Is the Hurewicz isomorphism the identity for $K(G,n)$?

Question

Let $X=K(G,n)$ be the Eilenberg-MacLane space with $\pi_n(X)=G$ and $\pi_i(X)=0$ for $i \neq n$. Is the Hurewicz isomorphism $h: \pi_n(X) \rightarrow H_n(X)$ the identity for such spaces? I mean, if we see $h$ as an element of $Aut(G)$, is it the identity? Because of the naturality of $h$ it seems to be plausible, but I can't see a direct proof (this is my opinion, it could be false as far as I know).

Answer 1

It's not the identity simply because $\pi_nK(G,n)$ and $H_nK(G,n)$ are different groups. Eg. homotopy classes of maps $S^n\rightarrow K(G,n)$ vs. say, equivalence classes of simplices $\Delta^n\rightarrow K(G,n)$ for singular homology. It doesn't even make sense to say that they are the same group. The Hureweicz map gives one way to compare them, and in this case we handily know that it is an isomorphism.

The identification $\pi_nK(G,n)\cong G$ is non-canonical and comes as extra structure with the Eilenberg-Mac lane space. Generally these spaces can be defined en masse using some nice method, for instance the bar construction, which will generate a distinguished choice of such isomorphism. Hence we can generally forget about the choice of the identification and carry on doing homotopy theory without thinking about it. If we choose a different method to generate our $K(G,n)$s, say with a Milnor classifying space, then we get a different choice of identification.

Furthermore these spaces are only defined up to homotopy type. Using the previous example we get a canonical homotopy equivalence between the bar construction $K(G,n)$ and the Milnor classifying space $K(G,n)$, and these is no reason that this homotopy equivalence this will induce an identity on $\pi_n$, let alone be compatible with the different identifications we have chose of this group with $G$. We simply know that it will be an automorphism.

As an example consider $K(\mathbb{Z},1)\simeq S^1$. Choosing a clockwise orientation we use the degree function to get $\pi_1S^1\cong\mathbb{Z}$. If we triangute $S^1$ with an anticlockwise orientation when we get a isomorphism $H_1S^1\cong \mathbb{Z}$. The Hurewicz map is an isomorphism $h:\mathbb{Z}\cong \pi_1S^1\xrightarrow{\cong}H_1S^1\cong \mathbb{Z}$, however with our definitions it acts as $1\mapsto -1$, the non-trivial automorphism of $\mathbb{Z}$.

The point is that nearly everything relies on us making some choices but generally the only time that they are important is when we ask question such as yours. For instance it is most convenient to define $H_nK(G,n)$ using the Hurewicz map $h$ to give a canonical set of generators. In this case the Hurewicz map itself is not an identity, but by defintion composes to the identity $id_G:G\stackrel{choice}{\cong}\pi_nK(G,n)\stackrel{h}{\cong}H_nK(G,n)\stackrel{def}{\cong}G$.