1
$\begingroup$

We let $X$ be a CW complex with a unique 0-cell, given by the basepoint $x_0$, and no $k$-cells for $0 < k < n$. I want to show that $X$ can be embedded into a $K(\pi_{n}(X,x_0),n)$ space. I am really not sure how to get started on this. I suspect that I should somehow make a use of the CW approximation theorem but in reality I don't really know what sort of construction I should be trying to build. Any help would be much appreciated.

$\endgroup$
11
  • $\begingroup$ seems like a straightforward obstruction theory argument, but it's been a while since I've done one of those so give me a minute $\endgroup$
    – ziggurism
    Nov 19, 2019 at 13:23
  • $\begingroup$ @ziggurism take your time! I am expecting it to be sth "simple" but I haven't seen a question of this type before (this is the first time I'm doing topology) so I'm not really sure how to approach this $\endgroup$
    – billy192
    Nov 19, 2019 at 13:33
  • $\begingroup$ Eilenberg-Maclane (E-M) space is not unique. And I don't think that $X$ embeds into all of them. Take $X=S^1\vee S^2$ and $n=1$ (note that $X$ satisfies your conditions). Then $\pi_1(X)=\mathbb{Z}$ and $S^1$ is $K(\mathbb{Z},1)$. But clearly $X$ does not embed into $S^1$. Did you mean that $X$ embeds into some E-M space? That question is interesting, although it is trivial for metrizable CW complexes, which embed into a contractible space (and so you take a product of that space and any E-M space). Maybe every CW complex embeds into a contractible space? Not sure. $\endgroup$
    – freakish
    Nov 19, 2019 at 14:57
  • $\begingroup$ Here’s a place to start. By hurewicz, the nth homology of X matches the nth homotopy group. And by UCT so does cohomology. But cohomology is represented by K(G,n). So this gives you your map. But I guess I don’t see why it would be an embedding. $\endgroup$
    – ziggurism
    Nov 19, 2019 at 15:04
  • $\begingroup$ Thanks @freakish. I guess I want to show that $X$ embed into the E-M space of its $n$th homotopy group for some $n$. $\endgroup$
    – billy192
    Nov 19, 2019 at 15:16

1 Answer 1

1
$\begingroup$

For any (connected?) CW complex, we may attach an $n$-cell, and the resulting subcomplex inclusion map $X\hookrightarrow X\coprod e_n$ is an isomorphism on all homotopy groups $\pi_k(X)$ for $k<n-1$, and a surjection for $k=n-1$, and has unknown or complex effect for $k>n-1.$

Basically $n$-cells are nullhomotopies of the generators of $\pi_{n-1}(X)$, and so attaching one can kill off a generator (depending on the attaching map), while leaving the lower homotopy groups unchanged.

Starting with $k=n+2$, we want to attach an $(n+2)$-cell for every element of $\pi_{n+1}(X)$ (maybe it's enough to just iterate over the generators of $\pi_{n+1}(X)$?) to kill them. This leaves all the lower homotopy groups unchanged, but can scramble the higher ones of the resulting constructed CW complex.

$$ Y^{(n+2)} = X\coprod_{i\in\pi_{n+1}(X)} e^i_{n+2}. $$

Then we ascend the skeleton, doing the same at each higher dimension, attach an $(n+3)$-cell to $Y^{(n+2)}$ for every element of $\pi_{n+2}(Y^{(n+2)})$ to kill all the elements of $\pi_{n+2}(Y^{(n+2)})$, and call the resulting complex

$$ Y^{(n+3)} = Y^{(n+2)}\coprod_{i\in\pi_{n+2}(Y^{(n+2)})} e^i_{n+3}. $$

Continue up all the dimensions of $X$, kill all the homotopy groups above $n$, and the resulting space $Y^{(\infty)} = \operatorname{colim} Y^{(k)}$ is a $K(G,n)$ and we have an inclusion $X \hookrightarrow K(G,n)$ as a subcomplex.

$\endgroup$
14
  • $\begingroup$ Thank you @ziggurism! So it is just the Postnikov Tower right? $\endgroup$
    – billy192
    Nov 19, 2019 at 22:13
  • $\begingroup$ @billy192 I didn't specifically invoke the Postnikov tower, but the Postnikov tower is constructed using the same lemma. The normal way the Postnikov tower works though gives you maps $K(\pi_n(X),n)\to X$ rather than the other way. $\endgroup$
    – ziggurism
    Nov 20, 2019 at 5:56
  • $\begingroup$ ohh I see. I hope you don't mind me asking, what do you mean when you say "unknown or complex effect"? Could you tell me what the lemma was (I have a copy of Hatcher on my desk so you can just tell me the number of the page if it's from there)? $\endgroup$
    – billy192
    Nov 20, 2019 at 9:18
  • $\begingroup$ @billy192 What did I mean by "unknown or complex effect"? Well if you start with a space that's just a 1-type (pi1 is nontrival, all the other pi_n are trivial), and you want to build a new space out of it with trivial pi1, then you will attach a 2-cell for every element of pi1. This may create some 2-spheres in your resulting space. A 2-sphere has nontrivial pi_n for every n≥2. So the cost of killing all the pi1 generators of our space was to add new generators in every dimension.... $\endgroup$
    – ziggurism
    Nov 20, 2019 at 14:22
  • $\begingroup$ ...And there's no simple formula for homotopy groups of spheres, and computing homotopy groups for complex spaces is hard. So our new 1-connected space has trivial pi_1, but likely nontrivial and hard to compute pi_n for every higher n. But adding an n-cell cannot change pi_k for k lower than n. And it can only kill pi_n, not add complexity. So as long as we start from the low dimensions and build upward inductively, we can build a complex with whatever homotopy groups we want. $\endgroup$
    – ziggurism
    Nov 20, 2019 at 14:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .