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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?

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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.

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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

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