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Fix $n > 1$ and let's say I have a $(n-1)$-connected space $X$ (not necessarily a CW-Complex) and an Eilenberg-Maclane space $K(\pi_n(X), n)$, I want to show (assuming that this is true in general of course) that there exists a continuous map $f : X \to K(\pi_n(X), n)$ such that the induced map $f_* : \pi_n(X) \to \pi_n(K(\pi_n(X), n)) \cong \pi_n(X)$ is (equivalent to) the identity map $1_{\pi_n(X)}$.

Now if the functors $\pi_n$ were full functors then this would be a trivial category theoretic proof. However I'm not sure if the functors $\pi_n$ are full functors and if they are not I guess I would have to construct such a map by hand.

I have two questions:

  1. Are the functors $\pi_n$ full functors?
  2. Are there any references where I can read up further about a proof of this?

One thing I will mention is that Lemma 4.31 in Hatcher's Algebraic Topology is quite similar to what I am looking for, however the problem with that is that the desired map works in the opposite way to what I'm looking for. (Though I guess that one could perhaps modify the proof of that to obtain the desired result I'm seeking).

Edit: I've added the assumption that $X$ is $(n-1)$-connected.

Perturbative
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    If $\pi_k(X)=0$ for $k<n$, then you can construct $K(\pi_n(X),n)$ by attaching cells of dimension $n+2$ or higher to $X$ to kill off the rest of its homotopy groups (with the caveat that $K(\pi_n(X),n)$ is now not necessarily a CW-complex). In this case, the inclusion map $X\hookrightarrow K(\pi_n(X),n)$ should do the trick. If $X$ has non-trivial homotopy in lower degrees, I suspect such an $f$ does not have to exist, but my intuition could be wrong. – Niven Aug 29 '19 at 07:49
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    There is no map in general if unless you assume that $X$ is $(n-1$)-connected. For instance there is no map $S^3\rightarrow K(\mathbb{Z}_2,4)$ inducing an isomorphism on $\pi_4$. – Tyrone Aug 29 '19 at 08:51
  • @Tyrone Okay we can assume that $X$ is $(n-1)$-connected. But don't we also need to assume that $X$ is a CW-Complex? – Perturbative Aug 29 '19 at 10:50
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    Yes, that will probably help. For instance there is no map from the pseudocircle to $S^1\simeq K(\mathbb{Z},1)$ which induces an isomorphism on $\pi_1$. See https://ncatlab.org/nlab/show/pseudocircle for instance. – Tyrone Aug 29 '19 at 11:02
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    The functors $\pi_n$ are not full. For example there are homomorphisms $\mathbb{Z}\rightarrow\mathbb{Z}$ that cannot be realised as the induced map on $\pi_4$ by a self-map of $\mathbb{H}P^n$ (quaternionic projective space). See, for instance, McGibbon's Self Maps of Projective Spaces. – Tyrone Aug 29 '19 at 11:11
  • @Tyrone Thanks for your comments! Is it still possible to recover such a map $f$, if $X$ is not a CW-Complex? – Perturbative Aug 29 '19 at 12:08
  • In general no, and you have my example of the pseudocircle for that. For an arbitrary space $X$ you can always choose a CW approximation $\overline X$ and work with that instead. – Tyrone Aug 29 '19 at 13:08
  • A thought is that the nth cohomology of $X$ With coefficients in $\pi_n(X)$ is $Hom(\pi_n(X),\pi_n(X))$ so maybe apply representability of cohomology. – Connor Malin Aug 29 '19 at 14:30
  • I think a relevant question is https://math.stackexchange.com/questions/3237596/when-do-elements-of-operatornamehomg-g-correspond-to-invertible-self-maps – Connor Malin Aug 29 '19 at 14:34
  • In particular, read the last paragraph. – Connor Malin Aug 29 '19 at 14:36
  • @ConnorMalin Thank you very much! I think this just might be what I was looking for since I don't need to assume $X$ is a CW-Complex! – Perturbative Sep 01 '19 at 09:52

1 Answers1

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Here are two approaches.

  1. If you believe that $K(A, n)$ represents the cohomology functor $H^n(-, A)$, then the set of homotopy classes of maps $[X, K(\pi_n(X), n)]$ can be identified with $H^n(X, \pi_n(X))$. By the universal coefficient theorem this is $\text{Hom}(H_n(X), \pi_n(X))$ (we use the assumption that $X$ is $(n-1)$-connected here), and by the Hurewicz theorem this is $\text{Hom}(\pi_n(X), \pi_n(X))$ (we use the $(n-1)$-connectedness assumption again here). Now we just take the element of this Hom corresponding to the identity.

  2. Start with $X$ and repeatedly attach cells to kill off every element of $\pi_{n+1}(X)$, then $\pi_{n+2}(X)$, and so on. This eventually produces a $K(\pi_n(X), n)$ equipped with a map from $X$ inducing the identity on $\pi_n$.

This construction produces the first nontrivial stage in the Postnikov tower of $X$.

Qiaochu Yuan
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  • In the second approach, does one need to assume $X$ is a CW-complex? I guess if not, then I'd need to take a CW approximation – Perturbative Sep 12 '19 at 13:41
  • I am basically always working up to weak equivalence in homotopy theory, so whatever helps you sleep at night! – Qiaochu Yuan Sep 12 '19 at 19:03