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From the path fibration we extract that $\pi_{i+1}(K(G,r))=\pi_{i}\Omega K(G,r)$, then for all $k$, $\pi_{k}(K(G,r-1))=\pi_{k}\Omega K(G,r)$.

How can we conclude that $K(G,r-1)\simeq \Omega K(G,r)$?

If there were a map $K(G,r-1))\rightarrow \Omega K(G,r)$ that induces this isomorphism on $\pi_k$, then we would conclude by the Whitehead theorem, but here I don't see this map.

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You haven't accepted any of the answers to your other questions yet. If you're not satisfied with them, you should point out what's missing so that it can be amended. Otherwise, it's good practice to accept an answer, not just for rewarding people for helping you, but also to mark the questions as answered. –  joriki May 15 '11 at 12:45
    
@joriki ok understood. thanks. –  studento May 15 '11 at 14:44
    
@Grigory M: i'm using the standard one but how is it obvious? –  studento May 15 '11 at 14:45
3  
An Eilenberg--MacLane space is characterized by vanishing of homotopy in all but one degree, where it is equal to some given $G$. In other words, if some given space has the same homotopy types as an Eilenberg--MacLane space, it is itself an Eilenberg--MacLane space (by definition). The question of whether two Eilenberg--MacLane spaces are necessarily homotopic is a separate one (and is perhaps what you really mean to ask). –  Matt E May 15 '11 at 14:55
    
@Matt E: I now understand the problem clearly. thanks. –  studento May 15 '11 at 15:32

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