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Mar 4, 2017 at 13:27 comment added Tom Goodwillie Yes. Just as a contractible space might fiber over a non-contractible space that is a subspace of a contractible space.
Mar 4, 2017 at 8:30 comment added მამუკა ჯიბლაძე Thanks for the clarification! Specifically with the case considered, if $f:A\to B$ is a continuous homomorphism between $K(\pi,n)$s inducing isomorphisms of homotopy groups, then from the long exact sequences of homotopy groups for the factorization$$K\rightarrowtail A\twoheadrightarrow I\rightarrowtail B\twoheadrightarrow C$$one gets $\pi_{n+1}(I)\cong\pi_n(K)$, $\pi_{n-1}(I)\cong\pi_n(C)$ and $\pi_n(I)\cong\pi\oplus\pi_{n-1}(K)\cong\pi\oplus\pi_{n+1}(C)$; seems like all these may be nontrivial (although all other homotopy groups of $K$, $I$ and $C$ are zero)...
Mar 3, 2017 at 19:25 comment added Tom Goodwillie Yes, the $E\pi$ that I was thinking of has an abelian group structure if $\pi$ is an abelian group. Think of the exact sequence of chain complexes $0\to (\pi\leftarrow 0) \to (\pi\leftarrow\pi)\to (0\leftarrow\pi)\to 0$.
Mar 3, 2017 at 19:23 comment added Tom Goodwillie What you say about contractible kernels and cokernels is not correct. You can make a chain map from a contractible chain complex to another contractible complex such that the kernel and the cokernel have nontrivial homology. (The image will have to also have nontrivial homology.)
Mar 3, 2017 at 19:18 comment added Tom Goodwillie (2) I don't know anything about that.
Mar 3, 2017 at 19:18 comment added Tom Goodwillie Sorry about (1). I seem to have been thinking about the casde $\pi=\mathbb Z$ when I wrote that part. I've edited it to correct that now.
Mar 3, 2017 at 19:16 history edited Tom Goodwillie CC BY-SA 3.0
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Mar 3, 2017 at 18:19 comment added მამუკა ჯიბლაძე ...by the way (and maybe this also should go into the main question), could it be that for $\pi$ abelian one can find a $E\pi$ which is an abelian group too? That is, can every abelian group be embedded as a subgroup into a contractible abelian group?
Mar 3, 2017 at 18:10 comment added მამუკა ჯიბლაძე Oh sorry one more thing. Say $\pi$ is not discrete; one can still form $E\pi$ as the geometric realization of the simplicial space $$\pi\begin{smallmatrix}\leftarrow\\\rightarrow\\\leftarrow\end{smallmatrix} \pi^2\begin{smallmatrix}\leftarrow\\\rightarrow\\\leftarrow\\\rightarrow\\\leftarrow\end{smallmatrix}\pi^3\begin{smallmatrix}\leftarrow\\\vdots\\\leftarrow\end{smallmatrix}\cdots;$$does this have chance to be something like infinite-dimensional sphere of sorts?? This would say something about $K(\pi,n)$s with $n>1$...
Mar 3, 2017 at 18:02 comment added მამუკა ჯიბლაძე Also a separate question concerning the last paragraph (although this probably should go into the main question). Consider the category with objects all $K(\pi,n)$ topological abelian groups and morphisms continuous homomorphisms which induce isomorphism of homotopy groups; I believe this means having contractible kernels and cokernels. Do you expect that this category does possess neither initial nor terminal object? Could it have weak terminals or initials (existence without uniqueness)?
Mar 3, 2017 at 18:02 comment added მამუკა ჯიბლაძე Some questions, please? I don't understand two things: (1) You name three simplicial abelian groups; I don't see how the third one depends on $\pi$. (2) At least when $\pi$ is very large, presumably $E\pi$ cannot be homeomorphic to $S^\infty$ by cardinality reasons. Are there some similar "large spheres", something like $S^{\text{some cardinal}}$ which would work?
Mar 3, 2017 at 16:19 history answered Tom Goodwillie CC BY-SA 3.0