Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why does delooping the classifying space of a topological group G return a space homotopy equivalent to G. In symbols, why $\Omega(BG) \cong G$, where $G$ is a topological group and $BG$ its classifying space?

share|cite|improve this question

This construction is found in Hatcher if I recall correctly.

Let $p: EG \to BG$ be a quotient of a contractile space $EG$ by a free action of $G$. This gives rise to a fibration $G \hookrightarrow EG \twoheadrightarrow BG$.

We also have the pathspace fibration of $BG$: Let $P(BG) = \{ \gamma : I \to BG : \gamma(0) = * \}$, then we have a fibration $\pi: P(BG) \twoheadrightarrow BG$ given by $\pi(\gamma) = \gamma(1)$. The fiber, $\pi^{-1}(*)$, is exactly $\Omega(BG)$.

Now $EG$ is contractile, so there is a homotopy $h_t : EG \to EG$ between $h_1 = id$ and $h_0 = *$ (the constant map). Now define $\Phi : EG \to P(BG)$ by $\Phi(x)(t) = p(h_t(x))$ (we verify that we have $\Phi(x)(0) = p(h_0(x)) = p(*) = *$). This map $\Phi$ plugs into the diagram (because the orbit $G * \subset EG$ gets sent to $\Omega(BG)$ by $\Phi$):

$$\begin{matrix} G & \to & EG & \to & BG \\ \Phi \downarrow && \Phi \downarrow && \downarrow = \\ \Omega(BG) & \to & P(BG) & \to & BG \end{matrix}$$

This gives rise to a morphism between the exact sequences of the two considered fibrations induced by $\Phi$. But the two spaces $EG$ and $P(BG)$ have all trivial homotopy groups, and the maps between the homotopy groups of $BG$ are all identities. An application of the five lemma now shows that $\Phi_* : \pi_k(G) \to \pi_k(\Omega(BG))$ is an isomorphism for all $k$, and therefore $\Phi : G \to \Omega(BG)$ is a weak homotopy equivalence.


Note: When $G$ is discrete the situation is much simpler. In this case, $BG$ is an Eilenberg-MacLane space $K(G,1)$. An immediate verification (just look at the definition) also shows that $\pi_k(\Omega X) = \pi_{k+1}(X)$ for a space $X$. So $\Omega(BG)$ is a $K(G,0)$. That is, it's $G$.


Edit: This is basically Proposition 4.66 in Hatcher's Algebraic Topology, adapted in the case of $BG$; he also makes the observation that this implies what I wrote here.

share|cite|improve this answer
    
Does a similar argument show $B(\Omega G) \cong G$? – Michael Albanese Mar 15 at 1:56
    
@MichaelAlbanese No, there is a problem with $\pi_0$, consider $G = \mathbb{Z}$. – Najib Idrissi Mar 15 at 6:52
    
I see. $\Omega\mathbb{Z} \cong \mathbb{Z}$ and $B\mathbb{Z} = S^1\not\cong \mathbb{Z}$. More generally, $K(G, 0) \xrightarrow{B} K(G, 1) \xrightarrow{\Omega} K(G, 0)$ but $K(G, 0) \xrightarrow{\Omega} K(G, 0) \xrightarrow{B} K(G, 1)$. – Michael Albanese Mar 15 at 12:45
    
@MichaelAlbanese No, $\Omega \mathbb{Z} = *$ is the trivial group actually, and $B* = * \not\simeq \mathbb{Z}$. – Najib Idrissi Mar 15 at 12:46
    
Gah, basepoints. Thanks. – Michael Albanese Mar 15 at 12:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.