The Groove of Math

or in other words, the Devil is in the Details

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

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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:

screen-shot-2016-12-15-at-14-29-39

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):

screen-shot-2016-12-15-at-14-51-45

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:

screen-shot-2016-12-15-at-14-56-46

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

Screen Shot 2018-04-27 at 11.20.06

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.

Screen Shot 2018-04-26 at 09.12.39

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.

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Author: RicPed

Young mathematician :)

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