I am reading Ravenel's "complex cobordism and stable homotopy groups of spheres".
I am a bit confused by the notion of homotopy fiber, which basically gives a functorial way of regarding any map as a fibration.
In Lemma 1.2.3, Ravenel says that if $F$ is the fiber of the map $f:S^3\longrightarrow K(\mathbb{Z},3)$, given by the canonical map from $S^3$ to the above Eilenberg MacLane space, that the fiber of the fiber is $K(\mathbb{Z},2)$, i.e. there exists a map: $$K(\mathbb{Z},2)\longrightarrow F\longrightarrow S^3$$
My question is why is that the case? Is there some technique to compute the fiber, or the fiber of the fiber which should make this clearer?
Thanks in advance!
I think, it is related to the long fiber sequence (Puppe sequence) of the principal fibration of the first classifying map of the Postnikov resolution of $S^3$. It looks like $$......\rightarrow\Omega E\rightarrow\Omega S^{3}\rightarrow F=\Omega K(\Bbb{Z},3)=K(\Bbb{Z},2)\rightarrow E\rightarrow S^{3}\rightarrow K(\Bbb{Z},3)$$
I think the fiber is $F=K(\Bbb{Z},2)$ and $E\rightarrow S^{3}$ is the (principal) fibration. $S^{3}\rightarrow K(\Bbb{Z},3)$ is called the classifying map of this principal fibration and it is determined by the cohomology class $1\in H(S^3,\Bbb{Z})$.
Is $S^{3}\rightarrow K(\Bbb{Z},3)$ fibration? I don't know. This map is very mysterious.
See Emery Thomas's Lecture notes for the technique: https://www.maths.ed.ac.uk/~v1ranick/papers/thomasfib.pdf