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Let $X$ be some sort of sufficiently nice space, e.g. a (connected) cell complex. Then $X$ has a universal cover $\tilde{X}$. This is simply connected by definition and it is easy to show that $\pi_n(\tilde{X})\cong\pi_n(X)$ for all $n>1$. We have, in effect, killed off a homotopy group of $X$ with a construction that seems fundamentally different than merely attaching cells to $X$ like crazy.

I wonder if there is a way to continue this process. Obviously, taking covering spaces won't get us anywhere, so we allow ourselves some more leeway. Specifically, I'm asking whether there exist spaces $X_n$ and maps $f_{n+1}\colon X_{n+1}\to X_n$ such that:

  • $X_0=X$ and $X_1=\tilde{X}$, the universal cover
  • $f_1\colon\tilde{X}\to X$ is the covering projection
  • $f_n$ is a (either Serre or Hurewicz) fibration for all $n$
  • $\pi_i(X_n)=1$ for $i\leq n$ and $\pi_i(X_n)\cong \pi_i(X)$ for $i>n$

I've look around a bit for answers to this, but I don't really know what to search for. I did find out about Postnikov towers, which seem related; I'm asking for some sort of "reverse Postnikov tower", if that makes any sense.

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    $\begingroup$ This isn't quite what you want, but for an $(n-1)$-connected space $X$ with $\pi_n(X) = G$ you can find a map $X \to K(G,n)$ that induces an isomorphism on $\pi_n$; then you can take the homotopy fiber $F$. Then $\pi_k(F) \cong \pi_k(X)$ for $k > n$, and $\pi_k(F) = 0$ for $k \le n$. (You can then iterate this, since $F$ will be $n$-connected.) $\endgroup$ Commented Nov 9, 2012 at 3:02
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    $\begingroup$ Yes, this is called the Whitehead tower. I'm pretty sure you can find out about it in Bott & Tu. $\endgroup$ Commented Nov 9, 2012 at 4:41
  • $\begingroup$ Thanks @ThomasBelulovich and @Aaron! This was exactly what I was looking for. If someone wants to write this as an answer, please do. $\endgroup$ Commented Nov 10, 2012 at 17:02

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As @Aaron Mazel-Gee pointed out, those are called whitehead towers (and can be found in hatcher book: algebraic topology, for example).

Now I feel like a parasite!..but since this answer were morally answered by aaron (through a comment) it is good that i become officially answered too! :)

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