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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?

Andrews
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ArthurStuart
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1 Answers1

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This construction is found in Hatcher if I recall correctly.

Let $p: EG \to BG$ be a quotient of a contractible 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 contractible, 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.

ziggurism
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Najib Idrissi
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    Does a similar argument show $B(\Omega G) \cong G$? – Michael Albanese Mar 15 '16 at 01:56
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    @MichaelAlbanese No, there is a problem with $\pi_0$, consider $G = \mathbb{Z}$. – Najib Idrissi Mar 15 '16 at 06:52
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    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 '16 at 12:45
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    @MichaelAlbanese No, $\Omega \mathbb{Z} = $ is the trivial group actually, and $B = * \not\simeq \mathbb{Z}$. – Najib Idrissi Mar 15 '16 at 12:46
  • Gah, basepoints. Thanks. – Michael Albanese Mar 15 '16 at 12:47
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    Why does $P(BG)$ have all trivial homotopy groups? – Daniel Teixeira Feb 05 '20 at 16:42
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    @big-lion It's contractible by $H(\gamma,t)(s) = \gamma(ts)$. – Najib Idrissi Feb 06 '20 at 18:13
  • In the discrete case, one can also argue that $p\colon EG\rightarrow BG$ is the universal covering, from which one deduces that $\Omega BG\cong p^{-1}(\star)=G$, or am I missing something? – Johannes Huisman Oct 02 '20 at 15:15
  • @JohannesHuisman How do you prove that if $p : X \to Y$ is the universal covering of a path-connected aspherical space then $\Omega Y \simeq p^{-1}(*)$? The proof would be pretty close to what I have written in my answer, at least conceptually... – Najib Idrissi Oct 02 '20 at 15:34
  • Isn't this just by lifting paths? – Johannes Huisman Oct 02 '20 at 15:37
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    @JohannesHuisman Well yes. Which is just a simpler, one-dimensional version of lifting cubes for the long exact sequence, as far as I can tell. There are many ways to write the same proof, I guess. – Najib Idrissi Oct 02 '20 at 15:39
  • I'm getting pretty tripped up by basepoint issues trying to understand this proof. I can't find any good proofs that $EG$ is contractible by a basepoint preserving homotopy, all the arguments appeal to some kind of "swindle" which must move the basepoint. So in your argument if $e$ is the basepoint of $EG$ then your map $EG\to PBG$ carries it to some loop $\omega$ in $\Omega G$, not necessarily the constant path at the basepoint. – Patrick Nicodemus Dec 16 '21 at 20:18
  • If $e$ is nondegenerate, which it should be in the Milnor construction, then we can homotope this map $EG\to PBG$ to a basepoint-preserving one, but I can't figure out a way to do this in a way that respects the fibers. Put another way I'm not sure why there is a path in $\Omega BG$ from $\Phi(e)$ to the constant map at the basepoint, what if one needs to move the basepoint. Some argument about how the fundamental group acts on itself by conjugation, maybe? – Patrick Nicodemus Dec 16 '21 at 20:20
  • After some time thinking about it I've figured out that $\Phi(e)$ is indeed homotopic to the constant path by a basepoint preserving homotopy, this is a corollary of the fact that the dunce cap is simply connected. This is good enough for me – Patrick Nicodemus Dec 16 '21 at 21:39