TITLE: What is a principal refinement of a Postnikov system?
QUESTION [8 upvotes]: I've been reading the book of Hilton, Mislin, and Roitberg on Localization of Nilpotent Groups and Spaces. In Section II.2 they define a principal refinement at stage $n$ of a Postnikov system $$\cdots \to X_n\overset{p_n}{\to} X_{n-1}\to \cdots$$
to be a factorization of $p_n$ into a finite sequence of fibrations 
$$X_n=Y_c \overset{q_c}{\to}\cdots\to Y_1\overset{q_1}{\to}Y_0=X_{n-1}$$
whose fibers are Eilenberg-MacLane spaces $K(G_i,n)$.
But isn't the point of a Postnikov system that $p_n$ is already a fibration whose fibers are Eilenberg-MacLane spaces? So I don't understand why the condition of the definition isn't satisfied trivially at every stage. Perhaps there's some subtlety involving the condition also given that each $q_i$ be induced by a map $g_i: Y_{i-1}\to K(G_i, n+1)$, but aren't all fibrations with fiber $K(G_i, n)$ induced this way since $K(G_i, n+1)$ is the base of a path-space fibration with fiber $K(G_i, n)$?

REPLY [7 votes]: A fibration with $K(G,n)$ fibers need not be principal. The base of the universal fibration with fiber $K(G,n)$ is not a $K(G,n+1)$, even if $G$ is abelian. For example, in the case $G=\mathbb Z$ and $n=1$ it is $BO(2)$, with $\pi_1=\mathbb Z/2$ and $\pi_2=\mathbb Z$. In fact, in general it has homotopy groups $\pi_1=Aut(G)$, $\pi_{n+1}=G$, all the rest trivial. That's if $n>1$; in the case $n=1$, $\pi_1$ is the outer automorphisms of $G$ (the automorphisms modulo inner automorphisms) and $\pi_2$ is the center of $G$.
You can see this by viewing that space (base of universal fibration with fiber $F$) as a delooping of the space of homotopy equivalences of $F$. The latter fibers over $F$ with fiber the pointed self-equivalences; in the case $F=K(G,n)$ the space of pointed self-equivalences is equivalent to the discrete space $Aut(G)$.