A classifying space $BG$ of a topological group $G$ is the quotient of a weakly contractible space $EG$ by a free action of $G$. The claim is that if $G$ is a discrete group then $EG/G$ is an Eilenberg-MacLane space, so that the fundamental group of $EG/G$ is $G$ and all higher homotopy groups are trivial.

Is it easy to prove such a claim, or does anyone have a reference for this?


1 Answer 1


For any topological group $G$, $EG \to BG$ is a principal $G$-bundle, so by the long exact sequence in homotopy, we have the long exact sequence

$$\dots \to \pi_{n+1}(BG) \to \pi_n(G) \to \pi_n(EG) \to \pi_n(BG) \to \pi_{n-1}(G) \to \dots$$

As $EG$ is weakly contractible, $\pi_n(EG) = 0$ for $n > 0$, so we see that $\pi_{n+1}(BG) \cong \pi_n(G)$. In particular, if $G$ is discrete, $\pi_n(G) = 0$ for $n > 0$ and $\pi_0(G) = G$, so $\pi_n(BG) = 0$ for $n \neq 1$, and $\pi_1(BG) = G$. Therefore, $BG$ is a $K(G, 1)$.

More generally, as $\pi_{n+1}(BG) \cong \pi_n(G)$, if $G$ is itself an Eilenberg-Maclane space, say a $K(H, m)$, then $BG$ is an Eilenberg-Maclane space, namely a $K(H, m + 1)$. We can see this as a generalisation of the above if we consider a discrete topological group $G$ as a $K(G, 0)$. Conversely, if $BG$ is an Eilenberg-Maclane space, $G$ is either an Eilenberg-Maclane space or discrete.

  • $\begingroup$ That's brilliant, thank you. I'd never thought of using a long exact sequence. Are we using anywhere that the action of $G$ is free? And is it easy to see that a discrete group $G$ has fundamental group $G$ and all higher groups trivial? I haven't come across homotopy groups of discrete groups before. $\endgroup$ Commented Jul 26, 2016 at 13:00
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    $\begingroup$ As the action of $G$ is free, $EG \to BG$ is a principal $G$-bundle. As for the homotopy groups of a discrete space, note that for a non-connected space, the homotopy groups are just the homotopy groups of a connected component (in particular, the connected component of the basepoint). As the connected components of a discrete space are singletons, the homotopy groups are zero. $\endgroup$ Commented Jul 26, 2016 at 13:18
  • $\begingroup$ I've been thinking about this for an hour or so, sorry if it's a silly question, but how do we know that $EG \to BG$ is a principal $G$-bundle if the action of $G$ is free? $\endgroup$ Commented Jul 26, 2016 at 15:23
  • $\begingroup$ This takes some work. See Theorem $11.2$ of Chapter $4$ of Husemoller's Fibre Bundles for example. $\endgroup$ Commented Aug 21, 2016 at 16:37

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