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



Induced fibration of Eilenberg-MacLane spaces
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 $|FB\_{\ast}|$. 


