Random Exercise #3: K(G,1) is a model for the classifying space BG via covering theory (NOT l.e.s. of a fibration)

I decided to write down this nice exercise since it is the kind of exercise which is readily solved with the right tools (i.e. l.e.s. of a fibration), but can become non-trivial if one doesn’t know them.

In particular, I want to prove that the Eilenberg MacLance space $K(G,1)$, for a discrete group $G$ is a model for the classifying space $BG$.

First of all, using l.e.s. this exercise is trivial: in fact we have the following fibration

$G \hookrightarrow EG \twoheadrightarrow BG$

gives us the long exact sequence of homotopy groups. Recall that $\pi_0(G) \cong G$ and $\pi_i(G)=0$ for $i>0$ ($G$ is discrete). By definition $EG$ is contractible, therefore the long exact sequences boils down to

$\pi_1(BG)=G$

$\pi_i(BG)=0$ for $i\neq 1$

This proves that $BG$ is a model for the $K(G,1)$.

what happens if we don’t know this? Ok, it’s an easy result everyone should know, but it’s good to have a proof which relies on a totally different approach, since it’s the perfect chance to revise some notions in covering theory.

Our approach will be the following: given $K(G,1)$, we will show that it has the universal property of the classifying space $BG$. Let us start with the following crucial observation: begin $G$ discrete, a principal $G$-bundle $H$ over $X$ is in particular a covering space  of $X$. As I said before, this is crucial and it’s a good sign that it is crucial otherwise we would be able to find counterexamples with the above proof. So, this is our situation so far:

we want to find a map $X \to K(G,1)$ which classifies $p \colon H \to X$. The first step to build such map is to observe that, as a covering, $p$ is classified by $p_*\pi_1H$ and such $p$ is automatically a normal covering of $X$. In fact, by definition of $G$ principal bundle, we have a fibre-wise transitive $G$-action on $H$, but this is exactly the requirements to be a normal covering (see [Hat] page 70).

Claim 1  $\pi_1X / p_*\pi_1H \cong G$

Proof:  first of all, the quotient we wrote above makes sense since we have shown a moment ago that $p$ was a normal covering of $X$. This means that, by Prop. 1.39.(a) page 71 in [Hat], that $p_*\pi_1H$ is a normal subgroup of the fundamental group of $X$. Again, by Prop. 1.39.(b), w have that the quotient is indeed $G$.

Claim 2 Let $X$ be a connected CW complex and let $Y$ be a $K(G,1)$. Then every homomorphism $\pi_1(X,x_0) \to \pi_1(Y,y_0)$ is induced by a map $(X,x_0)\to (Y,y_0)$ that it’s unique up to based homotopy.

Proof: This is Prop 1B.9 page 90 in [Hat]. This result can be generalised but since we need just this I will stick with this weaker version.

Now by Claim 1 we have the canonical quotient morphism $u_*\colon \pi_1(X)\to G$ which corresponds, thanks to Claim  2, to a map  $u \colon X \to K(G,1)$:

Notice that by construction

$(u\circ p)_*\pi_1H =0$

since $u_*$ is designed to kill $p_*\pi_1H$. Therefore there exists a lift (Prop 1.33 page 61 in [Hat] ) which we call $\tilde{u} \colon H \to \widetilde{K(G,1)}$ making the diagram commute:

Now the crucial observation is that $\tilde{u}$ is a $G$-map.

Claim 3 The map $\tilde{u}\colon H \to \widetilde{K(G,1)}$ is a $G$-map.

Proof:

We have to prove the following:

$\tilde{u}(g\bullet \tilde{x}) \overset{?}{=} g \bullet \tilde{u}(\tilde{x})$

for every $\tilde{x}\in H$ and every $g\in G$. Since $H$ is a $G$-bundle, hence (with our assumptions) a normal cover, it’s easy to observe that we can express the action of $G$ via lifts of appropriate loops on $X$. Therefore let $\alpha \colon S^1\to X$ be a loop representing $g\in \pi_1H/\pi_1X$. Let $\widetilde{\alpha}_{\tilde{x}}$ be the lift of $\alpha$ based at $\tilde{x}$. Similarly let $\beta\colon S^1\to K(G,1)$ be a loop representing $g\in \pi_1(K(G,1))$, and let $\widetilde{\beta}_{\tilde{u}(\tilde{x})}$ be the (unique) lift of it starting at $\tilde{u}(\tilde{x})$. With this notation what we need to prove is the following

Notice that $u$ is the map inducing the isomorphism

$\dfrac{\pi_1(H)}{\pi_1(X)}\cong G$

hence $[u\alpha]=[\beta]\in \pi_1(K(G,1))$. Notice now that since $\tilde{u}(\widetilde{\alpha}_{\tilde{x}})$ is a lift of $u\alpha$ and $\widetilde{\beta}_{\tilde{u}(\tilde{x})}$ is a lift of $\beta$ at the same basepoint (in fact by construction $\tilde{u}(\widetilde{\alpha}_{\tilde{x}})(0)=\tilde{u}(\tilde{x})=\widetilde{\beta}_{\tilde{u}(\tilde{x})}(0)$). Hence after lifting the homotopy on $K(G,1)$ we notice that the two lifts must be homotopy equivalent rel. end points, in particular their value at $1$ must coincide, which is precisely what we need here.

Claim 4 The pullback of $\pi$ along the map $u\colon X \to K(G,1)$ is isomorphic to the bundle $p$. In symbols:

$p \cong u^*(\pi)$

Proof: Consider the pullback of $\pi$ along $u$. It’s easy to observe that we have the following commutative diagram using the universal property of the pullback (in the category of $G$-bundles.

Notice that $\tilde{p}$ is a lift of $p$, and for the same reason given for $\tilde{u}$, it’s a $G$-map. By the classification of coverings of $X$ ([Hat] Thm. 1.38 page 67) $\tilde{p}$ is an isomorphism of bundle and a $G$-map, hence an isomorphism of principal $G$-bundles. Hence the claim

Claim 5. Isomorphic principal $G$-bundles $H,H'$ gives rise to homotopic maps $u,u'\colon X \to K(G,1)$ and homotopic maps $u,u'\colon X \to K(G,1)$ gives rise to isomorphic bundles.

Proof: The second assertion is easily seen to be true, thanks to the fact that homotopic maps gives rise to isomorphic pullback bundles. For the first half, let $H\cong H'$. Being isomorphic, we have by prop 1.37 page 67 in [Hat] that $p_*\pi_1H=p'_*(H')$, therefore the same map $u_*\colon \pi_1(X)\to G$ can be used to build $u$ which is unique up to homotopy.

This concludes the exercise, since for every (isomorphism class of) principal $G$-bundle we construct a (homotopy class of) map $u \colon X \to K(G,1)$ classifying it.

Reference:

[Hat] A. Hatcher, Algebraic Topology, Cambridge University Press.