z http://mathoverflow.net/feeds/question/53105 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53105/contractible-space-of-maps-between-eilenberg-mac-lane-spaces Jeff Strom 2011-01-24T19:16:39Z 2011-01-24T19:50:27Z

<p>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 $n, m\geq 1$.</p> <p>Unfortunately, I want to prove this without homology or (explicitly) cohomology. </p> <p>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 $k\geq 2$. </p> <p>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 $A$.</p> <p>Question: Is there such an "elementary" argument for the general case?</p>

http://mathoverflow.net/questions/53105/contractible-space-of-maps-between-eilenberg-mac-lane-spaces/53113#53113 John Klein 2011-01-24T19:50:27Z 2011-01-24T19:50:27Z

<p>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?</p>