Let $n \geq 2$ and $X$ be a $(n-1)$-connected simplicial complex. This means that all of the lower homotopy groups $\pi_{k}(X) = 0$ for $k \leq n-1$. My goal is to compute the higher homotopy groups of $X$ (in terms of homology of some spaces).
Summary of Serre's method
Case. $k=n$
By Hurewitz theorem, $\pi_n(X) \simeq H_n(X)$.
Case. $k=n+1$
Serre's method is to use Hurewitz theorem again [1], by producing another simplicial complex $E = E_X$ whose homotopy groups agree with those of $X$, except $\pi_n(E) = 0$. Indeed, with such $E$, Hurewitz theorem guarantees that $\pi_{n+1}(E) \simeq H_{n+1}(E)$. Hence $\pi_{n+1}(X) \simeq H_{n+1}(E)$.
Case. $k>n+1$
Repeat the previous prodecure, with $E$ being $X$, and so on.
Crux
How do we actually construct such $E$? I understand that if we construct $E$ as a fibre product $E \to X$, with fiber $F \simeq K(\pi_{n}(X), n-1)$, then by (writing down) the associated long exact sequence of homotopy groups, it is easy to see that the goal is equivalent to proving that the induced map in the long exact sequence
$$\Phi: \pi_{n}(X) \to \pi_{n-1}(K(\pi_{n}(X), n-1))$$
is an isomorphism.
Question
- How to build that bundle / fibre product so that $\Phi$ is an isomorphism?
- Can we build such $E$ so that the result is functorial, i.e. each $f: X \to Y$ has $f^E: E_X \to E_Y$ such that the composition
$$ \pi_{n+1}(X) \xrightarrow{b.iso^{-1}} \pi_{n+1}(E_X) \xrightarrow{h.iso} H_{n+1}(E_X) \xrightarrow{f^E} H_{n+1}(E_Y) \xrightarrow{h.iso} \pi_{n+1}(E_Y) \xrightarrow{b.iso} \pi_{n+1}(Y)$$ equals the map $\pi_{n+1}(X) \xrightarrow{f} \pi_{n+1}(Y)$? (Here, $h.iso$ denotes the Hurewitz isomorphism, while $b.iso$ denotes the isomorphism induced from the bundle data constructed in 1.)