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Is an Eilenberg-MacLane space $K(G,1)$ the same as the classifying space $BG$ for a group $G$ ?

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The answer (as I understand it) is that this is true for a discrete group $G$, but $BG$ applies to topological groups as well. – Qiaochu Yuan Jul 30 '12 at 17:39
To add to Qiaochu's comment and Henry's answer: Take $S^1$ thought of as a topological group. Then $BS^1 \simeq \mathbb{C}P^\infty$ which is $K(\mathbb{Z},2)$. It gets worse though - there are topological groups that are not even $K(G,n)$ for some $n$ – Juan S Jul 31 '12 at 3:11
I'd say the short answer is "no" they are two different kinds of constructions. $K(G,1)$ is a space with fundamental group $G$ and all other homotopy groups trivial. $BG$ is a classifying space for principal $G$ bundles, i.e., principal $G$ bundles over $X$ are in correspondence with maps $X \to BG$. If $G$ is given the discrete topology, then a little homotopy theory tells you that $BG$ is a $K(G, 1)$, but the point of view is totally different. – Justin Young Aug 1 '12 at 5:02
up vote 6 down vote accepted

Let $G$ be a discrete group (i.e. a group equipped with the discrete topology), $BG$ its classifying space, and $EG$ the universal $G$-bundle. Then we have a fibration $$G \hookrightarrow EG \rightarrow BG,$$ and the long exact sequence of this fibration reads $$\cdots \to \pi_{n+1}(BG) \to \pi_n(G) \to \pi_n(EG) \to \pi_n(BG) \to \cdots \to \pi_0(G) \to \pi_0(EG) \to 0.$$ The total space of the universal $G$-bundle is contractible and $G$ has the discrete topology, so $\pi_{n}(BG) \cong 0$ for all $n > 1$. Since $EG$ is connected, the tail end of the sequence tells us that $$\pi_1(BG) \cong \pi_0(G) \cong G,$$ since $G$ has the discrete topology. Therefore we see that $BG$ is a $K(G,1)$ when $G$ is a discrete group.

Note that since $EG$ is contractible, $\pi_{n+1}(BG) \cong \pi_n(G)$ for all $n \geq 1$, so if $G$ is equipped with a topology for which $\pi_n(G) \not\cong 0$ for $n \geq 1$, then $BG$ cannot be a $K(G,1)$.

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