It is well known that for any abelian group $G$, and any CW-complex $X$, the set $[X, K(G,n)]$ of homotopy classes of maps from $X$ to $K(G,n)$ is in natural bijection with the $n^{\mathrm{th}}$ singular cohomology group $H^n(X; G)$ with coefficients in $G$.

My question is, is there a similar bijection if the group is nonabelian? Notice that we only need to consider $[X, K(G,1)]$. In particular, I am trying to figure out what $[X, K(G,1)]$ looks like if $G$ is a finite group.


Assume $X$ connected.

Yes, this is known. For pointed homotopy classes this is $\text{Hom}(\pi_1 X, G)$. This is 1B.9 of Hatcher, usually the first place one sees obstruction theory, and requires less work than the case of $n > 1$.

For unpointed homotopy classes of maps, one quotients by conjugacy of elements of $G$. As you point out in the comments below, this is 4A.2 of Hatcher.

  • $\begingroup$ What does it mean by "quotients by conjugacy of $\pi_1 X$ and $G$? From Hatcher's book 4A.2, we only need to quotient the pointed homotopy by $G$. $\endgroup$ – Totoro Feb 10 at 20:39
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    $\begingroup$ @Totoro Here was the thought process (though you are right). In what follows I probably have some handedness wrong of my actions. Both $H$ and $G$ act on $\text{Hom}(H, G)$: let $$(h \cdot \rho \cdot g)(h') = g^{-1} \rho(h h' h^{-1}) g.$$ This is a new homomorphism and one can modify the homomorphism by either of these actions following an unbased homotopy. But because $\rho$ is a homomorphism, this is the same as conjugating by $\rho(h^{-1})g$, and hence the quotient by $H$ may be subsumed into the quotient by $G$. $\endgroup$ – user98602 Feb 10 at 20:58
  • $\begingroup$ BTW, @Totoro, thank you for the reference in Hatcher's book where he pins down the difference between homotopy classes of maps and based homotopy classes of maps. I always forget where that is (which is why it wasn't included in this answer). I will add it now. $\endgroup$ – user98602 Feb 10 at 21:13

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