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As an exercise, I'm trying to show that for an $(n-1)$-connected space $L$ with $\pi=\pi_n(L)$, the map $\iota_L:L\rightarrow K(\pi,n)$ associated to the fundamental class $\iota_L\in H^n(L;\pi)$ induces an isomorphism $\pi_n(L)\rightarrow \pi_n(K(\pi,n))$. After playing with this for a while, I've found that this feels so tautological that it has to be true, although that doesn't quite constitute a proof. I'm supposed to use only basic definitions / first principles. Presumably, these are:

  1. The universal coefficient theorem yields $H^n(L;\pi)\cong Hom(H_n(L),\pi)$. The Hurewicz homomorphism $h:\pi \rightarrow H_n(L)$ is an isomorphism, and its inverse $h^{-1}\in Hom(H_n(L),\pi)$ corresponds to $\iota_L\in H^n(L;\pi)$.

  2. We have a canonical bijection $H^n(L;\pi) \cong [L,K(\pi,n)]$. Any $[f]\in [L,K(\pi,n)]$ corresponds to $f^*\iota\in H^n(L;\pi)$, where $\iota\in H^n(K(\pi,n);\pi)$ is the fundamental class of $K(\pi,n)$ (which is associated to its identity map). This is actually a group isomorphism if we add maps on the right side by using the fact that $K(\pi,n)=\Omega K(\pi,n+1)$ (or at least $\simeq$, although what does $K(\pi,n)$ even mean really). I'd imagine that this is canonical too, but I don't know for sure.

  3. We have a map $[L,K(\pi,n)]\rightarrow Hom(\pi,\pi)$ given by $[f]\mapsto f_\#$. Presumably the idea is to show that the image of $\iota_L$ is an isomorphism, but I can't tell if I'm just complicating the question by phrasing it in these terms.

I think my problem is that I don't really understand the defining way Eilenberg-MacLane spaces work. I have an intuitive picture of the Hurewicz homomorphism, and so I guess I have an intuitive picture of its inverse: it takes a homology class in degree $n$ and realizes it as the image of a bunch of based $n$-spheres (which is made possible by the Hurewicz theorem). We can look at this more or less as a cochain in $C^n_{cell}(L;\pi)$, and this is the element $\iota_L\in H^n(L;\pi)$. But then I have no idea how to actually turn this into a map $\iota_L : L\rightarrow K(\pi,n)$, or whether I'm even supposed to.

Because I'd still like to work this out myself to whatever extent I can, a good hint (if one exists) is worth more to me than a straight-up answer. Of course, I'm happy with either. Thanks!

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Try a relative version of the Hurewicz Theorem, (known as the Whitehead Theorem) - A map is an isomorphism on homology up to degree $n$ if and only if it is an isomorphism on homotopy up to degree $n$. – Jeffrey Giansiracusa Sep 16 '10 at 7:27
up vote 1 down vote accepted

Here is a solution, based on Jeffrey's comment:

Let $f:L\rightarrow K(\pi,n)$ be the map classifying $\iota$. We will show that the induced map $f_*:H_i(L)\rightarrow H_i(K(\pi,n))$ is isomorphic for $i \lt n+1$ and epimorphic for $i=n+1$. This will imply by Whitehead's theorem that the same is true for $f_\# :\pi_i(L)\rightarrow \pi_i(K(\pi,n))$ (assuming $n \gt 1$).

First, $f_*$ is automatically an isomorphism for $i \lt n$, since in that case by the Hurewicz theorem $H_i(L)$ and $H_i(K(\pi,n))$ are both trivial.

Next, in dimension $n$, we use the fact that $f^*:H^n(K(\pi,n);\pi)\rightarrow H^n(L;\pi)$ satisfies $f^*\iota_n = \iota$, where $\iota_n$ is the fundamental class of $K(\pi,n)$. By the universal coefficient and Hurewicz theorems, we may regard $H^n(K(\pi,n))=Hom(H_n(K(\pi,n)),\pi)$ and $H^n(L)=Hom(H_n(L),\pi)$, and the fundamental classes both correspond to isomorphisms (namely, the inverses of the respective Hurewicz homomorphisms). In general, if a group homomorphism $\varphi:B\rightarrow A$ induces $\varphi^\#:Hom(A,C)\rightarrow Hom(B,C)$ and $\varphi^\#$ takes an isomorphism $\psi_1:A\rightarrow C$ to an isomorphism $\psi_2:B \rightarrow C$, then $\varphi$ itself must be an isomorphism, because $\psi_2 = \varphi^\# (\psi_1)=\psi_1\varphi$ and so $\varphi = \psi_1^{-1}\psi_2$. Here, since $f^*:Hom(H_n(K(\pi,n)),\pi)\rightarrow Hom(H_n(L),\pi)$ is induced from $f_*:H_n(L)\rightarrow H_n(K(\pi,n))$ and takes an isomorphism to an isomorphism, then $f_*:H_n(L)\rightarrow H_n(K(\pi,n))$ itself is an isomorphism.

Lastly, by the Hurewicz theorem, since $K(\pi,n)$ is $n$-connected, the Hurewicz homomorphism $h:\pi_{n+1}(K(\pi,n))\rightarrow H_{n+1}(K(\pi,n))$ is an epimorphism, but $\pi_{n+1}(K(\pi,n))=0$ by definition so $H_{n+1}(K(\pi,n))=0$ as well. Hence $f_*: H_{n+1}(L)\rightarrow H_{n+1}(K(\pi,n))$ is automatically epimorphic.

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