How does the inclusion $\mathbb Z\rightarrow \mathbb Q$ induce a fibration $K(\mathbb Z,n)\rightarrow K(\mathbb Q,n)$ with fibre $\Omega K(\mathbb Q/\mathbb Z,n)$?

Probably the most functorial approach is to use the Dold-Kan equivalence $$F:\{\text{chain complexes}\} \to \{\text{simplicial abelian groups}\}. $$ Let $A_{\ast}$ denote the chain complex with just $\mathbb{Q}/\mathbb{Z}$ in dimension $n-1$, let $B_{\ast}$ be the one with a surjective differential from $\mathbb{Q}$ in dimension $n$ to $\mathbb{Q}/\mathbb{Z}$ in dimension $n-1$, and let $C_{\ast}$ be the one with just a $\mathbb{Q}$ in dimension $n$. There is an evident short exact sequence (and therefore fibration) $A_{\ast}\to B_{\ast}\to C_{\ast}$, which gives a fibration $|FA_{\ast}|\to |FB_{\ast}|\to |FC_{\ast}|$ of topological abelian groups. Here $|FA_{\ast}|$ and $|FC_{\ast}|$ are $K(\mathbb{Q}/\mathbb{Z},n-1)$ and $K(\mathbb{Q},n)$ essentially by definition, and it is easy to produce a weak equivalence from the corresponding model for $K(\mathbb{Z},n)$ to $|FB_{\ast}|$.

undergraduatestudent. $\endgroup$ – Fernando Muro Jun 8 '11 at 11:11