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Consier the fibration $$\Omega K(\mathbb Q / \mathbb Z, n)\rightarrow K(\mathbb Z , n) \stackrel{f}{\rightarrow} K( \mathbb Q, n)$$

The context is that this fibration is a crucial step in the proof of the rational hurewicz theorem in this paper:

< http://old.mfo.de/organisation/institute/klaus/HOMEPAGE/Publications/Publications/P13/preprint13.pdf>

(the last proof on page 7).

1) what is $f$

2) is this a special case of a fibration $$\Omega K(G / N, n)\rightarrow K(N , n) \stackrel{f}{\rightarrow} K( G, n)$$ where $G$ is an abelian group and $N$ is a subgroup of $G$.

3) how does this relate to the path fibration $$\Omega K(\mathbb Z, n)\simeq K(\mathbb Z, n-1)\rightarrow PK(\mathbb Z , n) \stackrel{f}{\rightarrow} K( \mathbb Z, n)$$

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1 Answer 1

$f$ is presumably the canonical map (unique up to homotopy) that is constructed from the inclusion $\mathbb Z\to\mathbb Q$ (One way to realize this is to construct $K(A,n)$ using an iterated classifying space construction $B\cdots B(A)$) But you unless you give us more context, it is really impossible to know! That the fiber is what you wrote follows from the long exact sequence for homotopy for the fibration. (2) is exactly the same.

Why would there be any relation in (3)?

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The context is that this fibration is a crucial step in the proof of the rational hurewicz theorem in this paper: old.mfo.de/organisation/institute/klaus/HOMEPAGE/.../preprint13.pdf (the last proof on page 7). I only need to understand the map $f$. –  palio May 30 '11 at 18:48
1  
@palio: do add that information to the question so that it is easily found. –  Mariano Suárez-Alvarez May 30 '11 at 18:54

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